# Prove projection is self adjoint if and only if kernel and image are orthogonal complements

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.

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

$\pi$ self-adjoint

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

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

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

-

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

-

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

-

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

• This was exactly what I was looking for, thank you so much! Nov 28, 2013 at 10:03
• 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, 2017 at 4:33
• Never mind I got it. Just let $y=-x_U+x_W$.
– Vim
May 31, 2017 at 4:39
• @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. Jul 4, 2017 at 15:45
• @Mickey You can but it only proves the $\Leftarrow$ direction of the last step. Jul 4, 2017 at 17:51

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 $$=0$$.

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

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

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