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Let $V$ be a real inner product space, and let $v_1,v_2, \dots ,v_k$ be an orthonormal set of vectors. How do you prove that

$$\sum_{i=1}^k | \langle x,v_i \rangle \langle y,v_i\rangle| \leq \|x\|\cdot\|y\|?$$

When does the equality hold?

I've been trying to do this with the Bessel and Cauchy-Schwarz inequalities, but I can't make it work yet. Any help would be greatly appreciated.

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    **Hint**: $\langle x, v_i\rangle$ are *coordinates*.2011-11-20
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    Use cardinal's *excellent* hint, and look again at [Cauchy-Schwarz](http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality).2011-11-20
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    @robjohn: I'm sorry, but I am still very confused!2011-11-20
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    @cardinal: How do you calculate the coefficient of $v_i$ in the unique linear combination of $v_1,\dotsc,v_k$ which equals $v$?2011-11-20
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    I have used cardinal's hint in my answer (although I don't mention it explicitly). I hope that that clears things up.2011-11-20

1 Answers 1

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We don't necessarily know that $\{v_i:1\le i\le k\}$ is a basis of $V$ since we don't know the dimension of $V$, but we are given that $\{v_i\}$ are orthonormal; that is $\left=0$ when $i\not=j$ and $\left=1$.

Consider the vector $$ x^\perp=x-\sum_{i=1}^k\leftv_i\tag{1} $$ $x^\perp$ is perpendicular to $\{v_i\}$: $$ \begin{align} \left &=\leftv_i,v_j\right>\\ &=\left-\left\left\\ &=0\tag{2} \end{align} $$ Therefore, $x^\perp$ is perpendicular to $x-x^\perp=\sum\limits_{i=1}^k\leftv_i$.

Next consider $$ \begin{align} \left &=\left<\sum_{i=1}^k\leftv_i,\sum_{j=1}^k\leftv_j\right>\\ &=\sum_{i=1}^k\left\left\left\\ &=\sum_{i=1}^k\left\left\tag{3} \end{align} $$ Note that since $x^\perp$ is perpendicular to $x-x^\perp$, $$ \|x-x^\perp\|^2+\|x^\perp\|^2=\|x\|^2\tag{4} $$ which implies that $\|x-x^\perp\|\le\|x\|$.

Now $(3)$, Cauchy-Schwarz, and $(4)$ yield $$ \begin{align} \left|\sum_{i=1}^k\left\left\right| &=\left|\left\right|\\ &\le\|x-x^\perp\|\|y-y^\perp\|\\ &\le\|x\|\|y\|\tag{5} \end{align} $$ To finish off the proof (thanks to cardinal), consider the vector $$ x^+=\sum_{i=1}^k\left|\left\right|v_i\tag{6} $$ Note that $\|x^+\|^2=\sum\limits_{i=1}^k\left^2=\|x-x^\perp\|^2$.

Plugging $x^+$ and $y^+$ into $(5)$ gives $$ \begin{align} \sum_{i=1}^k\left|\left\left\right| &=\left|\sum_{i=1}^k\left\left\right|\\ &\le\|x^+\|\|y^+\|\\ &\le\|x\|\|y\|\tag{7} \end{align} $$

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    ...and, to conclude, note that this holds for *arbitrary* $x$ and $y$, hence, it holds for $x^+ = \sum_{i=1}^k |\langle x, v_i \rangle | v_i$ and similarly defined $y^+$. Noting that $\|x^+\| \leq \|x\|$ and $\|y^+\| \leq \|y\|$ completes the proof. (+1)2011-11-20
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    @cardinal: isn't $(5)$ exactly what was asked? In fact, $x^+=x-x^\perp$ so I think your concerns are already handled.2011-11-20
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    The modulus is on the inside of the sum and $x^+ \neq x - x^\perp$ in general.2011-11-20
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    @cardinal: Oh, I missed that. Thanks. Do you mind if I incorporate the fix?2011-11-20
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    Sure. Please do. Cheers. :)2011-11-20
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    @robjohn Nice. Sorry if my "answer" was confusing. I didn't read your question as carefully as I should have.2011-11-20
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    Thanks! This is brilliant! However i'm slightly unsure about line $(4)$, could you explain why this holds?2011-11-20
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    @LHS: Pythagoras. (Or, just check $\langle x^\perp, x-x^\perp \rangle = 0$ by the definitions.)2011-11-20
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    @cardinal: Oh so you can just say that by Pythagoras, interesting. Thank you and robjohn so much for your help, it's been invaluable!2011-11-20
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    @robjohn: Right below (6) you have $\|\cdot\|$ in two places where it should be $\|\cdot\|^2$. :)2011-11-20
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    @cardinal: fixed. Thanks again for your comments!2011-11-20
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    This is a very nice answer.2011-11-20