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What more can be said about the eigenvalues (especially the spectrum) of the $N \times N$ matrix ${\bf M} = {\bf A} + {\bf A}^T$ in terms of $\bf A$ if $\bf A$ is not symmetric and its eigenvalues are not necessarily positive ($\bf A$ is not necessarily positive semi-definite)?

  1. The eigenvalues of $\bf M$ are real since $\bf M$ is symmetric.
  2. $\sum_{i=1}^{N} \lambda_i({\bf M}) = {\rm trace}({\bf M}) = 2~{\rm trace}({\bf A}) = 2\sum_{i=1}^{N} \lambda_i({\bf A})$

The problem originates from principal component analysis. The principal components are the largest eigenvalues of the sample covariance matrix (SCM) $\bf C = ({\bf X}+{\bf W})^T({\bf X}+{\bf W})$ where $\bf X$ is a data matrix and $\bf W$ is a noise (white Gaussian) matrix.

Usually it is assumed that $\bf X$ and $\bf W$ are uncorrelated (when many samples are used) and the SCM can then simply be written as ${\bf C} = {\bf X}^T{\bf X} + {\bf W}^T{\bf W}$. The eigenvalue spectrum can under these conditions $quite~easily$ be determined using Weyl's inequalities since ${\bf X}^T {\bf X}$ and ${\bf W}^T{\bf W}$ are both Hermitian (or real symmetric).

However, when the number of samples is limited, the error (cross) terms ${\bf E} = {\bf X}^T{\bf W} + {\bf W}^T{\bf X}$ begin to influence the eigenvalues significantly. Since $\bf E$ is Hermitian, Weyl's inequalities can again be used to determine the eigenvalue spectrum of $\bf C$. Though, the eigenvalue spectrum of $\bf E$ is required to do so.

Setting ${\bf A} = {\bf X}^T{\bf W}$, it is clear that ${\bf A}^T = {\bf W}^T{\bf X}$, such that ${\bf M} = {\bf E}$.

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If this comes from some larger problem, maybe the singular values will be more useful to you. –  Felix Goldberg May 23 '12 at 11:54
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2 Answers

up vote 13 down vote accepted

You cannot do much, for the following reason. Take any real symmetric matrix $M$. It is a consequence of the Toepliz-Hausdorff Theorem about the Numerical Range that $M$ is unitarily similar to a (symmetric) matrix $N$ whose diagonal is constant: $$n_{jj}=\frac1n{\rm Tr} M.$$ Then $N=B+B^T$, where $B$ is upper triangular and has a constant diagonal. Coming back to the original basis, you deduce that there exist an $A$ such that $M=A+A^T$ and $A$ has only one eigenvalue, namely $\lambda=\frac{1}{2n}{\rm Tr}M$.

In this situation, you see that $A$ is very poor from the point of view of its spectrum, whereas $M$ can be as rich as possible: no constraints appart from the reality of its eigenvalues and their sum.

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On the other hand, there are some very special cases where some information flows form $M$ to $A$. For instance, if $M$ is positive definite, then $A$ is positive stable.

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