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We define :

  • $n \in \mathbb{N}$
  • $\mathcal{B}=(e_1 \dots e_n) $ is the initial basis orthonormal
  • $A$ is the matrix symmetric real of $f$ in $\mathcal{B}$
  • $\mathcal{B'}=(\epsilon_1 \dots \epsilon_n)$ is an orthonormal basis
  • $\forall ~ 1\leq i \leq n, ~ f(\epsilon_i)= \lambda_i \epsilon_i$
  • $\lambda_n \leq \lambda_{n-1} \dots \leq \lambda_1$

We want to prove that : $$\forall k \leq n, \quad \sum_{j=1}^{k} a_{j,j} \leq \sum_{j=1}^{k} \lambda_j $$

Here are the following steps : $\forall ~ 1 \leq j \leq n$

  1. $ a_{j,j}\leq \sum_{i=1}^{k} \lambda_i <e_j,\epsilon_i> ^2 + \sum_{i=k+1}^{n} \lambda_k <e_j,\epsilon_i> ^2$
  2. $ a_{j,j} \leq \sum_{i=1}^{k} ( \lambda_i - \lambda_k )<e_j,\epsilon_i> ^2 + \lambda_k $
  3. We deduce the claim

My attempt :

  1. Let $P$ the matrix of transition from $\mathcal{B}$ à $\mathcal{B'}$ $e_j= \sum_{k=1}^{n} p_{jk} \epsilon_k \Rightarrow p_{j,k} = <e_j,\epsilon_k>$ for all $k$

$ \begin{align*} u(e_j) &= u( \sum_{i=1}^{n} <e_j,\epsilon_i> \epsilon_i ) \\ &= \sum_{i=1}^{n} <e_j,\epsilon_i> u( \epsilon_i ) \\ &= \sum_{i=1}^{n} <e_j,\epsilon_i> \lambda_i \epsilon_i \\ &= \sum_{i=1}^{n} <e_j,\epsilon_i> \lambda_i \epsilon_i \end{align*} $

$ \begin{align*} a_{j,j}&=<u(e_j),e_j> \\ &= <e_j, \sum_{i=1}^{n} <e_j,\epsilon_i> \lambda_i \epsilon_i > \\ &= \sum_{i=1}^{n} <e_j,\epsilon_i> \lambda_i <e_j, \epsilon_i > \\ &= \sum_{i=1}^{n} \lambda_i <e_j,\epsilon_i> ^2 \\ &= \sum_{i=1}^{k} \lambda_i <e_j,\epsilon_i> ^2 + \sum_{i=k+1}^{n} \lambda_i <e_j,\epsilon_i> ^2 \\ &\leq \sum_{i=1}^{k} \lambda_i <e_j,\epsilon_i> ^2 + \sum_{i=k+1}^{n} \lambda_k <e_j,\epsilon_i> ^2 \\ \end{align*} $

2.

$ \begin{align*} 1&=\| e_j\| \\ &= \sum_{i=1}^{n} <e_j,\epsilon_i> ^2 \\ \end{align*} $

$ \begin{align*} a_{j,j} &\leq \sum_{i=1}^{k} \lambda_i <e_j,\epsilon_i> ^2 + \sum_{i=k+1}^{n} \lambda_k <e_j,\epsilon_i> ^2 \\ &= \sum_{i=1}^{k} \lambda_i <e_j,\epsilon_i> ^2 + \lambda_k ( 1- \sum_{i=1}^{k} <e_j,\epsilon_i> ^2 \\ &= \sum_{i=1}^{k} ( \lambda_i - \lambda_k )<e_j,\epsilon_i> ^2 + \lambda_k \\ \end{align*} $

  1. $ \sum_{j=1}^{k} a_{j,j} \leq \sum_{j=1}^{k} \sum_{i=1}^{k} ( \lambda_i - \lambda_k )<e_j,\epsilon_i> ^2 + \sum_{j=1}^{k} \lambda_k $ $ \begin{align*} \sum_{j=1}^{k} a_{j,j} &\leq \sum_{j=1}^{k} \sum_{i=1}^{k} ( \lambda_i - \lambda_k )<e_j,\epsilon_i> ^2 + \sum_{j=1}^{k} \lambda_k \\ &\leq \sum_{j=1}^{k} \sum_{i=1}^{k} ( \lambda_i - \lambda_k )<e_j,\epsilon_i> ^2 + k \lambda_k \\ &= \sum_{i=1}^{k} \sum_{j=1}^{k} ( \lambda_i - \lambda_k )<e_j,\epsilon_i> ^2 + k \lambda_k \\ &= \sum_{i=1}^{k}( \lambda_i - \lambda_k ) \sum_{j=1}^{k} <e_j,\epsilon_i> ^2 + k \lambda_k \\ &\leq \sum_{i=1}^{k}( \lambda_i - \lambda_k ) \sum_{j=1}^{n} <e_j,\epsilon_i> ^2 + k \lambda_k \\ &\leq \sum_{i=1}^{k}( \lambda_i - \lambda_k ) \| e_i\|^2 + k \lambda_k \\ &\leq \sum_{i=1}^{k}( \lambda_i - \lambda_k ) + k \lambda_k \\ &=\sum_{i=1}^{k-1}\lambda_i - (k-1) \lambda_k + k \lambda_k \\ &= \sum_{i=1}^{k-1} \lambda_i + \lambda_k \\ \end{align*} $
zestiria
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  • The question has typo errors, but the answer is the Ky - Fan maximum principle... – Toni Mhax Aug 22 '20 at 03:42
  • I have solved the question. Could you explicit the reference with the Ky-Fan principle ? – zestiria Aug 22 '20 at 15:35
  • You may google it (plenty of references) Trace maximum principle or Ky-Fan variational maximum p. Anyway you proved what Ky-Fan has almost done in his paper on this ( variational maximum p.) If you want a precise litterature repost that. Way to go. (even for the notations are messy the proof can be more compact) – Toni Mhax Aug 23 '20 at 00:55
  • This is a standard majorization result and has extremely short proofs e.g. using Cauchy Eigenvalue Interlacing. For many different proofs see: https://math.stackexchange.com/questions/3637453/maximize-mathrmtrqtcq-subject-to-qtq-i/ and consider $Q$ having the first k standard basis vectors as its columns vs the orthonormal eigenvectors associated with $k$ largest eigenvalues of A as columns. – user8675309 Aug 28 '20 at 00:22

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