Moore-Penrose equals original matrix

Question

I'm a bit stuck with my homework in a subject called "Matrices in Statistics".
The task is as follows:

Prove, that if $A$ is symmetric ($A=A^{T}$) and idempotent ($A=A^2$). Then
$$ A^{+} = A $$ Where $A^{+}$ is called the Moore-Penrose generalized inverse matrix.

Can you give me any ideas/tips, how to get started with this one? I would be very thankful.

Answer 1

Just show directly that the Moore-Penrose identities hold. For instance, $$AAA=A^2A=AA=A^2=A. $$

Answer 2

We write $B$ instead of $A^{+} $. $B$ is uniquely determined by the conditions

$(*) \quad ABA=A, BAB=B, (AB)^T=AB$ and $(BA)^T=BA$.

Now it is easy to see that $(*)$ holds with $B=A$, since $A^2=A$ and $A^T=A$.

You are done !