# 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?

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 $P^*P=\left(\begin{matrix}1 & 1\\0 & 0\end{matrix}\right)\left(\begin{matrix}1 & 0\\1 & 0\end{matrix}\right)=\left(\begin{matrix}2 & 0\\0 & 0\end{matrix}\right)$.
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).