Suppose V and W are finite-dimensional. Let $v \in V$. Let $E=\{T \in \mathscr{L}(V,W)\ |\ Tv=0\}.$ Show E is subspace of $\mathscr{L}(V,W)$ Suppose V and W are finite-dimensional. Let $v \in V$. Let $$E=\{T \in \mathscr{L}(V,W)\ |\ Tv=0\}.$$
a.) Show E is subspace of $\mathscr{L}(V,W)$
b.) Suppose that $v \neq 0$, what is dim E?
Here is what I have so far. 
Proof: Let $v,u \in V$ then $$T(u+v)=T(u) +T(v) = 0+0=0.$$
For some $\alpha \in \mathbb{R}$ we have $$T(\alpha v)=\alpha T(v)=\alpha* 0=0.$$
I have a couple questions. How do I show E contains the identity? I also have no clue how to approach part b.
 A: (a) The zero element of $\mathscr L = \mathscr L(V,W)$ is the function $T: V \rightarrow W$ given by $T(v) = 0$ for ALL $v$.  So certainly the zero element is in $E$.
Here is a hint for (b). Try extending $v$ to a basis for $V$.  Let's say that $v_1, ... , v_n$ is a basis for $V$ with $v = v_1$.  Let $V_0$ be the $(n-1)$th dimensional vector space with basis $v_2, ... , v_n$.  
Now, if we consider only those $T \in \mathscr L$ which send $v_1$ to $0$, then we are really talking about the vector space of linear transformations from $V_0$ to $W$.  So what should the dimension be?
A: I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
This exercise is Exercise 3.D.7 on p.88 in this book.
I solved this exercise as follows:

(b)
Let $n:=\dim V$.
Let $m:=\dim W$.
Let $v,v_2,\dots,v_n$ be a basis of $V$.
Let $w_1,\dots,w_m$ be a basis of $W$.
Let $T_{i,j}$ be a linear map from $V$ to $W$ such that
$T_{i,j}v=0$,
$T_{i,j}v_i=w_j$,
$T_{i,j}v_k=0$
for $i\in\{2,\dots,n\}$, $j\in\{1,\dots,m\}$ and $k\in\{2,\dots,n\}-\{i\}.$
Then, $T_{i,j}\in E$.
Let $T\in E$.
Let $Tv_i=a_{1i}w_1+\dots+a_{mi}w_m$ for $i\in\{2,\dots,n\}$.
Then, $(\sum_{j=1}^{m}\sum_{l=2}^{n}a_{jl}T_{l,j})v_i=a_{1i}w_1+\dots+a_{mi}w_m$ for $i\in\{2,\dots,n\}$.
So, $T=\sum_{j=1}^{m}\sum_{l=2}^{n}a_{jl}T_{l,j}$.
And,
if $\sum_{j=1}^{m}\sum_{l=2}^{n}a_{jl}T_{l,j}=0$, then $0=0v_i=(\sum_{j=1}^{m}\sum_{l=2}^{n}a_{jl}T_{l,j})v_i=a_{1i}w_1+\dots+a_{mi}w_m$ for $i\in\{2,\dots,n\}$.
So, $a_{ji}=0$ for $j\in\{1,\dots,m\}$ and $i\in\{2,\dots,n\}$.
So, $T_{2,1},\dots,T_{2,m},\dots,T_{n,1},\dots,T_{n,m}$ is a basis of $E$.
So, $\dim E=(n-1)m=(\dim V-1)\dim W$.

