Is $T : V → V$ not an isomorphism if T doesn't have eigenvalues?

Question

Let$ T: V→V $ be a linear transformation $$$$Prove/Disprove: If T is not an isomorphism, then T has no eigenvalues. $$$$I think that by proving that if T has eigenvalues then we can prove that T is inverse thus proving it is isomorphism. However I'm not sure how to continue. What do you think of this way of proof?

Answer 1

Here, you assume the statement is true. Thats why you prove it via contrapositive. But however it is false.

Any $n \times n$ matrix can be viewed as a linear transformation from $\Bbb F^n$ to $\Bbb F^n$. So in particular, take $T$ to be any nilpotent matrix, which is not invertible, but has eigenvalues, namely all of zeros!

Answer 2

If $T$ is no isomorphism its kernel is non-trivial, hence $0$ is an eigenvalue of $T$.