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F I R S T___ A N S W E R

(***upvote*** or ***downvote*** my 1rst answer only. My 2nd,3rd,4th and 5th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

F I R S T___ A N S W E R

(***upvote*** or ***downvote*** my 1rst answer only. My 2nd,3rd,4th and 5th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.
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F I R S T___ A N S W E R

(***upvote*** or ***downvote*** thismy 1rst answer only. My 2nd,3rd,4th and 4th5th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

F I R S T___ A N S W E R

(***upvote*** or ***downvote*** this 1rst answer only. My 2nd,3rd and 4th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

F I R S T___ A N S W E R

(***upvote*** or ***downvote*** my 1rst answer only. My 2nd,3rd,4th and 5th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.
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F I R S T___ A N S W E R

(***upvote*** or ***downvote*** this 1rst answer only. My 2nd,3rd and 4th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

F I R S T___ A N S W E R

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

F I R S T___ A N S W E R

(***upvote*** or ***downvote*** this 1rst answer only. My 2nd,3rd and 4th answers are addenda to it)

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) \begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation} \begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}

\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}

(details on product -states,-spaces,-transformations are given in my SECOND, THIRD and FOURTH ANSWERS)


A representation : \begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation} Now \begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align} and \begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align} From equations (05) and (06) \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation} If for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have \begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation} because, as could be proved(1), the product 4-dimensional Hilbert space is the _direct sum_ of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation} In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation} Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the _direct sum_ of $\:n\:$ _mutually orthogonal_ subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n) $ \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation} where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension \begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation} with \begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align} Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : \begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation} Equation (11) is a special case of equation (16) : \begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}
(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : \begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation} since for \begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation} \begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation} we have \begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation} and consequently \begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align} From \begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation} we have finally \begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation} For its eigenvalues $\lambda$ \begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation} So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.
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