Let $A$ be a projection such that $A^2=A$ then I want to prove that $A=A^* \Leftrightarrow \ker(A) \perp \operatorname{im}(A)$.
The implication $A=A^*\Rightarrow \ker(A) \perp \operatorname{im}(A)$ is easy, but I have troubles with the converse. Is there anybody who knows how to do this?
Best Answer
Let $P$ be a projection matrix in a vector space $V$. Let us also assume that the kernel of $P$ is a proper subspace of $V$: $\ker(P)\neq\{0\}$ and $\ker(P)\neq V$, as in these cases the result is trivial.
Any projection (self-adjoint or not) has eigenvalues in $\{0,1\}$. Moreover, one always has $\mathrm{im}(P)=\ker(P-I)$ (because if $y=Px$ then $Py=y$).
If $P$ is normal/self-adjoint (the two things are equivalent for projections), then its eigenspaces are orthogonal, that is, $\ker(P-I)\perp\ker(P)$. It follows that, if $P$ is self-adjoint, then $\mathrm{im}(P)\perp\ker(P)$.
For the other implication, let us assume $\mathrm{im}(P)\perp\ker(P)$. Again, this is equivalent to $\ker(P-I)\perp\ker(P)$ because $P$ is a projection. But this means that the eigenspaces of $P$ are orthogonal, which is equivalent to saying that $P$ is normal, and thus self-adjoint.