Invertibility

**lllll** · Mar 3rd 2008, 06:52 PM

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

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

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

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

**perwiradua** · Mar 3rd 2008, 08:03 PM

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

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

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