Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $A^T$.

share|improve this question
    
If $A$ is normal, the eigenspaces are the same. –  Igor Rivin Jan 24 '11 at 18:36
    
A is not normal in my case :( –  Abhishek Kumar Jan 25 '11 at 0:28
add comment

1 Answer 1

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 $A$.

share|improve this answer
    
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. –  ARupinski Jan 25 '11 at 6:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.