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Let $W$ denote a subspace of a finite dimensional inner product space $V$, and let

$$\beta = \{w_1,w_2,\dots,w_r\}$$

denote an orthogonal basis for $W$. For any $v\in V$ define

$$proj_{\beta}v = \sum \limits_{j=1}^r \frac{\langle v,w_j\rangle}{\langle w_j,w_j\rangle}w_j$$

Prove that $proj_\beta v$ is independent of the choice of orthogonal basis for $W$, i.e. if $\gamma$ is any orthogonal basis for $W$ then

$$proj_\beta v = proj_\gamma v$$

Are these answers right? I don't get it



Now here I don't really see what $\gamma$ is. Am I just making: $$\gamma = \{y_1,y_2,\dots,y_r\}?$$

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Yeah, that's right. $\gamma$ is an orthogonal basis. –  ChocolateAndCheese May 29 at 3:39
    
@ChocolateAndCheese So I just need to change the basis, and prove that it is the same? I don't know how to do this, but I have heard of the Gram-Schmidt process, is that the way to go? –  user142198 May 29 at 5:10
1  
For pretty angle brackets use \langle and \rangle. –  Marc van Leeuwen May 30 at 4:18

3 Answers 3

I think the best approach for such questions, showing independence of a given definition from choices made in it, is to give a different characterisation of the notion in which no choices are made at all, and show that it is equivalent to the given definition.

In the current case there is a fairly easy way though to prove directly that the projections defined using two different orthonormal bases of $W$ give the same result, if you are willing to use the standard fact that for any orthonormal basis $b_1,\ldots,b_n$ of$~V$ and any $v\in V$ the coordinates of $v$ w.r.t. the basis are given by inner products: $v=\sum_{i=1}^n\langle v,b_i\rangle b_i$. First prove your result explicitly for the case where $\gamma$ is the orthonomal basis obtained by normalising $\beta$ (this is easy). This allows to reduce the problem to the case for orthonormal bases, for which one can drop the denominators.

Now given two orthonormal bases $\beta,\gamma$ of $W$, choose an orthonormal basis $\delta=[\delta_1,\ldots,\delta_{n-r}]$ of $W^\perp$; now one can combine $\beta$ and $\delta$ to form an orthonormal basis $\beta'$ of $V$, and similarly combine $\gamma$ and $\delta$ to obtain an orthonormal basis $\gamma'$ of $V$. Using the mentioned standard fact for $\beta'$ and $\gamma'$ on can now write $$ \operatorname{proj}_\beta(v) =v-\sum_{i=1}^{n-r}\langle v,\delta_i\rangle\delta_i =\operatorname{proj}_\gamma(v). $$

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Let $P$ be the projector upon $W$. Then, for any orthonormal basis $\{y_{j}\} $ for $W$, \begin{equation*} P=\sum_{j=1}^{r}<\;,y_{j}>y_{j} \end{equation*} Thus \begin{equation*} Pv=\sum_{j=1}^{r}<v,y_{j}>y_{j} \end{equation*} In particular we can take $\{y_{j}\}=\{w_{j}\}$.

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You are just assuming the result (that projections are well defined independently of the basis); this answer adds nothing to justify that fact. –  Marc van Leeuwen May 30 at 4:22
    
The concept of a subspace and the projector upon it is more fundamental than that of a basis (see, for instance, Kato, Perturbation Theory of Linear Operators). If we have a separable Hilbert space, then, given the subspace, we can use the Gram-Schmidt construction to arrive at an orthonormal basis. –  Urgje May 30 at 6:45
    
That may be true, but is irrelevant in the setting of the current question. OP manifestly does not use a basis-free definition of projection; if you want to use such a definition instead, then answering the question essentially comes down to showing your definition equivalent the the one given by OP (which you do not do). –  Marc van Leeuwen May 30 at 7:03

Let $w:=\text{proj}_\beta v$. Obviously $w\in W$, and by direct computation $\langle v-w,w_j\rangle=0$ for all $j$. Thus, $v-w$ is orthogonal to each $w_j$, and hence to every vector in $W\!$. These two conditions, $w\in W$ and $v-w\perp W$, are basis independent and it remains to show that they produce a unique vector $w$ given by the same formula in any orthogonal basis.

Let $u_1,\dots,u_r$ be another orthogonal basis of $W$, then $w=a_1u_1+\dots a_ru_r$ for some coefficients $a_j$. Taking the inner product with $u_j$ and using that $\langle u_i,u_j\rangle=0$ for $i\neq j$ we get $\langle v,u_j\rangle-a_j\langle u_j,u_j\rangle=0$. Solving for $a_j$ gives the expected formula in this new basis, and also proves uniqueness of $w$ since all $a_j$ are uniquely determined.

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