A projection which has $||E\alpha|| \le ||\alpha||$ is an orthogonal projection

Question

this problem screwed me up for a long time. Please help me solve this.

$V$ is an inner product space and $W$ is a finite-dimensional subspace of $V$. Prove that if $E$ is a projection with range $W$, then $E$ is the orthogonal projection on $W$ iff $||E\alpha|| \le ||\alpha||$ for all $\alpha$ in $V$.

I can prove that if $E$ is an orthogonal projection then $||E\alpha|| \le ||\alpha||$ using property: $(E\alpha | \beta) = (\alpha|\beta)$ for all $\alpha$ in $V$ and $\beta$ in W. But the converse seems more difficult.

score 6 · Accepted Answer · edited Apr 12 '19 at 19:12

6

Let $V = W \oplus \ker(E)$. If $E$ is not orthogonal, then there is $b\in ker(E)$ and $a\in W$ such that $(a|b)\neq 0$. Let us consider $\alpha=\lambda a+b$.

$$\|\alpha\|^2= \|\lambda a\|^2+2\lambda\operatorname{Re}(a|b)+\|b\|^2 = \|E\alpha\|^2+2\lambda\operatorname{Re}(a|b)+\|b\|^2$$

using $\lambda = -\dfrac{\|b\|^2}{\operatorname{Re}(a|b)}$ yields $\|\alpha\|^2 = \|E\alpha\|^2-\|b\|^2 < \|E\alpha\|^2$

so if it's not orthogonal, the relation is not verified. If it is verified, the projection must be orthogonal.

edited Apr 12 '19 at 19:12

user0

3,567

answered Mar 06 '13 at 14:14

imj

1,460

Thanks so much. It's a simple and brief solution for such a problem – le duc quang Mar 07 '13 at 01:34
1

You can use the same method for the converse : If $E$ is orthogonal, and $\alpha=a+b$ then $|\alpha|^2= |\ a|^2+|b|^2 = |E\alpha|^2+|b|^2 \geq |E\alpha|^2$ – imj Mar 07 '13 at 01:46
It's more simple than my method. so great ^^.thanks – le duc quang Mar 07 '13 at 02:31
Why $(a|b)=(b|a)$ in your solution? – Majid May 31 '18 at 16:25
@Majid It's not (unless it is a real number). But they are conjugates, so the imaginary parts cancel out and you're left with twice the real part (Re) (sry I'm late) – imj Jul 04 '21 at 10:52

A projection which has $||E\alpha|| \le ||\alpha||$ is an orthogonal projection

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