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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 an 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 $k \ll n$. 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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    $\begingroup$ 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? $\endgroup$
    – drbobmeister
    Commented Jan 20, 2011 at 6:22
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    $\begingroup$ 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. $\endgroup$
    – BS.
    Commented Jan 20, 2011 at 10:53
  • $\begingroup$ 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? $\endgroup$
    – Abhishek Kumar
    Commented Jan 21, 2011 at 0:21
  • $\begingroup$ Related: math.stackexchange.com/questions/3709480/… $\endgroup$
    – valle
    Commented Jun 8, 2020 at 0:13

3 Answers 3

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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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    $\begingroup$ Can you cite some sources for this for the proofs? $\endgroup$
    – valle
    Commented Jun 8, 2020 at 0:13
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    $\begingroup$ @becko. See my book Matrices, GTM216, Springer-Verlag (2nd edition). The inequalities for the numerical radius are in Proposition 5.11. In addition, the operator norm dominates the spectral radius (Proposition 7.6). $\endgroup$
    – Denis Serre
    Commented Jun 8, 2020 at 5:41
  • $\begingroup$ Thanks. I had not seen this statement anywhere else so I'll definitely try to check out your book. Just one more question. As you say, every eigenvalue of $M$ satisfies $\mathrm{Re}(\lambda)\in[\lambda_-,\lambda_+]$. Can we say something in the converse direction? E.g., if $M$ has all eigenvalues with positive real part, then $\lambda_- > 0$? The reason I ask is because I am trying to see the connection with math.stackexchange.com/questions/3709480/…. $\endgroup$
    – valle
    Commented Jun 8, 2020 at 10:44
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    $\begingroup$ @becko. The answer is No. Just take $M=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ whose eigenvalues are $0$ (double), but those of $N$ are $\pm\frac12$. $\endgroup$
    – Denis Serre
    Commented Jun 8, 2020 at 16:15
  • $\begingroup$ Ok. In the particular case of math.stackexchange.com/questions/3709480/…, though, I'd need a counter-example where $N$ has a positive eigenvalue but all eigenvalues of $M$ are negative (or vice-versa). A zero eigenvalue is not enough. $\endgroup$
    – valle
    Commented Jun 8, 2020 at 16:31
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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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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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