Suppose $A$ is a non-symmetric matrix (also, not a normal matrix) with all non-negative eigenvalues. Is there a relation between eigenspace (subspace spanned by eigenvectors) of $A$ and eigenspace of $(A+A^T)$? Is there an overlap? One obvious observation is that row space of $A$ is same as column space of $A^T$.
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asked Jan 24, 2011 at 17:18
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$\begingroup$ If $A$ is normal, the eigenspaces are the same. $\endgroup$– Igor RivinJan 24, 2011 at 18:36
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$\begingroup$ A is not normal in my case :( $\endgroup$– Abhishek KumarJan 25, 2011 at 0:28
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1 Answer
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Trivially one has that the rank of $A+A^t$ cannot be larger than 2*rank($A$) since rank($A$) = rank($A^t$).
Without knowing additional constraints on $A$, I don't see that much can be said about the eigenvalues of $A+A^t$. For example, let $A = [[2,2],[6,k]]$, it is easy to check that $A$ is not normal for any value of $k$.
Consider the cases $k=7,8,9$. In each case $A$ has 2 positive eigenvalues, but:
For k=7, $A+A^t$ has 1 positive, 1 negative eigenvalue
For k=8, $A+A^t$ has one positive and one zero eigenvalue
For k=9, $A+A^t$ has 2 positive eigenvalues
So for $k\neq 8$, $A+A^t$ has the same eigenspace as $A$ (although the eigenvectors are different) while when $k=8$, the eigenspace of $A+A^t$ is a strict subspace of the eigenspace of $A$.
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$\begingroup$ Incidentally, right after I finished writing this, I noticed your other similar question which is a more restrictive setting, so perhaps this answer is not as enlightening to your particular situation. $\endgroup$ Jan 25, 2011 at 6:21