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$ \newcommand{\ip}[2]{\left\langle #1,#2 \right\rangle} $

Here is the statement of the problem:

Suppose that $V$ is a real inner product space with an inner product $\langle\cdot,\cdot\rangle$, and let $\left\{ e_1,\dots,e_k\right\}$ be an orthonormal set of vectors in $V$. For any vector $v\in V$, show that

$$ \sum_{i=1}^k \ip{v}{e_i}^2 \le ||v||^2 $$

when does the equality hold?

My attempt: I'm bascially stuck from the beginning. All of my approaches indirectly had the (incorrect) assumption that $V$ can be expressed as a direct sum of subsapce of $V$ spanned by $\{e_1,\dots,e_k\}$ and its orthogonal complement, which I know is true for finite dimensional vector spaces. But I spotted that there were no assumption about $V$ being finite dimensional.

But then I couldn't think of any effective approach afterwards. Any help would be appreciated!

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First we prove Pythagoras Theorem for inner product space $$ \left\|\sum \limits_{i=1}^{k}a_i e_i\right\|^2=\sum \limits_{i=1}^{k}|a_i|^2 \|e_i\|^2 \hspace{5 mm} $$ $e_i$ is the orthonormal base.

\begin{align} \left\|\sum \limits_{i=1}^{k}a_i e_i\right\|^2&=\langle\sum \limits_{i=1}^{k}a_i e_i,\sum \limits_{i=1}^{k}a_i e_i \rangle \\ &=\sum \limits_{i=1}^{k}a_i \langle e_i,\sum \limits_{j=1}^{k}a_j e_j\rangle \\ &= \sum \limits_{i=1}^{k}a_i \overline{\langle \sum \limits_{j=1}^{k}a_j e_j,e_i\rangle} \\ &=\sum \limits_{i=1}^{k}a_i \langle \sum \limits_{j=1}^{k}\overline{a_j} \overline{e_j},\overline{e_i}\rangle \\ &=\sum \limits_{i=1}^{k}a_i \sum \limits_{j=1}^{k}\overline{a_j} \langle\overline{e_j},\overline{e_i}\rangle \\ &=\sum \limits_{i=1}^{k} \sum \limits_{j=1}^{k}a_i\overline{a_j} \langle e_i,e_j\rangle \\ &=\sum \limits_{i=1}^{k}\overline{a_i}a_i \langle e_i,e_i\rangle=\sum \limits_{i=1}^{k}|a_i|^2 \|e_i\|^2 \end{align} Next \begin{align} \left\|v-\sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i\right\|^2&=\langle v-\sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i\,,v-\sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i\ \rangle \\ &=\langle v,v\rangle+\langle \sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i,\sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i\rangle-\langle v,\sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i\rangle-\langle \sum \limits_{i=1}^{k}\langle v,e_i\rangle e_i,v\rangle \hspace{7 mm} \text{(by Pythagoras Theorem)} \\ &=\|v\|^2+\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2\|e_i\|^2-\sum \limits_{i=1}^{k}\overline{\langle v,e_i\rangle}\langle \overline{e_i},\overline{v}\rangle-\sum \limits_{i=1}^{k}\langle v,e_i\rangle\langle e_i,v\rangle \\ &=\|v\|^2+\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2-\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2-\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2 \\ &=\|v\|^2-\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2 \\ &\geqslant 0 \end{align}

So $\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2\leqslant \|v\|^2$

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  • $\begingroup$ Yes, I see it now. My problem was little bit simpler than this since it was real vector space (no need for conjugacy). Thanks for detailed answer! $\endgroup$ – user160738 May 30 '15 at 4:48
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By Pythagoras, $\|v\|^2 = \|\sum_i \langle v,e_i\rangle e_i\|^2 + \|v - \sum_i\langle v,e_i\rangle e_i\|^2$.

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  • $\begingroup$ So I guess you are assuming that $\sum_i \langle v,e_i\rangle e_i$ and $v-\sum_i \langle v,e_i\rangle e_i$ are orthogonal? But isn't that equivalent to saying, that for any $v\in V$ there is a component of $v$ lying in the space spanned by $\{e_1,\dots,e_k\}$ and a component that is orthogonal to that space, which means $V$ can be expressed as direct sum of these two spaces? But that is not true for infinite dimensional $V$ in general, is it? $\endgroup$ – user160738 May 29 '15 at 5:40
  • $\begingroup$ We don't have to assume; just compute their inner product and you'll see. This is essentially a proof that $V$ can be so decomposed. The difficulty with infinite-dimensional $V$ would only be if there were infinitely many $e_i$, so that the sums there were actually limits, and maybe the space is not complete. $\endgroup$ – user21467 May 29 '15 at 5:52

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