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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 ?

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wantToLearn
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1 Answers1

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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: $P_{U}(\sum\alpha_ie_i +w):=\sum_j \langle \sum_i\alpha_ie_i +w,e_j\rangle e_j$

rschwieb
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  • ohh I I think I understand. assuming v=$\sum_{i=1}^n ,a_{i}e_{i}$. when looking at 0=$\sum_{i=1}^n (\langle v,e_{i}\rangle-a_{i}) e_{i}$ then because $e_{1},...,e_{n}$ is a basis then $(\langle v,e_{i}\rangle-a_{i})$=0 and therefore $\langle v,e_{i}\rangle$=$a_{i}$. and from there the way is easier. right ? – wantToLearn May 20 '13 at 20:59
  • @wantToLearn Well, kind of, but you can't assume that $v$ is in the span of the $e_i$ (they only span $U$). Just try to compute the last line of my solution. But yes, it does turn out that $\langle v,e_i\rangle=a_i$ – rschwieb May 20 '13 at 21:05
  • in the first comment I ment $u$ and not $v$ . as far as I know the line is $\sum$$a_{i}e_{i}$. thanks for the hint but I still missis something. but thank you very much – wantToLearn May 20 '13 at 21:25
  • anyone, somebody know how to do it ? – wantToLearn May 21 '13 at 05:43
  • @wantToLearn After rereading it, I just noticed I left out a summation. It's fixed now. I'm not sure what else you are looking for. You have shown that $P_{U,U^\perp}(v)=P(v)$. Do you mean the self-adjoint part? – rschwieb May 21 '13 at 10:20
  • I do not mean yet for the adjoint I know that the what should be in the line is $\sum a_{i}e_{i}$ so basicly what I need to show that $a_{i}=$(v,$e_{i}$) for each i. so how do I do it – wantToLearn May 21 '13 at 11:38
  • @wantToLearn Apologies again, I discovered I was transposing the maps. I've added another hint line. The result you are seeking follows from the properties of the inner product. – rschwieb May 21 '13 at 12:06