Show that if $A$ is any square matrix such that $A^n = 0$ for some positive intiger $n$, then $A$ is not invertible. (answer check)

Question

Show that if $A$ is any square matrix such that $A^n = 0$ for some positive integer $n$, then $A$ is not invertible.

I'm not sure if my proof is good enough, or enough "work" as my teacher put it after my last test.

This is what I have:

$$A^n = 0$$

or $A$ to be invertible, $$A^n A^{-n} = I$$

Then:

$$A^{n} \cdot A^{n} =0 \cdot A^{-n} $$

$$I \neq 0$$

this is not true, so $A$ is not invertible.

Answer 1

$0=det(A^n)=(det(A))^n \implies det(A)=0$ thus $A$ is not invertible.

Answer 2

You can also do this: If $n$ is the small natural number such that $A^n=0,$ then let $v$ be such that $A^{n-1}v\ne 0,$ then $A(A^{n-1}v)=0$ and so $A$ isn't invertible.

Answer 3

Hint:If $A$ is nilpotent operator, then its only eigenvalue is zero.