Axler exercise 3.D.7 I am trying to solve the following problem in Axler's book.

Suppose $V$ and $W$ are finite-dimensional. Let $v \in V$. Let $E = \{T \in \mathcal{L}(V,W) : Tv = 0\}$. Suppose $v \neq 0$. What is $\dim E$?

I know that the answer must be $m(n-1)$, but I am having difficulty formalizing the argument. This is what I have.

If $v \neq 0$, we can extend to a basis $v, \ldots, v_n$ of $V$. Let $w_1, \ldots, w_m$ be a basis for $W$. The definition of $\mathcal{M}(T)$ implies
$$Tv = A_{1,1} w_1 + \ldots + A_{1,1} w_m = 0.$$
The $\{w_k\}$ are lineaely independent, so $A_{i,j} = 0$ for all $i,j$. So the first column of $\mathcal{T}$ is all zeroes.

There are two conflicting theorems I want to cite here. The first says that there is an isomorphism between $\mathcal{L}(V,W)$ and $\mathbb{F}^{m,n}$, the set of $m \times n$ matrices over $\mathbb{F}$. If the first column of $\mathcal{T}$ is all zeroes, then a basis for $\mathbb{F}^{m,n}$ is not $mn$, but $m(n-1)$ (matrices with $1$ in a single entry not in column $1$ and zeroes elsewhere, of which there are $m(n-1)$). But I feel that I need to biject with a subspace of $\mathbb{F}^{m,n}$ and then biject that with $\mathcal{L}(V,W)$ to get the appropriate isomorphism and then conclude equality of dimensions.
Any help would be appreciated.
 A: Begin as they do: extend $\;\{v\}\;$ to a basis $\;\mathcal B:=\{v=v_1, v_2,..., v_n\}\;$ of $\;V\;$ . Any $\;T\in\mathcal L(V,W)\;$ is uniquely and completely determined by its action on $\;\mathcal B\;$.
Let now $\;\mathcal L_v\subset \mathcal L\;$ be defined as $\;\mathcal L_v:=\{T\in\mathcal L\;|\;Tv=0\;$} , then
$$T\in\mathcal L_v\iff Tv=0\iff \forall x=\sum_{k=1}^n a_kv_k\;,\;\;Tx=\sum_{k=2}^na_kTv_k$$
and the above means that we can see $\;T\;$ as a a linear map from $\;\text{Span}\,\mathcal B\setminus\{v\}\to W\;$ and, of course: any linear map $\;S: \text{Span}\,\mathcal B\setminus\{v\}\to W\;$ can easily be extended to $\;\overline S\in\mathcal L(V,W)\;$ simply by defining $\;\overline Sv:=0\;$.
Thus, we get that
$$\;\mathcal L_v\cong \mathcal L(\text{Span}\,\mathcal B\setminus\{v\},\,W)\implies \dim \mathcal L_v=(n-1)m\;$$
with $\;m=\dim W\;$ .
A: With your bases $(v_i)$ and $(w_j)$, define $\varphi_j : V \to W$, $j=1,\cdots,m$, by $$ \varphi_j(v_1)=w_j \text{ and } \forall i=2,\cdots,n, \, \varphi_j(v_i)=0 .$$ Then show that $\mathcal{L}(V,W)=E \oplus \mathbb{K} \varphi_1 \oplus \cdots \oplus \mathbb{K} \varphi_m$ ($\mathbb{K}$ is your base field) and use the dimensions of the two members of the equality
