Prove that a self adjoint and idempotent matrix is a orthogonal projection matrix.

Question

$\newcommand{\R}{\operatorname{Ran}} \newcommand{\K}{\operatorname{Ker}}\newcommand{\b}{\mathbf}$

Prove that a self adjoint and idempotent matrix $P$ is an orthogonal projection matrix.

I was given the following hint:

Consider the decomposition $\b x = \b x_1 + \b x_2$, where $\b x_1 \in \R P$, and $\b x_2 ⊥ \R P$. Now show that $P \b x_1 =\b x_1$, and $P\b x_2 = \b 0$.

---

I proved the $2$nd condition of the hint as follows.

Since $\b x_2 \in (\R P)^\perp = \K P^* = \K P$, $P \b x_2 = 0$.

I think the first hint is not correct. I think it should be $\b x_1 - P \b x_1 \in \K P = (\R P)^\perp$, which is easy enough to prove. Am I right ?

Answer 1

If $P$ is an orthogonal projection matrix, it is with respect to its column space (range). Let $U$ be the column space of $P$: $$ U=\{Px:x\in\mathbb{R}^n\} $$ Take any $u\in U$; then $u=Pv$ for some $v$ and therefore $u=Pv=P^2v=P(Pv)=Pu$.

If $w\in U^\perp$, then $\langle w,u\rangle=0$, for every $u\in U$; it follows that, for every $u\in U$, $$ \langle Pw,u\rangle=w^TP^Tu=w^TPu=w^Tu=0 $$ so that $Pw\in U^\perp$.

Now write $x\in\mathbb{R}^n$ as $x=y+z$, with $y\in U$ and $z\in U^\perp$; then, by definition, $y$ is the orthogonal projection of $x$ onto $U$ and $$ Px=Py+Pz=y+Pz $$ with $Pz\in U^\perp$, as showed before. Since the decomposition of a vector as sum of a vector in $U$ and a vector in $U^\perp$ is unique, from the fact that $Px\in U$ it follows that $y=Px$ and $Pz=0$.

Therefore the orthogonal projection of $x$ onto $U$ is $y=Px$.

Answer 2

Let $p$ be the linear morphism represented in $n \in \mathbb{N}^{*}$ dimension. You know that $P^2=P$ then we have $p \circ p=p$ which means $p$ is a projection. That's the first step to it.

Now comes the hint, we need to prove that it is not a simple projection but a projection that satisfies $$\text{Ran } P =\left(\text{Ker }P\right)^{\perp}$$

Let $\displaystyle \langle x, y\rangle$ be the scalar product of the vectors $x$ and $y$. For $x \in \text{Ran } P$ and $y \in \text{Ker } P$ you know that $p(x)=x$ and $p(y)=0$ so first $$\displaystyle \langle x, y\rangle=\langle p\left(x\right),y\rangle$$ Then $p$ is self-adjoint then it is the same as $$\langle x,p\left(y\right)\rangle =\langle x, 0\rangle =0$$ And we proved that $p$ is an orthogonal projection and $P$ is the matrix of $p$ in "ladite"-base. I hope it helped you !