1. 'Diagonalizable' proof

Let $A \in M_{n}(R)$ such that $A^{m}=0$ for some positive integer $m$. If $A$ is a nonzero matrix, show that $A$ is not diagonalizable.

I proved that if $A^{m}=0$ for some positive integer $m$, then the only eigenvalue of $A$ is $0$. I think it can be proven by contradiction. But I'm still struggle.

2. You've almost got it. If, by contradiction, A is diagonalizable and all its eigenvalues are 0, then what's A's corresponding diagonal matrix? Do you see why that is a contradiction ( $A \neq 0$)?

3. Originally Posted by Defunkt
You've almost got it. If, by contradiction, A is diagonalizable and all its eigenvalues are 0, then what's A's corresponding diagonal matrix? Do you see why that is a contradiction ( $A \neq 0$)?

Thank you. I got it.