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.


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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.