2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.