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Given a set of equations with unknowns matrixes X and Y expressed as: $$XBX^{T} +YCY^{T} =A$$ where

  • A is symmetric and positive definite (i.e. in my particular case it is a covariance matrix),

  • C and B are diagonal and positive definite (in my particular case they are the matrixes of eigenvalues of A where the first k eigenvalues are stored on the diagonal of B and the remaining are on the diagonal of C),

  • and the unknowns are X and Y (in my case I can put X and Y equal to the eigenvectors corresponding to the eigenvalues in B and C respectively),

the question 1 is: is the solution for X and Y unique subject to the added constraints that $1^{T}X=1$ and $1^{T}Y=1$?

This question may be rephrased as: provided that we are working with normalized eigenvectors, is it true that the equation above is solved only putting X and Y equal to the normalized eigenvectors corresponding to the eigenvalues in B and C? That is equivalent to saying: are normalized eigenvectors unique for A? Assume A is a matrix whose eigenvalues are all distinct (i.e. all the eigenvalues have multiplicity 1).

The question 2 is I have read somewhere that it is rare to find estimated covariance matrixes in real-life applications (for example we have a sample of data for random variables and estimate their variance covariance matrix) where some eigenvalues have multiplicity which is not 1. Can you confirm?

I add this link because it is helpful to the subject of uniqueness of eigenvectors for each eigenvalue with multiplicity 1.

Fr1
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  • @loupblanc help if you can, many thanks in advance – Fr1 Aug 17 '19 at 18:44
  • I assume that the considered matrices are real. You say that $B=diag(\lambda_1,\cdots,\lambda_k,?,?),C=diag(\lambda_{k+1},\cdots,\lambda_n,?,?)$ where $spectrum(A)=(\lambda_i)_i$ is composed with $n$ positive distinct eigenvalues. But what are the $?,?$ (the other elements of the diagonals); are they arbitrary $>0$ reals ? –  Aug 18 '19 at 15:38
  • @loupblanc I am so sorry for the misspelling of the problem, I just changed the expression fixing the transpose.. so sorry.. you are right.. god bless me and my lack of focus! Too many hours spent on this!! :-) So sorry thanks a lot – Fr1 Aug 18 '19 at 15:43
  • I don't understand one word of your problem. Give the dimensions of $X,A,B,C$. Give explicitly $B,C$. Are the considered equations $XBX^T+Y^CY^T=A, 1^TX=1^T,1^TY=1^T$ ? You ask if a solution of this system, using eigenvectors (of $A$ ?), is unique. What solution do you think? –  Aug 18 '19 at 16:12
  • @loupblanc yes I am saying the following: given the set of equations above, they are clearly solved by putting X=normalized eigenvectors corresponding to the eigenvalues on B and Y=normalized eigenvectors corresponding to the eigenvalues on C. Are there any other possible solutions? – Fr1 Aug 18 '19 at 16:18

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