$\textbf{Exercise 6.}$ Suppose that $W$ is finite-dimensional and $S, T \in \mathcal{L}(V, W)$. Prove that $\operatorname{null} S = \operatorname{null} T$ if and only if there exists an invertible $E \in \mathcal{L}(W)$ such that $S = ET$. $\textbf{Solution 6.}$ First suppose that there exists an invertible $E \in \mathcal{L}(W)$ such that $S = ET$. Let $v \in V$. Then $\begin{align*} v \in \operatorname{null} S &\iff Sv = 0 \\ &\iff E(Tv) = 0 \\ &\iff Tv = 0 \\ &\iff v \in \operatorname{null} T, \end{align*}$ where the second equivalence holds because $E$ is invertible (and hence injective). Thus $\operatorname{null} S = \operatorname{null} T$, as desired. Now suppose that $\operatorname{null} S = \operatorname{null} T$. Define $E_1: \operatorname{range} T \to W$ by $\begin{equation*} E_1(Tv) = Sv \end{equation*}$ for each $v \in V$. To show that this definition makes sense, we must show that if $u, v \in V$ and $Tu = Tv$, then $Su = Sv$. But this is true, because if $Tu = Tv$ then $u - v \in \operatorname{null} T = \operatorname{null} S$ and hence $Su = Sv$. Now that $E_1$ is well defined, it is easy to verify that $E_1$ is a linear map from $\operatorname{range} T$ to $W$. Suppose $w \in \operatorname{range} T$ and $E_1w = 0$. There exists $v \in V$ such that $w = Tv$. Now $\begin{equation*} 0 = E_1w = E_1(Tv) = Sv, \end{equation*}$ and thus $v \in \operatorname{null} S = \operatorname{null} T$. Hence $w = Tv = 0$. Thus $E_1$ is injective. By Exercise 5, we can extend $E_1$ to an invertible linear map $E$ from $W$ to itself. The definition of $E_1$ shows that $S = ET$. --- ![[sol-5.pdf#page=5]]#