Is every linear projection normal? Let $V$ be a vector space. For simplicity, say finite-dimensional over the reals or the complex. A linear transformation $T \in \text{End}\,(V)$ is a projection if $T^2=T$. If $V$ is in addition an inner-product space then we can talk about the operator $T^*$ adjoint to $T$, defined by 
$$\forall v,w \in V, \langle Tv,w\rangle =\langle v,T^*w\rangle$$
An adjoint exists and is unique, at least for finite-dimensional spaces (compare here). $T$ is called an orthogonal projection if in addition it is self-adjoint, $T=T^*$.
Recall that a linear operator is normal if $T^*T=TT^*$. A self-adjoint operator is clearly normal.
Is every linear projection normal? How about orthogonal projections?
 A: An orthogonal projection is by definition self-adjoint and hence normal.
A non-orthogonal projection need not be normal, as the following example demonstrates. 
Let $P=\left(\begin{matrix}1 & 0\\1 & 0\end{matrix}\right)$ on the real plane. A direct computation shows that $P$ is a projection, $P^2=P$. The matrix of the adjoint is $\bar{P}^t=\left(\begin{matrix}1 & 1\\0 & 0\end{matrix}\right)$, which can also be verified directly to satisfy the definition of an adjoint.
However, $P$ is not normal as
$PP^*=\left(\begin{matrix}1 & 0\\1 & 0\end{matrix}\right)\left(\begin{matrix}1 & 1\\0 & 0\end{matrix}\right)=\left(\begin{matrix}1 & 1\\1 & 1\end{matrix}\right)$
while on the other hand 
$P^*P=\left(\begin{matrix}1 & 1\\0 & 0\end{matrix}\right)\left(\begin{matrix}1 & 0\\1 & 0\end{matrix}\right)=\left(\begin{matrix}2 & 0\\0 & 0\end{matrix}\right)$.
A: There is no need to talk about normal projections, because if $P^2=P$ then $P$ is normal iff $P$ is self-adjoint.
Indeed, assume $P^\dagger P=PP^\dagger$. Then $P$ is unitarily diagonalisable. But the minimal polynomial of any projection is $z\mapsto z(1-z)$ (unless $P=I$, in which case it is $z\mapsto z-1$), and thus the eigenvalues of $P$ are only $0$ and $1$. These are reals, and therefore $P$ must be Hermitian (and positive semi-definite as well).
