Dtjytyk

Suppose $dim V = dim W$. Given $phi:Vrightarrow V$ and $psi:Wrightarrow W$ when do we have bijective linear maps $a,b:Vrightarrow W$ 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 $aneq b$ allowed, it seems this is much weaker.

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 $V=V_1opluskerphi$, so $phi$ induces an injective linear map $V_1to V$, with the same image as $phi$. Write $V=operatorname{im}phioplus V_2$.

Do similarly for $psi$, writing $W=W_1opluskerpsi=operatorname{im}psioplus W_2$.

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 $V_1tooperatorname{im}phi$ and $W_1tooperatorname{im}psi$. However this is not really different from choosing bases.

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

To show this, take a basis $e_i^V$, $iin I_1$ of the kernel of $phi$ and extend it to a basis $e_i^V$, $iin I_1cup I_2$ of $V$. Then, the elements of the basis that are not in the kernel are mapped to a basis $f_i^V$, $iin J_1$ of the range of $phi$. Let's assume that we index them such that $phi(e_i^V)=f_i^V$ for $iin I_1$. Extend also this basis to a basis $f_i^V$, $iin J_1cup J_2$ of $V$ (but the $V$ that is the codomain of $phi$).

Do the same business with $W$ and $psi$ to obtain bases $e_i^W$, $iin I_1cup I_2$ and $f_i^W$, $iin J_1cup J_2$. Assume that we also repeated the construction such that $psi(e_i^W)=f_i^W$, for $iin I_1$. 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.