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.
 A: $\pi$ 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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 $\iff \forall y_U\in U, \forall x_W \in W, \langle x_W\mid y_U\rangle=0$

A: 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 $\pi=\pi^{*}$.
