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Suppose we have a matrix $M$ such that $M$ is non-symmetric real and has positive eigenvalues. Do we have a relation between eigenvalues/eigenvectors of $(M+M^T)$ and those of $M$? What if $M$ and $(M+M^T)$ both are of low rank?

Suppose, $M = AP$ where $A$ is a positive semi-definite matrix and $P$ is a orthogonal projection matrix of the form $UU^T$ ($U$ being an orthogonal matrix). $M$, $A$, $P$ are all of size $n\times n$, $U$ is of size $n\times k$, where $k << n$. We know that $M$ will have real and non-negative eigenvalues. The question is, how are the eigenvectors/eigenspaces of $M$ and $(M+M^T)$ are related?

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If I am not mistaken, the requirement that all eigenvalues of $M$ are positive implies $det(M)$ is positive and hence $M$ has maximal rank, since the rows must be linearly independent. So what gives? How can $M$ have low rank? Or am I missing something? –  drbobmeister Jan 20 '11 at 6:22
    
See my answer to this related question : mathoverflow.net/questions/31238/a-signature-inequality Any symmetric matrix with positive trace is of the form $AB+BA$ for symmetric positive definite $A,B$ (and any $M$ in your question is an $AB$). I realize that this doesn't answer your question, but methods in Ballantine's paper might help. –  BS. Jan 20 '11 at 10:53
    
There are some relations on eigenvalues as pointed out by Denis Serre. I am also looking for a relation between eigenvectors of $M$ and $(M+M^T)/2$. Can we write the eigenvectors of $M+M^T$ in terms of eigenvectors of $M$? or any relation between the subspaces the eigenvectors span? –  Abhishek Kumar Jan 21 '11 at 0:21

3 Answers 3

up vote 4 down vote accepted

Let $N:=(M+M^T)/2$. besides the obvious equality $Tr(N)=Tr(M)$ which is an equality of the sums of eigenvalues, you have the following. Let $\lambda_\pm$ be the smallest/largest eigenvalues of $N$. Then every eigenvalue of $M$ satisfies $\Re\lambda\in[\lambda_-,\lambda_+]$. In addition, if $w(M):=\max\{\lambda_+,-\lambda_-\}$ is the numerical radius of $M$, then $$w(M)\le\|M\|\le2w(M),$$ in operator norm. This implies that the singular values, hence the moduli of the eigenvalues of $M$, are not greater than $2w(M)$.

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Let $Q$ be any symmetric matrix with 2's on the diagonal. This allows a variety of choices for the rank and eigenvalues of Q. Now let M be the upper triangular matrix which is 1 on the diagonal and equal to Q above it. Then all eigenvalues of M are 1 and it has full rank.

Of course the diagonal does not have to be constant.

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Yes. If $N=(M+M^t)/2,$ then $tr\ M = tr\ N,$ while for any vector $v,$ $(v, M v) = (v, N v).$

An additional remark: if $M$ is normal, then the rank of $N$ is at most twice the rank of $M,$ and the eigenvectors of $N$ are the same as those of $M.$

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I doubt this. In fact, you must consider complex unit vectors $v$ to obtain the numerical range (=field of values), so that you only have $W(N)=\Re W(M)$. –  BS. Jan 20 '11 at 10:48

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