Minimal polynomial and diagonalization. How would I be able to prove that a linear endomorphism $T$ is not diagonalizable if it's minimal polynomial is $x^2(x-1)(x+1)$?
I thought I could try to put this into Jordan blocks and say that since $2$ is the maximum value of the eigenvalue of $0$, then the characteristic polynomial for $T$ will have an exponent value equal to or greater than $2$ for the eigenvalue of $0$. So the eigenvalues are not distinct and therefore it is not diagonalizable.
But then for $1$ and $-1$ it could also mean that the characteristic polynomial has an exponent greater than or equal to $1$ and this doesn't really seem like a concrete proof at all to me. So I was just wondering of a better way I should prove this. Thanks.

Hint: What does it mean if the minimal polynomial contains a factor of $x^2$ instead of $x$? 
Alternate Hint: Show that if $T$ is diagonalizable then all distinct factors of the minimal polynomial appear with exponent of $1$, e.g. no squares like $x^2$. (Using the diagonalization)
