Show there exists an isomorphism

Question

$T: V\rightarrow \!\,V$ is a linear operator on a finite-dimensional vector space $V$, and the map $ψ : W\rightarrow \!\,W$ is a linear operator on $W$.

$\dim(W) = \dim(V)$ and $W$ is a subspace of $V$,

The question is, show that there is an isomorphism $φ:V\rightarrow \!\,W$ such that $φ(T(v)) = ψ(φ(v))$

I thought if $\dim(W) = \dim(V)$ and $W$ is a subspace of $V$, then $W=V$. I worked my way to the point that I have $T(v)=ψ(v)$. Then I am stuck. From $T(v)=ψ(v)$, how can I show there is an isomorphism? I want to show there is an isomorphism $φ$ such that $φψ=ψφ$. I think I am missing something very fundamental. Thank you!

Answer 1

This is not true for $T=0$ if $\psi \ne 0$.

Answer 2

As you found out, $W=V$ follows. But then the question is equivalent to asking if $T$ and $\psi$ (both $V\to V$) are similar. This is clearly not true for arbitrary $T,\psi$.

Answer 3

You cannot always find such an isomorphism. Like you noted, if $W \subseteq V$ and $\dim W = \dim V$ then $W = V$. Thus, you are given two linear maps $T,\psi \colon V \rightarrow V$ and you are asked to find a linear isomorphism $\varphi \colon V \rightarrow V$ such that $\varphi \circ T = \psi \circ \varphi$, or, equivalently, that $\varphi \circ T \circ \varphi^{-1} = \psi$. This can be done if and only if $T$ and $\psi$ are similar.