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Let $ A $ be a compact positive-definite linear operator on a Hilbert space $ \mathcal{H} $. Let $ \{ v_{1},v_{2},\ldots,v_{n} \} $ be an orthonormal $ n $-subset of $ \mathcal{H} $. Let $ \lambda_{1} \geq \lambda_{2} \geq \lambda_{3} \geq \ldots $ denote the non-increasing sequence of eigenvalues of $ A $. I want to show that $$ \sum_{i = 1}^{n} \langle A v_{i},v_{i} \rangle \leq \sum_{i = 1}^{n} \lambda_{n}. $$

So far, I know the Min-Max Principal, which solves the case for $ n = 1 $: $$ \lambda_{1} = \sup \{ \langle A x,x \rangle \mid \| x \| = 1 \}. $$

The extension of this principal to general values of $ n $ is $$ \lambda_{n + 1} = \min_{x_{1},\ldots,x_{n} \in \mathcal{H}} \sup \left\{ \langle A x,x \rangle ~ \Big| ~ \| x \| = 1 ~ \text{and} ~ x \in \{ x_{1},\ldots,x_{n} \}^{\perp} \right\}, $$ but I can’t seem to find a way to use this to solve the general case.

EDIT:

Developing an argument based on the comment below yields \begin{align} \sum_{i = 1}^{n} \langle A v_{i},v_{i} \rangle & = \sum_{m = 1}^{\infty} \sum_{i = 1}^{n} \lambda_{m} |\langle v_{i},e_{m} \rangle|^{2} \\ & = \sum_{m = 1}^{n - 1} \lambda_{m} \sum_{i = 1}^{n} |\langle v_{i},e_{m} \rangle|^{2} + \sum_{m = n}^{\infty} \lambda_{m} \sum_{i = 1}^{n} |\langle v_{i},e_{m} \rangle|^{2}. \end{align} As $ \{ v_{i} \}_{i = 1}^{n} $ is orthonormal and $ \| e_{m} \| = 1 $, the first term in the last line is bounded by $ \displaystyle \sum_{i = 1}^{n - 1} \lambda_{i} $. To finish the proof, I need to show that the second term is at most $ \lambda_{n} $. By the Min-Max Principal stated above, it suffices to show that for some $ x \in \text{Span} \! \left( \{ e_{m} \}_{m = n}^{\infty} \right) $, we have $$ \langle A x,x \rangle = \sum_{m = n}^{\infty} \lambda_{m} \sum_{i = 1}^{n} |\langle v_{i},e_{m} \rangle|^{2}. $$ (Alternatively, we can replace $ = $ by $ \leq $ and try to find an $ x $ that satisfies the resulting inequality.) My first guess for $ x $ was $ \displaystyle x = \sum_{m = n}^{\infty} \sum_{i = 1}^{n} \langle v_{i},e_{m} \rangle e_{m} $, but that only gives $$ \langle A x,x \rangle = \sum_{m = n}^{\infty} \lambda_{m} \sum_{i = 1}^{n} \sum_{j = 1}^{n} \langle v_{i},e_{m} \rangle \langle e_{m},v_{j} \rangle, $$ which isn’t exactly what I want (also, it doesn’t give a bound as some of the $ \langle v_{i},e_{m} \rangle \langle e_{m},v_{j} \rangle $-terms could be negative).

I feel that I’m close, but I’m not sure how to continue from here.

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Let $\{ e_{m} \}_{m=1}^{\infty}$ be the orthonormal subset of eigenvectors of $A$ with non-zero eigenvalues. That is, $Ae_{m}=\lambda_{m}e_{m}$, where the $\lambda_{m}$ are repeated according to multiplicity. If $P$ is the orthogonal projection of the Hilbert space $H$ onto the closure $\mathcal{R}(A)^{c}$ of the range of $A$, then $$ Py = \sum_{m}(y,e_{m})e_{m}. $$ Therefore, because $PA=AP=A$, $$ (Av_{i},v_{i})=(APv_{i},Pv_{i}) $$ Because this is homework, I won't go further.

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  • $\begingroup$ Thanks for the comment. I edited my post with some progress, but I'm still having some difficulties with the last step (assuming my direction is correct). $\endgroup$ – user1656308 Jun 6 '14 at 8:22
  • $\begingroup$ Further hint: $Pv_{i}=\sum_{n}(v_{i},e_{n})e_{n}$ and $APv_{i}=\sum_{n}\lambda_{n}(v_{i},e_{n})e_{n}$. You won't get the cross terms that you say 'can be negative.' $\endgroup$ – DisintegratingByParts Jun 6 '14 at 8:40
  • $\begingroup$ I'm not sure if I'm missing something trivial. I already used the last hint to get $\sum_{i=1}^n<Av_i,v_i> = \sum_m \sum_{i=1}^n \lambda_m |<v_i,e_m>|^2 $. One way to continue is to argue that since $v_i$ are orthonormal, then $ \sum_{i=1}^n |<v_i,e_m>|^2 \leq ||e_m||^2 = 1$ but this leaves us with the sum $ \sum_m \lambda_m $ but I only want the sum for $ m \leq n$ The other way I see is to split the sum for $ m \leq n-1 $ and $ m \geq n$ the way I did above, but then I don't see how I can used the given hint. $\endgroup$ – user1656308 Jun 6 '14 at 10:08
  • $\begingroup$ another thing I tried was to argue that since for $ m \geq n $ we have $\lambda_m \leq \lambda_n$ then $\sum_{m\geq n} \lambda_m \sum_{i=1}^n |<v_i,e_m>|^2 \leq \lambda_n \sum_{m\geq n} \sum_{i=1}^n |<v_i,e_m>|^2 = \lambda_n \sum_{i=1}^n \sum_{m\geq n} |<v_i,e_m>|^2 \leq \lambda_n \sum_{i=1}^n ||v_i||^2 = n\lambda_n$ which is not $\lambda_n$ as I wanted. $\endgroup$ – user1656308 Jun 6 '14 at 11:17
  • $\begingroup$ Ok, I got it. the last step was actually in the right direction. What was missing was to split the first term as well by $\lambda_m = \lambda_n + (\lambda_m - \lambda_n) $ $\endgroup$ – user1656308 Jun 6 '14 at 11:33

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