I am reading "Introduction to Linear Algebra" (in Japanese) by Kazuo Matsuzaka.
The following exercise (Exercise 6 on p.353) is in this book:
Let $V$ be a finite dimensional inner product space over $\mathbb{C}$.
Let $W$ be a subspace of $V$.
Let $W^{'}$ be a subspace of $V$ such that $V=W\oplus W^{'}$.
Let $P:V\to W$ be a projection on $W$ along $W^{'}$.
Prove the following:
If $||P(v)||\leq ||v||$ for all $v\in V$, then $P$ is the orthogonal projection from $V$ to $W$.
I solved this exercise.
I guess this exercise is not difficult, but my solution is very long.
I am not sure my solution is ok.
Please give me a right solution if my solution is wrong.
Please give me a better (more natural) solution even if my solution is ok.
We need to prove that $W^{'}=W^{\bot}$.
Assume that $W^{'}\neq W^{\bot}$.
If $(W^{'})^{\bot}=W$, then $W^{'}=((W^{'})^{\bot})^{\bot}=W^{\bot}$.
But $W^{'}\neq W^{\bot}$ by assumption.
So, $(W^{'})^{\bot}\neq W$.
$\dim (W^{'})^{\bot}=\dim V-\dim W^{'}$.
$\dim W=\dim V-\dim W^{'}$.
So, $\dim (W^{'})^{\bot}=\dim W$.
So, $W \not\subset (W^{'})^{\bot}$.
So, there exists $w$ such that $w\in W$ and $w\notin (W^{'})^{\bot}$.
Let $P^{'}:V\to (W^{'})^{\bot}$ be the orthogonal projection from $V$ to $(W^{'})^{\bot}$.
Since $w\in V$ and $V=(W^{'})^{\bot}\oplus ((W^{'})^{\bot})^{\bot}$, we can write $w=w_1+w_2$ ($w_1\in (W^{'})^{\bot}, w_2\in ((W^{'})^{\bot})^{\bot}$).
Since $w\notin (W^{'})^{\bot}$, $w_2\neq 0$.
$P^{'}(w)=w_1$ holds by the definition of $P^{'}$.
Let $\{v_1,\dots,v_r\}$ be an orthonormal basis for $(W^{'})^{\bot}$.
Let $\{v_1,\dots,v_r,v_{r+1},\dots,v_n\}$ be an orthonormal basis for $V$.
Then,
$w=\sum_{i=1}^{n} (v_i|w)v_i$,
$w_1=\sum_{i=1}^{r} (v_i|w)v_i$,
$w_2=\sum_{i=r+1}^{n} (v_i|w)v_i$.
$||P^{'}(w)||^2=||w_1||^2=(w_1|w_1)=\sum_{i=1}^{r} |(v_i|w)|^2$.
$||w||^2=(w|w)=\sum_{i=1}^{r} |(v_i|w)|^2+\sum_{i=r+1}^{n} |(v_i|w)|^2$.
Since $w_2\neq 0$, $||P^{'}(w)||^2<||w||^2$.
So, $||P^{'}(w)||=||w_1||<||w||$.
Since $w=w_1+w_2$, $w_1=w+(-w_2)\in W\oplus W^{'}$.
So, $P(w_1)=w$.
So, $||w||=||P(w_1)||\leq ||w_1||$.
This is a contradiction.
So, $W^{'}=W^{\bot}$.