Prove $A^n = 0$ for some number n can never be invertible I would like to prove that a matrix $A$ with $A^n=0$ for some number $n$ can never be invertible.
My professor said to prove it with contradiction. Any hints?
 A: For contradiction, suppose $A^{-1}$ exists such that $AA^{-1}=A^{-1}A=I$. Consider $A^n(A^{-1})^n$.
$$A^n(A^{-1})^n=0$$
since $A^n=0$. However,
$$A^n(A^{-1})^n=\underbrace{A\dots A}_n\underbrace{A^{-1}\dots A^{-1}}_n=I$$
since there are as many $A$ as there are $A^{-1}$, which all cancel out. This is a contradiction so $A^{-1}$ cannot exist.
A: We know that $\det AB = \det A \det B$. Therefore, $\det A^n = \left( \det A \right)^n$.
Clearly, $\det A^n = 0$ would imply $\det A = 0$, meaning $A$ is not invertible.
A: Let $A^N = 0$. If also $A=0$, you're done. Otherwise, there exists a maximal $n$ such that $A^n\neq 0$. Choose a vector $x$ such that $y := A^nx\neq 0$. Then $Ay = A(A^nx) = A^{n+1}x = 0$ since $A^{n+1}=0$. So, you have $Ay =0$ and $y\neq 0$. That is, $A$ cannot be invertible.
A: If $A$ is invertible then so is $A^n$: it isn't hard to verify that $(A^n)^{-1} = (A^{-1})^n$. By contrapositive, if $A^n$ is not invertible neither is $A$.
If $A^n = 0$ it will fail to be one-to-one since $A^nx = A^ny$ for all columns $x$ and $y$. 
Not one-to-one means not invertible.
Thus if $A^n = 0$, $A^n$ is not invertible. Consequently neither is $A$.
A: A matrix $M$ is invertible iff the linear map $f_M$ associated to it is surjective. If $f_A$ is surjective, so it is $f_M\circ f_M\circ \ldots\circ f_M$, associated to $M^n$, and the invertibility of $M$ implies the invertibility of $M^n$.
