When can two endomorphisms on different vector spaces be identified

Question

Suppose $\dim V = \dim W$. Given $\phi:V\rightarrow V$ and $\psi:W\rightarrow W$ when do we have bijective linear maps $a,b:V\rightarrow W$ such that $$\require{AMScd} \begin{CD} V @>\phi>> V \\ @VaVV @VVbV \\ W @>\psi>>W \end{CD} $$ commutes?

My intuition says that if the rank of $\phi$ and $\psi$ match there should exist $a,b$. Of course with $a=b$, this requires that $\phi,\psi$ are similar (after identifying $V$ and $W$ in any way you like), but with $a\neq b$ allowed, it seems this is much weaker.

Answer 1

This is essentially the change of bases.

Suppose that $\phi$ and $\psi$ have the same rank $k$. Take a basis $\{v_1,\dots,v_n\}$ of $V$ such that $\{\phi(v_1),\dots,\phi(v_k)\}$ is a basis of the range of $\phi$.

Choose similarly a basis $\{w_1,\dots,w_n\}$ for $W$.

Now complete $\{\phi(v_1),\dots,\phi(v_k)\}$ to a basis $\{\phi(v_1),\dots,\phi(v_k),v_{k+1}',\dots,v_n'\}$ of $V$. Similarly, complete $\{\psi(w_1),\dots,\psi(w_k)\}$ to a basis $\{\psi(w_1),\dots,\psi(w_k),w_{k+1}',\dots,w_n'\}$ of $W$.

Define $a$ by $a(v_i)=w_i$; define $b$ by $b(\phi(v_i))=\psi(w_i)$ for $i=1,2,\dots,k$ and $b(v_i')=w_i'$ for $i=k+1,\dots,n$.

Then $a$ and $b$ induce bijective linear maps and, for $i=1,2,\dots,n$, $$ b(\phi(v_i))=\psi(w_i)=\psi(a(v_i)) $$

Can you show the converse?

Without bases, you can use complements. Write $V=V_1\oplus\ker\phi$, so $\phi$ induces an injective linear map $V_1\to V$, with the same image as $\phi$. Write $V=\operatorname{im}\phi\oplus V_2$.

Do similarly for $\psi$, writing $W=W_1\oplus\ker\psi=\operatorname{im}\psi\oplus W_2$.

By assumption $\dim V_1=\dim W_1$ and $\dim V_2=\dim W_2$. Choose suitably the isomorphisms, noting that $\phi$ and $\psi$ induce isomorphisms $V_1\to\operatorname{im}\phi$ and $W_1\to\operatorname{im}\psi$. However this is not really different from choosing bases.

Answer 2

Then that condition is just equivalence of $\phi$ and $\psi$, which is equivalent to having the same rank.

To show this, take a basis $e_i^V$, $i\in I_1$ of the kernel of $\phi$ and extend it to a basis $e_i^V$, $i\in I_1\cup I_2$ of $V$. Then, the elements of the basis that are not in the kernel are mapped to a basis $f_i^V$, $i\in J_1$ of the range of $\phi$. Let's assume that we index them such that $\phi(e_i^V)=f_i^V$ for $i\in I_1$. Extend also this basis to a basis $f_i^V$, $i\in J_1\cup J_2$ of $V$ (but the $V$ that is the codomain of $\phi$).

Do the same business with $W$ and $\psi$ to obtain bases $e_i^W$, $i\in I_1\cup I_2$ and $f_i^W$, $i\in J_1\cup J_2$. Assume that we also repeated the construction such that $\psi(e_i^W)=f_i^W$, for $i\in I_1$. Note that I used the same index sets, since the ranks are assumed equal. Therefore, the sets of indexes required have the same cardinality. So, we can just take them equal.

Finally define $a(e_i^V)=e_i^W$ and $b(f_i^V)=f_i^W$ and extend them by linearity.