Let $V$ be a finite dimensional complex inner product space, and $P$ be a projection. Show that $P$ is normal if and only if it is an orthogonal projection.
My work:
For the statement "$P$ is normal implies it is an orthogonal projection", we need to show that for all $u, v$, $$ \langle Pu, v\rangle =\langle u, Pv\rangle $$
But we only know that $P^2=P$ and $P^*P=PP^*$. How to prove that?
We may assume that $P\neq 0.$ If $P^2=P$ then $(P^*)^2=P^*.$ The matrix $PP^*$ is also a projection as $$(P^*P)^2=P^*PP^*P=(P^*)^2P^2=P^*P.$$ Moreover $P^*P$ is orthogonal, hence $\|P^*P\|=1.$ Thus $$\|P\|^2=\|P^*P\|=1$$ i.e. $\|P\|=1.$ It is well known that a contractive projection is orthogonal, therefore $P$ is orthogonal (see)
By the spectral theorem the matrix of the transformation with respect to some orthonormal basis is diagonal. Because the transformation is a projection the square of the matrix is equal to itself. Therefore, the entries on the diagonal are either 0 or 1, which is the matrix of an orthogonal projection.
