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Let A be a set of similar (symmetric) matrices, sharing the same eigenvalues. I understand that their eigenvectors would be different. Let us focus on one eigenvector (e.g. corresponding to the lowest eigenvalue).

Can we find the matrix with the most compact eigenvector, without diagonalizing all of them? What I mean is that the eigenvector elements, if sorted in decreasing order based on their absolute values, decay fastest. An ideal situation would be to have the eigenvector [1, 0, ..., 0].

Thanks in advance!

Edit: replaced PSD with symmetric to make the question more general. Edit: made the question clearer. Sorry for any confusion.

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  • $\begingroup$ Decay with respect to what? Is your set of similar matrices actually a sequence? $\endgroup$
    – LSpice
    Commented Apr 27, 2021 at 13:36
  • $\begingroup$ 1. What I meant by decay is that the eigenvector elements approach zero or a very small number if we sort them in decreasing order based on their absolute values. 2. My question was for a general PSD matrix set. $\endgroup$
    – twofiveone
    Commented Apr 27, 2021 at 13:44
  • $\begingroup$ As the question is written now, the trivial answer is "the eigenvector basis, in which the sought eigenvector is [1,0,0,...,0]". Maybe you need to formulate your request in a different way. $\endgroup$
    – Federico Poloni
    Commented Apr 27, 2021 at 14:08
  • $\begingroup$ @FedericoPoloni, although the question is very unclear to me, I think we are meant to look at different eigenvectors corresponding to different, but similar, matrices. (I still can't tell in what sense decay is meant—decay of the finite sequence of entries, whatever that means, for a fixed matrix, or decay as we vary the matrix?) If that is so, then picking an eigenbasis for a single matrix is not enough. $\endgroup$
    – LSpice
    Commented Apr 27, 2021 at 14:22
  • $\begingroup$ I edited the post to make it clearer. Thanks! $\endgroup$
    – twofiveone
    Commented Apr 27, 2021 at 14:29

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