Prove that a projection $A^2=A$ is self-adjoint if and only $\ker(A) \perp \operatorname{im}(A)$

Question

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?

Answer 1

Hint: Take an orthonormal basis $B_1$ of $\ker A$ and an orthonormal basis $B_2$ of $\operatorname{im}A$ and prove that $\langle Av,u\rangle=\langle v,Au\rangle$ for all $v,u\in B=B_1\cup B_2$.

Answer 2

For any projection $P=P^2$ on a Hilbert space $\mathcal{H}$, the identity $$ x=Px+(I-P)x $$ gives the decomposition $$ \mathcal{H}=\mathcal{R}(P)\oplus\mathcal{R}(I-P). $$ This is because any $y\in\mathcal{R}(P)\cap\mathcal{R}(I-P)$ must satisfy $$ (I-P)y=0=Py \implies y=0. $$ This is an orthogonal decomposition iff $$ \langle Px,(I-P)y\rangle=0,\;\; x,y\in\mathcal{H}, $$ which is equivalent to $P^*(I-P)=0$ or $P^*=P^*P$, which is also equivalent to $$ P=(P^*)^* = (P^*P)^* = P^*P. $$ And that implies $P=P^*$, which is also equivalent to the above.

Answer 3

This is true in any inner product space and the forward implication (the “only if” part) does not depend on the condition that $A^2=A$.

Suppose $A^\ast=A$. For any $x\in\ker(A)$ and $y\in\operatorname{im}(A)$, we have $y=Au$ for some vector $u$. Therefore $ \langle x,y\rangle =\langle x,Au\rangle =\langle A^\ast x,u\rangle =\langle Ax,u\rangle =\langle 0,u\rangle =0 $ and $\ker(A)\perp\operatorname{im}(A)$.

Now suppose $A^2=A$. Then $\operatorname{im}(I-A)\subseteq\ker(A)$. If we are also given that $\ker(A)\perp\operatorname{im}(A)$, then $ \langle A^\ast(I-A)x,y\rangle =\langle (I-A)x,Ay\rangle =0 $ for any vectors $x$ and $y$. Hence $A^\ast(I-A)=0$ and $A^\ast=A^\ast A$. Taking Hermitian adjoint on both sides, we also have $A=A^\ast A$. Therefore $A^\ast=A$.