Endomorphism is normal and idempotent iff it is an orthogonal projection.

Question

I've searched for answers for this question here for some time but haven't found an applicable answer because I could only find related questions, but not this one in particular.

Suppose $V$ is a finite dimensional (real or complex) inner product space, and $f: V \rightarrow V$ an endomorphism. Then $f$ is an orthogonal projection on $im(f) = U$ iff $f$ is normal and idempotent. Here orthogonal projection means that $f(v) = u$ with the unique decomposition $v = u + u'$, where $u \in U, u' \in U^{T}$.

The first direciton is straight forward, because you immediately get an orthonormal basis of $U$, fill it up with an orthonormal basis of $U'$ and use that all these vectors are eigenvectors.

The problem is the other direction, how do I proceed here? I know that essentially I have to show that $\forall u' \in U^{T}: f(u') = 0$, because an idempotent implies identity on it's image, but I don't know where to use that $f$ is normal. I have somehow gotten an argument without using that $f$ is normal, but this one has to be wrong I guess.

Any help is appreciated!

Answer 1

Let $f^*$ denote the adjoint of $f$. Let $u' \in U^\bot$. Then

$$\langle f(u'), f(u')\rangle = \langle u', f^*(f(u'))\rangle = \langle u', f(f^*(u'))\rangle = 0.$$

The second equality follows from normality, the third holds because $f(f^*(u')) \in U$ and $u' \in U^\bot$.

Answer 2

Being idempotent implies that $f$ is a projection, without the normality assumption. You already know it implies $f$ is the identity on its image; it follows that $v = fv + (1-f)v$ (valid for any linear map) splits $v$ into $u\in\operatorname{im} f$ and $u'\in\ker f$. (Note that $(1-f)v\in \ker f$ as $f(1-f) v = fv-f^2v = fv-fv=0$.) It follows that every element of $V$ is represented as a sum of an element of $U=\operatorname{im}f$ and $U'=\ker f$, and that $f$ sends $v$ to the former, forgetting the latter. The representation is unique because $\ker f \cap \operatorname{im} f = 0$ since if $v$ is in the image then $v=fv$, but if also in the kernel then $fv=0$.

You need the normality to conclude that the projection is orthogonal, i.e. that $U = \ker f$ and $U' = \operatorname{im} f$ are orthogonal to each other. You can get this from McFry's answer.