## Similarity among nonsquare matrices

Let $F$ be a field, let $V$ be an $n$-dimensional vector space over $F$, and let $W$ be an $m$-dimensional vector space over $F$. Let $B_1, B_2 \subseteq V$ and $E_1, E_2 \subseteq W$ be bases, let $\varphi : V \rightarrow W$ be a linear transformation, and let $A = M^{E_1}_{B_1}(\varphi)$ and $B = M^{E_2}_{B_2}(\varphi)$. Let $P = M^{B_1}_{B_2}(1)$ be the matrix of the identity map on $V$ with respect to the source basis $B_2$ and the target basis $B_1$, and likewise let $Q = M^{E_1}_{E_2}(1)$ be the matrix of the identity map on $W$ with respect to the source basis $E_2$ and the target basis $E_1$.

Prove that $Q^{-1} = M^{E_2}_{E_1}(1)$ and that $Q^{-1}AP = B$.

Using Theorem 12 in D&F, we have $Q \times M^{E_2}_{E_1}(1) = M^{E_1}_{E_2}(1) \times M^{E_2}_{E_1}(1)$ $= M^{E_1}_{E_1}(1 \circ 1)$ $= M^{E_1}_{E_2}(1) = I$. Similarly, $M^{E_2}_{E_1}(1) \circ Q = M^{E_2}_{E_2}(1) = I$. By the uniqueness of inverses, $M^{E_2}_{E_1}(1) = Q^{-1}$.

Now $Q^{-1}AP = M^{E_2}_{E_1}(1) \times M^{E_1}_{B_1}(\varphi) \times M^{B_1}_{B_2}(1)$ $= M^{E_2}_{B_2}(1 \circ \varphi \circ 1)$ $= M^{E_2}_{B_2}(\varphi) = B$.