# Concerning the Matrix of $T\in\mathcal{L}(V,W)$

Is the Following Proof Correct ?

Theorem. Given that $V$ and $W$ are finite dimensional vectors spaces and $T\in\mathcal{L}(V,W)$ i.e. a linear transformation from $V$ to $W$ then there exists bases for $V$ and $W$ such that all entries of $\mathcal{M}(T)$ i.e. the matrix of $T$ are $0$ except that the entries in row $j$ and column $j$ equal 1 for $1\leq j\leq\dim\operatorname{range}T$.

Proof. Since $W$ is finite dimensional and $\operatorname{range}T\leqslant W$ it follows that $\operatorname{range}T$ must have a basis, we may therefore invoke the existence of vectors $r_1,r_2,...,r_l\in\operatorname{range}T$ where $l=\dim\operatorname{range}T$ such that they act as a basis for $\operatorname{range}T$.

Consider now the list of vectors $v_1,v_2,...,v_l$ such that $\forall j\in\{1,2,...,l\}(Tv_j=r_j)$, furthermore it is not difficult to see that the linear independence of $r_1,r_2,...,r_l$ implies that linear independence of $v_1,v_2,...v_l$.

Extending the list $r_1,r_2,...,r_l$ to form a basis for $W$ we have $$r_1,r_2,...,r_l,w_1,w_2,...,w_m\in W$$

furthermore by letting $u_1,u_2,...,u_n$ be a basis for $\operatorname{null}T$ and invoking the rank-nullity theorem we may form a basis for $V$ i.e.

$$v_1,v_2,...,v_l,u_1,u_2,....u_n\in V$$

Given the above choice of basis we see that $\mathcal{M}(r_k)$ i.e the matrix of $r_k$ is a $(l+m)\times 1$ column vector with the $k$th entry equal to $1$ and all others equal to $0$, consequently $\mathcal{M}(T)$ is as specified in the statement of the theorem.

$\blacksquare$

Since you didn't specify any condition for the $u_j$, they may not be in the kernel of $T$ (so no, it's not "evident" that they span it). What you need to do is take $u_1,\ldots,u_n$ to be a basis for the kernel of $T$, and then use rank-nullity to argue that $v_1,\ldots,v_l,u_1,\ldots,u_n$ is a basis.
After this, you get $Tv_j=r_j$, $Tu_k=0$, which guarantees that $T$ has the desired form.
• So all i need to do is to add the clause that $u_1,u_2,...u_n$ is basis for $\operatorname{null}T$? Nov 20, 2017 at 8:01