Thread: Invertibility

Let $\displaystyle A$ be an $\displaystyle n \times {n}$ matrix
suppose that $\displaystyle A^2 = 0$. Prove that $\displaystyle A$ is not invertible.

I know that for $\displaystyle A$ to be invertible it must satisfy:
1) $\displaystyle T:A \rightarrow B$, such that $\displaystyle UT = I_{v}$ which would shows that it's 1-1.
2) $\displaystyle U:B \rightarrow A$, such that $\displaystyle TU = I_{w}$, showing that it's onto.

But since $\displaystyle A^2 =0$ even though $\displaystyle A \neq 0$, stipulating that the cancellation property for multiplication is not valid.

Is this right, or am I missing a large part?

Let us suppose that A is non zero and is invertible and arrive at a contradiction.

If $\displaystyle A $ is invertible then there exists a non zero matrix B such that
$\displaystyle AB = I_n = BA $ where $\displaystyle I $ is the identity matrix. But this would imply
$\displaystyle A = AI = A(AB) = A^2 B = 0B = 0, $
a contradiction.