How to get the expectation value of the spin of a generate spin triplet states? For two spin-1/2 electrons, the general spin triplet states is a linear combination of the three basis states: $\left.|\uparrow\uparrow\right>, \left.|\uparrow\downarrow\right>+\left.|\downarrow\uparrow\right>, \left.|\downarrow\downarrow\right>$, which are the simultaneously eigenstates of $S^z=S_1^z+S_2^z$ and $\bf{S}^2=(S_1+S_2)^2$.
When denote these three states in the matrix form respectively(named as direct product of states?): $\chi^{1}=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \chi^{0}=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \chi^{-1}=\begin{pmatrix}0&0\\0&1\end{pmatrix}$, the most generate form of a spin triplet states can be written as a linear combination of them, explicitly:
$$
\chi= \sum_i \alpha^i\chi^i
$$
Question is, how can I get the expectation value of $\bf{S}$ of this generate spin state using its matrix form? I don't know how to express the operator $\bf{S}$ in matrix form, it seems that $\bf{S}=\bf{S_1}\otimes1+1\otimes S_2$ is a $4\times 4$ matrix and I don't know how to apply this on this $2\times 2$ state. 
ps: I know the method using the operator form, which requires that $\bf S_1$ only operate on the first spin and $\bf S_2$ only operate on the second spin, etc... I just want to see how the matrix form of spin state can be used to do the calculation... 
 A: The expressions you wrote for the $\chi^j$ are not exactly what you want. You'd like to define a composite state in the basis 
$$ \vert\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix},\quad \vert\downarrow\rangle = \begin{pmatrix}0\\1\end{pmatrix} $$
which is
$$ \vert\uparrow\uparrow\rangle\equiv\vert\uparrow\rangle \otimes \vert\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$ by the common definition of the Kronecker product.
As you noted correctly, by adopting this convention operators like $S^1\otimes 1$ are represented by $4\times 4$ matrices. Again, their representations in any given basis is by the Kronecker Product, e.g 
$$ S^3\otimes 1 = \frac{1}{2}\begin{pmatrix} 1 &&& \\ & 1 && \\ && -1 & \\ &&& -1\end{pmatrix} $$
Instead of this more canonical choice you wrote e.g 
$$ \chi^1 = \begin{pmatrix}1&0\\0&0\end{pmatrix}$$
which is also possible, but uncommmon. One would rather identify this expression with $$ \vert\uparrow\rangle \otimes \langle\uparrow\vert \in \mathbb{C}^2\otimes (\mathbb{C}^2)^*$$
Since $\mathbb{C}^4 \simeq \mathbb{C}^{2\times 2} \simeq \mathbb{C}^2\otimes (\mathbb{C}^2)^* \simeq\mathrm{End}(\mathbb{C}^2)$, both choices are equivalent. Work out the isomorphism relating the two representations! From there it should not be hard to derive the action of the composite operators on $\chi^j$. Be guided by what you already know, namely that by definition 
$$ (A\otimes B)\cdot (u\otimes v) = (Au)\otimes(Bv) $$
It then just comes down to choosing a basis for the tensor product space as either $\mathbb{C}^4$ or $\mathbb{C}^{2\times 2}$.
