Find Basis of $V$ based on $V/W$ and $W$ $W$ is nonzero proper subspace of $V$. Linear $T: V \to V$ satisfies $T(W) \subseteq W$. $T_{1}: W \to W, w \mapsto T(w)$ and $T_{2}: V/W \to V/W, v+W \mapsto T(v)+W$ are linear and well defined. 
I was given ordered bases $\alpha = \{w_{1},...,w_{m}\}$ of $W$ and $\beta = \{v_{1}+W,...,v_{n}+W\}$ of $V/W$. I am supposed to show the following two things. 
a. Prove that basis of $V$ is $\gamma=\{w_{1},...,w_{m},v_{1},...,v_{n}\}$ 
b. Prove that $[T]_{\gamma}$ = $\begin{bmatrix}[T_{1}]_{\alpha} & *\\0 & [T_{2}]_{\beta} \end{bmatrix}$
$[T]_{\gamma}$ is a matrix representation of linear transformation $T$ on basis $\gamma$
Now, part a looks similar to the proof that was required to show $dim(V/W)=dim(V)-dim(W)$ but I am not sure how to apply this to prove a. As for part b, I have no clue how to approach this. Could anyone help me? Thank you. 
 A: You just need to use the fact that
$$ [T]_{\gamma} = \begin{bmatrix} [T(w_{1})]_{\gamma} & \cdots & [T(w_{m})]_{\gamma} & [T(v_{1})]_{\gamma} & \cdots & [T(v_{n})]_{\gamma} \end{bmatrix} $$
Since $T(W) \subseteq W$, you can express each $T(w_{j})$ in terms of the $w_{i}$. This gives you that for $1 \leq j \leq m$,
$$[T(w_{j})]_{\gamma} = \begin{bmatrix} [T(w_{j})]_{\alpha} \\ 0 \end{bmatrix}$$
The second part is a little tricky: you need to use the fact that for each $v_{j}$, you can write $T(v_{j}) = w+T_{2}(v_{j})$ (where in a slight abuse of notation here I am considering the codomain of $T_{2}$ to simply be the span of the $v_{i}$). Now since $w$ is a linear combination of the $w_{i}$, you have that
$$[T(v_{j})]_{\gamma} = \begin{bmatrix} * \\ [T(v_{j})]_{\beta}    \end{bmatrix}$$
for each $1 \leq j \leq n$, where the $*$ represents the coefficients on the $w_{i}$ that are needed to obtain $w$.
A: The basis part is rather easy.
Let $x\in V$; then $x+W=b_1(v_1+W)+\dots+b_n(v_n+W)$. Consider $x'=b_1v_1+\dots+b_nv_n$; then, by definition, $x-x'\in W$ and so $x-x'=a_1w_1+\dots+a_mw_m$.
This proves that $\{w_1,\dots,w_m,v_1,\dots,v_n\}$ is a spanning set for $V$.
Suppose now that $a_1w_1+\dots+a_mw_m+b_1v_1+\dots+b_nv_n=0$. Then
$$
b_1(v_1+W)+\dots+b_n(v_n+W)=0+W
$$
Can you go on?
For the second part, just write down what the matrix does on each vector in the basis and apply the assumption that $T(W)=W$.
