Deriving The Spin-1 Matrices from Spin-1/2 In quantum mechanics, we know that the spin 1/2 matrices are:
$$S_x = \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, 
\quad S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, 
\quad S_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$
While I am pretty sure I understand how we got these, it is still fuzzy for me. Thus, as an application of this (and as part of homework), I am trying to understand how to get the matrices for higher spin levels.
Thus, with the spin 1/2 matrices, how do we obtain the spin 1 or greater matrices?
 A: You don't really derive higher-spin matrices from lower-spin ones. Rather, they are derived from the algebra of spin operators and from how they act oon the chosen spin basis.
The basis chosen is usually the basis of eigenvectors of operator $S_z$, and are often denoted $|j,m\rangle$, $2j\in\mathbb{N}$, $m\in\{-j,-j+1\dots,j\}$. Operator $S_z$ act on them as follows:
$$ S_z|j,m\rangle = \hbar m |j,m\rangle $$
We also have operators $S_+=S_x+iS_y$ and $S_-=S_x-iS_y$. Using the algebra of spin operators it can be proven that
$$ S_+|j,m\rangle = \hbar \sqrt{(j-m)(j+m+1)}|j,m+1\rangle $$
$$ S_-|j,m\rangle = \hbar \sqrt{(j+m)(j-m+1)}|j,m-1\rangle $$
(try to do it yourself, or ask if you need help).
These relations are enough to write down the matrices of $S_z$, $S_+$ and $S_-$ in the basis of vectors $|j,m\rangle$, and from $S_+$ and $S_-$ you can obtain $S_x$ and $S_y$.
A: Hints :
REFERENCE : Total spin of two spin- 1/2  particles.
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
We could derive the spin-1 matrices adding two spin-1/2. But this addition is not so easy as you may expect since you must study first about product spaces and product transformations in detail (see SECOND ANSWER and THIRD ANSWER in above REFERENCE respectively).
For the present in the main answer of above  REFERENCE you could watch from equation (01) to equation (10) how to derive the $S^{\left(j\boldsymbol{=}1\right)}_z$ matrix
\begin{equation}
S^{\left(j\boldsymbol{=}1\right)}_z\boldsymbol{=}\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
     \:1\: & \:0\: & \:0\: \\
     \:0\: & \:0\: & \:0\: \\
     \:0\: & \:0\: & \!\!\!-1
   \end{bmatrix} 
\tag{01}\label{01}   
\end{equation}
combining (adding) two $S^{\left(j\boldsymbol{=}1/2\right)}_z$ matrices
\begin{equation}
S^{\left(j\boldsymbol{=}1/2\right)}_z\boldsymbol{=}\tfrac{1}{2}
\begin{bmatrix}
     \:1\: & \:0\: \\
     \:0\: & \!\!\!-1
   \end{bmatrix}
\tag{02}\label{02}   
\end{equation}
As an other example you could watch how to derive the  $S^{\left(j\boldsymbol{=}3/2\right)}_k$ matrices combining (adding) the  $S^{\left(j\boldsymbol{=}1/2\right)}_k$ matrices with the $S^{\left(j\boldsymbol{=}1\right)}_k$ ones, see equations (Ex-25),(Ex-25.1),(Ex-25.2) and (Ex-25.3) in FIFTH ANSWER of above REFERENCE.
A: You don't build the spin $1$ matrices from the spin $\frac 12$ matrices.
You rather repeat the whole procedure, which you learned with
$2\times 2$ matrices for spin $\frac 12$.
But now you do it with $3\times 3$ matrices for spin $1$.
As you already know from spin $\frac 12$ the 3 matrices are not unique.
So there is much freedom in choosing a possible set of 3 matrices.
Begin with $S_z$. You know it has 3 eigenvalues: $+\hbar$, $0$, $-\hbar$.
So you can choose $S_z$ with these eigenvalues along the diagonal
and $0$ everywhere else.
$$S_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \tag{1}$$
To find possible $S_x$ and $S_y$ matrices, you need to make sure
they are Hermitian and satisfy the commutation relations.
$$\begin{align}
[S_x,S_y]&=i\hbar S_z \\
[S_y,S_z]&=i\hbar S_x \\
[S_z,S_x]&=i\hbar S_y
\end{align} \tag{2}$$
(By the way: these relations are required for every
set of angular momentum operators, not only for spin $\frac 12$ or spin $1$.)
You may find $S_x$ with $S_y$ by using high sophisticated math
as given in the other answers.
But actually it is not too difficult to find them just by try and error.
Get some inspirations from the $S_x$ and $S_y$ from spin $\frac 12$.
The $3\times 3$ matrices for spin $1$ don't look so much different from these.
Just put some $1$s into the $S_x$ matrix,
and some $\pm i$ into the $S_y$ matrix.
Then calculate the commutators and check the relations (2).
If they are not yet satisfied, then modify and try again.
