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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.

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    $\begingroup$ 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 $u_1,u_2,w_1,w_2$. $\endgroup$ Nov 28, 2013 at 9:52
  • $\begingroup$ Why is that the case? $\endgroup$
    – Mathlete
    Nov 28, 2013 at 9:52
  • $\begingroup$ Actually, I understand why that resulting in zero would prove it is self adjoint but why have you chosen those particular vectors? $\endgroup$
    – Mathlete
    Nov 28, 2013 at 9:53
  • $\begingroup$ How have you proven that $U$ and $W$ are orthogonal complements? $\endgroup$
    – Mathlete
    Nov 28, 2013 at 9:55
  • $\begingroup$ 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. $\endgroup$ Nov 28, 2013 at 9:55

2 Answers 2

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$\pi$ self-adjoint

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

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

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

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

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

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

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  • $\begingroup$ This was exactly what I was looking for, thank you so much! $\endgroup$
    – Mathlete
    Nov 28, 2013 at 10:03
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    $\begingroup$ Why $$\forall y_U\in U, \forall x_W \in W, \langle x_W\mid y_U\rangle=0\impliedby \forall x_U,y_U\in U, \forall x_W,y_W \in W, \langle x_W\mid y_U\rangle=\langle x_U\mid y_W\rangle$$? $\endgroup$
    – Vim
    May 31, 2017 at 4:33
  • $\begingroup$ Never mind I got it. Just let $y=-x_U+x_W$. $\endgroup$
    – Vim
    May 31, 2017 at 4:39
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    $\begingroup$ @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. $\endgroup$
    – xavierm02
    Jul 4, 2017 at 15:45
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    $\begingroup$ @Mickey You can but it only proves the $\Leftarrow$ direction of the last step. $\endgroup$
    – xavierm02
    Jul 4, 2017 at 17:51
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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 $\pi=\pi^{*}$.

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