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given a symmetric matrix A, of dimension $n \times n $ and an an arbitrary unitary matrix V of dimension $ n \times n$ the the trace of $V^T A V$ is an orthogonal basis of the eigenspace associated with the largest eigenvalues.\

In particular, it is achieved for the eigenbasis itself: if eigenvalues are labeled decreasingly and $ u_1,...,u_d$ are eigenvectors associated with the first d eigenvalues $\lambda_1, ... ,\lambda_d$ and $U=[u_1,....,u_d]$ with $U^T U=I$, then $$ \left\{\begin{array}{l} \max\limits_{V \in \mathbb{R}^{n \times d}} \operatorname{Tr}\left[V^{T} A V\right]=\operatorname{Tr}\left[U^{T} A U\right]=\lambda_{1}+\cdots+\lambda_{d} \\ V^{T} V=I \end{array}\right. $$

The proof is an immediate consequence of the courant-Fisher characterization. But i don't see how we suppose to use it?

Yan Peng
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mouad lmazini
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  • They don't make use of the Courant-Fischer characterization, but some proofs can be found here and here. – Ben Grossmann Mar 25 '21 at 17:57
  • more proofs here: https://math.stackexchange.com/questions/3637453/maximize-mathrmtrqtcq-subject-to-qtq-i/ – user8675309 Mar 25 '21 at 18:17
  • @BenGrossmann . In the article, the authors said " While this result is seldom explicitly stated on its own in standard textbooks, it is an immediate consequence of the Courant-Fisher characterization''. – mouad lmazini Apr 22 '21 at 10:31

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