Sum of the Eigenvalues of a Compact Positive-Definite Linear Operator on a Hilbert Space

Question

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.

Answer 1

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.