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Let $A$ be a compact, self-adjoint operator, $A \geq 0$. We need to prove that for any orthonormal system $\{e_i\}_1^{\infty}$ and for any $N$, $$\sum_1^N \langle Ae_i,e_i \rangle \leq \sum_1^N \lambda_i $$

Where $\lambda's$ are the eigenvalues ordered by decreasing order (as promised by the minimax principle).

Now, I have reached a line of inequalities, but am not sure about one of the stages. I'll mark it by $\leq^?$:

Take any $N$. Order $e_1,...,e_N$ such that $\langle Ae_1,e_1 \rangle \geq ... \geq\langle Ae_N,e_N \rangle $.

Then $\forall k$:

$$\langle Ae_k,e_k \rangle = \min_{1,...,k} \langle Ae_i,e_i \rangle \leq^? \min_{x \in span\{e_1,...,e_k\}} \langle Ax, x \rangle \leq \max_{V_k} \min_{x\in V_k} \langle Ax,x \rangle = \lambda_k$$

The last equality taken from the minimax principle. The $V_k$ are linear $k$-dimensional subspaces.

But obviously "$\leq^?$" can't be "$\leq$", since I'm taking minimum over a larger set in the RHS, so the minimal value can only decrease. Can "$\leq^?$" be a "$=$", though?

Also, if the statement is not true, what can I do about the question?

I also think that there may be a solution related to one of my previous questions, but I can't connect the two exactly.

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Let $A=\sum_k \lambda_k P_k$ be the orthogonal decomposition of $A$ where $u_k$ are its eigenvectors. Let $P=1-\sum_{k=1}^{N-1} P_k$ be the projection on the orthogonal complement of the space spanned by the first $N-1$ eigenvectors. We then have $$\begin{align}\langle Ae_i,e_i\rangle&= \langle \sum_{k=1}^\infty\lambda_kP_k e_i,e_i\rangle\\ &=\langle \sum_{k=1}^{N-1}\lambda_kP_k e_i,e_i\rangle+\langle \sum_{k=N}^\infty\lambda_kP_k e_i,e_i\rangle\\ &\leq \langle \sum_{k=1}^{N-1}\lambda_k P_k e_i,e_i\rangle+\langle \lambda_{N}P e_i,e_i\rangle \\ &=\sum_{k=1}^{N-1}\lambda_k|c_{ki}|^2+\lambda_N\left(1-\sum_{k=1}^{N-1}|c_{ki}|^2\right), \end{align}$$ where $c_{ki}=\langle u_k,e_i\rangle$. Hence $$\begin{align}\sum_{i=1}^N\langle Ae_i,e_i\rangle&\leq \sum_{i=1}^N \left(\sum_{k=1}^{N-1}\lambda_k|c_{ki}|^2+\lambda_N\left(1-\sum_{k=1}^{N-1}|c_{ki}|^2\right)\right)\\ &=\sum_{k=1}^{N-1}\lambda_k\sum_{i=1}^N|c_{ki}|^2+\lambda_N\left(N-\sum_{k=1}^{N-1}\sum_{i=1}^N|c_{ki}|^2\right)\\ &\leq\sum_{k=1}^{N-1}\lambda_k+\lambda_N\left(N-(N-1)\right)\\ &=\sum_{k=1}^{N}\lambda_k, \end{align}$$ where I used the fact that $$\sum_{i=1}^N|c_{ki}|^2\leq 1,$$ since $\{e_i\}$ is orthonormal.

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  • $\begingroup$ Very nice. Could you please share the intuition behind the proof? Or what lead you to prove it this way? I ask, because the proof above seems very technical to me - it reads like a series of formal manipulations that in the end brings us from point A to B, but I do not see what beforehand led us to choose this particular path. Thanks! $\endgroup$ – Aahz Sep 24 '14 at 15:18
  • $\begingroup$ The idea is that we can have an equality, only if the vectors $e_i$ are in the subspace spanned by the first eigenvectors. So to prove the inequality we must realize that if the vectors are in the orthogonal complement of this subspace, $\langle A e_i, e_i\rangle$ will be bounded by eigenvectors smaller than the first ones. But as they can have components on both spaces, we need to use the spectral projections to deal with each part separately. $\endgroup$ – Mateus Sampaio Sep 24 '14 at 15:33

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