Definition: Let $V$ be vector space, and $U$, $W$ be two subspaces such that $V=U\oplus W$.
We know that there exists for each $v \in V$ only one $u \in U$ and only one $w \in W$ such that $v=u+w$. Using this, we define a projection $P_{U,V}\colon V\longrightarrow V$ to be: $P_{U,W}(v)=u$
Now my question is this:
Let $V$ be an inner product space, and let $U$ be subspace of $V$. Let $\left\{e_{1},\ldots ,e_{n}\right\}$ be an orthogonal basis for $U$.
Let us define the orthogonal projection $P_{U}\colon V\longrightarrow V$ as
$$P_{U}(v)=\sum_{i=1}^n \langle v,e_{i}\rangle e_{i}$$
I need to prove that $P_{U}=P_{U,U^{\perp}}$.
$P_{U,U^{\perp}}$ is according to the definition in beginning.
How do I do it? I am sitting 1 hour on that and I have no clue.
plus I need to proof that $P_{U}$ is self adjoint
Basically, I have to prove that if $v=u+w$, then $P_{U}(v)=\sum_{i=1}^n \langle v,e_{i}\rangle e_{i}$ and
$P_{U,U\perp}(v)=u$ meaning I have to show $u=\sum_{i=1}^n \langle v,e_{i}\rangle e_{i}$.
But how do I do it ?
Best Answer
Hint: you can prove $P_U=P_{U,U^\perp}$ by checking that $P_U(v)=P_{U,U^\perp}(v)$ for every $v\in V$.
Added: You've got the start of the right strategy, but let me modify it it a bit. Let's start with your $v=u+w$ with $u\in U$ and $w\in U^\perp$ (I think you might be forgetting about this last fact.)
You know that $P_{U,U^\perp}(v)=u$
Since the $e_i$ are a basis for $U$, you can write $u=\sum \alpha_ie_i$ so that $v=\sum \alpha_ie_i +w$ where $w\in U^\perp$. Now, compute $P_{U}(v)=P_{U}(\sum\alpha_ie_i +w)=\_\_\_\_$.
Next hint: $P_{U}(\sum\alpha_ie_i +w):=\sum_j \langle \sum_i\alpha_ie_i +w,e_j\rangle e_j$