13

Let $V$ be an IPS and suppose $\pi : V \to V$ is a projection so that $V = U \oplus W$ (ie $ V = U + W$ and $U \cap W = \left\{0\right\}$) $ \ $ where $U = \ker(\pi)$ and $W = \operatorname{im}(\pi)$, and if $v = u + w \ $ (with $u \in U, \ w \in W$) then $\pi(v) = w$.
Prove $\pi$ is self adjoint if and only if $U$ and $W$ are orthogonal complements.

I'm hoping someone can give me a few hints on how to begin this question.

PinkyWay
  • 4,565
Mathlete
  • 1,347
  • 3
    Compute $\langle \pi (u_1+w_1), u_2+w_2\rangle - \langle u_1+w_1,\pi(u_2,+w_2)\rangle$. The projection is self-adjoint if and only if that results in $0$ for all $u_1,u_2,w_1,w_2$. – Daniel Fischer Nov 28 '13 at 09:52
  • Why is that the case? – Mathlete Nov 28 '13 at 09:52
  • Actually, I understand why that resulting in zero would prove it is self adjoint but why have you chosen those particular vectors? – Mathlete Nov 28 '13 at 09:53
  • How have you proven that $U$ and $W$ are orthogonal complements? – Mathlete Nov 28 '13 at 09:55
  • They aren't particular, that says "for all $v_1,v_2 \in V$. I just chose a particular representation in the hope that would help. – Daniel Fischer Nov 28 '13 at 09:55
  • Can anyone provide an example of an orthogonal projection such that its kernell and image are not orthogonal? I can´t come up with any... if it is not possible this would mean that orthogonal projection is always self adjoint. Is this it? – astro Oct 30 '23 at 23:23
  • Added: The example probably comes from a function´s space , maybe. I mean, there has to be counterexamples but I could not come up with one yet. – astro Oct 31 '23 at 19:34

2 Answers2

15

$\pi$ self-adjoint

$\iff \forall x, y \in V, \langle \pi(x)\mid y\rangle=\langle x\mid \pi(y)\rangle$

$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle \pi(x_U+x_W)\mid y_U+y_W\rangle=\langle x_U+x_W\mid \pi(y_U+y_W)\rangle$

$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U+y_W\rangle=\langle x_U+x_W\mid y_W\rangle$

-

$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle+\langle x_W\mid y_W\rangle=\langle x_U\mid y_W\rangle+\langle x_W\mid y_W\rangle$

-

$\iff \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle=\langle x_U\mid y_W\rangle$

-

$\iff \forall y_U\in U, \forall x_W \in W, \langle x_W\mid y_U\rangle=0$

xavierm02
  • 7,495
  • This was exactly what I was looking for, thank you so much! – Mathlete Nov 28 '13 at 10:03
  • 1
    Why $$\forall y_U\in U, \forall x_W \in W, \langle x_W\mid y_U\rangle=0\impliedby \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle=\langle x_U\mid y_W\rangle$$? – Vim May 31 '17 at 04:33
  • Never mind I got it. Just let $y=-x_U+x_W$. – Vim May 31 '17 at 04:39
  • @xaviermo2 is the last transition because $x_U,y_W$ are orthogonal therefore the inner product is $0$? or is there any other reason? I didn't quite understand it, thanks! – Mickey Jul 04 '17 at 15:01
  • 2
    @Mickey The $\Rightarrow$ is because you can set $x_U=0=y_W$. The $\Leftarrow$ is because by the hypothesis, both sides will be equal to zero, so that both sides are indeed equal. – xavierm02 Jul 04 '17 at 15:45
  • @xavierm02 how can you set them to be $0$? Don't you need to prove it for all $x,y$? Can't you "stop" at the last transition and just show that because they are orthogonal then both of the sides equal to $0$ and that's why we have an equality? – Mickey Jul 04 '17 at 15:51
  • @Mickey The very last statement is exactly saying that $V$ and $W$ are orthogonal. So you can't use it: you're trying to prove it. I can set them to be $0$ because the property is true for all $x_U$ and for all $y_W$, and is therefore true in the particular case where both are $0$. – xavierm02 Jul 04 '17 at 16:28
  • @xavierm02 , okay, I think I didn't explain myself well enough. If I assume that the linear transformation is a projection, and I want to prove that it is self-adjoint, then - can I use the orthogonality to prove the equality between the inner products? – Mickey Jul 04 '17 at 16:39
  • 1
    @Mickey You can but it only proves the $\Leftarrow$ direction of the last step. – xavierm02 Jul 04 '17 at 17:51
  • @xavierm02 thanks, that's what I wanted to prove! :) – Mickey Jul 04 '17 at 18:13
  • What is shown is that $U$ and $V$ are orthogonal, not orthogonal complements of each other. Can you clarify? – Atom Oct 16 '22 at 05:29
0

If a projection $\pi$ is self adjoint on finite dimensional inner product space $V$, we need to show $\pi$ orthogonal projection.

Take $y \in$ Range$(\pi)$ and $x \in$ Null$(\pi)$, we need to show $<y, x>=0$.

$<y,x>=<\pi(y),x>=<y, \pi(x)>=<y,0>=0$.

Null$(\pi) \subseteq$ Range$(\pi)^{\perp}$. Dim(Null$(\pi))$=Dim(Range$(\pi)^{\perp})$. So Null$(\pi)=$ Range$(\pi)^{\perp}$. For converse, take orthonormal basis, $\{e_{1}, e_{2},...e_{k},..., e_{n}\}$.

Range$\pi=\{e_{1}, e_{2},...e_{k}\}$. For any $y \in V$, $y=a_{1}e_{1}+a_{2}e_{2}+...+a_{n}e_{n}$, $<\pi(e_{i}), y>=<\pi(e_{i}),a_{1}e_{1}+a_{2}e_{2}+...+a_{n}e_{n}>=\overline{a_{i}}=<e_{i}, \pi(y)>$, for $i=1,...,k$.

For $\{e_{k+1},..., e_{n}\}$, $<\pi(e_{j}), y>=0=<e_{j}, \pi(y)>=<e_{j}, a_{1}e_{1}+a_{2}e_{2}+...+a_{k}e_{k}>0$. So $\pi=\pi^{*}$.

Supriya Saha
  • 51