If $A$ is nilpotent, why is $A$ non-invertible?

Question

Here $A$ is a square matrix. I'm confused as to why $A$ would be non-invertible as well.

Answer 1

If $A$ is invertible, $A^n$ is invertible for all $n$. But $0$ is not invertible.

Answer 2

If $\,BA = 1\,$ and $\,n\,$ is least with $\,A^n = 0\overset{\large\, B \times_{\phantom I}}\Rightarrow A^{n-1} = 0\,\Rightarrow\!\Leftarrow$

Answer 3

Suppose there were an inverse matrix $B$ such that, where $I$ is the identity: $$AB=I$$ It follows that $$A^nB^n= A(A(\cdots A(AB)B\cdots)B)B=I$$ But if $A^n = 0$ then $$0B^n=I$$ $$0=I$$ Hey wait...

Answer 4

Since $A$ is square and nilpotent $$ \det (A^n) = \left( \det(A) \right) ^n = \det(0) = 0, $$ so $\det(A) = 0$ and $A$ is singular.