Suppose that $W$ is finite-dimensional and $S, T \in \mathcal{L}(V, W)$. Prove that null $S=$ null $T$ if and only if there exists an invertible $E \subset \mathcal{L}(W)$ such that $S=E T$.
This is my proof:
This is the demonstration only in one direction
I will use this theorem for defining E:
Theorem
(Suppose $v_1, \ldots, v_n$ is a basis of $V$ and $w_1, \ldots, w_n \in W$. Then there exists a unique linear map $T: V \rightarrow W$ such that
$$
T v_j=w_j
$$
for each $j=1, \ldots, n$.)
proof
Let define $E:$ range $T \rightarrow W$ $u_1, \ldots, u_m, v_{m+1}, \ldots, v_n$ – basis of $V$ where $u_1, \ldots, u_m$ basis null $T$
T $v_{m+1}, \ldots$. T $v_n$ – span range $T$ it can be restricted to a basis T$v_1$…, T $v_k$
null $T=$ null $S \Rightarrow$ dim range $T=\operatorname{dim}$ range $S$
$\exists E: \operatorname{range} T \rightarrow W . E\left(T v_k\right)=S v_k$
$$
\begin{array}{l}
\forall v \in V: E(T v))=E\left(T\left(\sum_{k=1}^m u_k+\sum_{j=m+1}^n v_j\right)\right) \\
\left.=\sum_{j=1}^k S v_j=S\left(\sum_{j=M+1}^n v_j+\sum_{t=1}^m u_t\right)\right)=S v
\end{array}
$$
E- can be extended to an e
injective linear map E: W -> W
Dim W = dim W => E – injective <=> E – invertible
I would like to receive any suggestions. Maybe my proof is wrong, or made some mistakes or it can be improved. I am a self learner, so this would be very helpful
Best Answer
This is a proof assuming that $V$ is finite dimensional.
Also, my notations are slightly different from yours.
Let $\{u_1, u_2, ..., u_m, v_1,...,v_n\}$ be a basis of $V$ obtained upon extending $\{u_1, u_2, ..., u_m\}$, a basis of null $T$= null $S$.
Let $\{Tv_1, ...,Tv_k\}$ be a basis of Im $T$(WLOG). Extend this to $\{Tv_1, ...,Tv_k, w_1', w_2',..., w_r'\}$, a basis of $W$.
Now, $\{Sv_1, ...,Sv_k\}$ is a basis of Im $S$(Verify). Extend this to $\{Sv_1, ...,Sv_k, v_1', v_2',..., v_r'\}$, a basis of W.
Now, consider the following map:
$E: W \rightarrow W$, $E(Tv_i)= S(v_i)$ for $1 \leq i \leq k$, and $E(w_j')=v_j'$, for $1 \leq j \leq r$
Verify that the map $E$ is injective and surjective, and thus invertible.