5
$\begingroup$

Specifically, I am interested in the case where one eigenvalue is exactly $0$. Given an $n \times n$ symmetric matrix, I would like to find the closest $n\times n$ symmetric matrix that has one eigenvalue that is equal to $0$.

Although the $n\times n$ matrices form a Hilbert space, the set of symmetric matrices with a zero eigenvalue is probably not convex, because the eigenvalues of a sum of two matrices has a complex relationship with the original matrices. Thus, the classic projection theorem does not guarantee a unique solution. However, I only need a solution. Since the eigenvalues of a matrix are continuous in the entries, the set of $n\times n$ matrices with a zero eigenvalue should be closed, and so the set of solutions shouldn't be totally insane.

In any case, how can I find the closest $n\times n$ symmetric matrix(es) with a zero eigenvalue to an arbitrary given $n\times n$ symmetric matrix?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ How do you define "closest"? $\endgroup$
    – Michael Engelhardt
    Commented Oct 3, 2019 at 17:45
  • 1
    $\begingroup$ @MichaelEngelhardt with whatever arbitrary metric on matrices. A handful can be gotten from the various matrix norms. For context, I am using this in a constrained optimization problem, so I perturb my original matrix a bit and it no longer has a zero eigenvalue, I want to get it back to having a zero eigenvalue, but not be too far away from where I started. If a particularly convenient solution happens to correspond to a particular metric, then that's the correct metric. :) $\endgroup$
    – Him
    Commented Oct 3, 2019 at 18:17
  • $\begingroup$ Ah, then it seems the answer that has already been given below is just the thing - great! $\endgroup$
    – Michael Engelhardt
    Commented Oct 3, 2019 at 19:31

1 Answer 1

Reset to default
8
$\begingroup$

Any (real) symmetric matrix may be diagonalised by an orthogonal matrix. So let your matrix be $A$ and let $O$ be an orthogonal matrix such that $D=OAO^{-1}$ is diagonal. Note that the transformation $A\mapsto OAO^{-1}$ is distance preserving. Now get $D'$ by replacing whichever diagonal entry of $D$ has least absolute value, and let $A'=O^{-1}D'O$.

(Nathaniel Johnston's comment let me find the related paper "Low-Rank Positive Approximants of Symmetric Matrices" by Achiya Dax.)

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ I'll just note (for the sake of adding Google-able terms) that this is essentially the Eckart-Young-Mirsky theorem restricted to symmetric matrices. $\endgroup$
    – Nathaniel Johnston
    Commented Oct 4, 2019 at 12:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .