let $V$ be a finite-dimensional vector space and $T:V \to V$ a linear transformation. If $T$ is diagonalizable, is it always the case that $T$ is an isomorphism? I think I've shown (by example) that the converse is false. I.e., there are isomorphisms that are not diagonalizable.
I've gotten some ways into the problem but am stuck at a step.
Since by assumption $T$ is diagonalizable, there exists a basis of eigenvectors {$v_1, …, v_n$} for $V$. Then to show that $T$ is an isomorphism it would be enough to show that $T$ is injective.
So suppose $v\in V$. Then we can write $v = a_1v_1 + … + a_nv_n$.
Not suppose $T(v) = 0$:
$\to T(a_1v_1 + … + a_nv_n) = 0$
$\to a_1T(v_1) + … +a_nT(v_n) = 0 $
$\to (a_1\lambda_1)v_1 + … +(a_n\lambda_n)v_n = 0$, since the $v_i$'s are eigenvectors. The $\lambda_i$'s need not be distinct.
Now to show that $T$ is an isomorphism it would have to be the case that no $\lambda_i$ were $0$. So I'm not sure what to do from here.
Best Answer
Singular transformations, for instance the one given by $T(v) = 0$ for all $v$, can be diagonalizable.
For the other direction, over $\Bbb R$, rotations are isomorphisms that are not (generally) diagonalizable.