Eigenvalues of sum of a non-symmetric matrix and its transpose $(A+A^T)$ 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?
 A: 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 $2w(M)$.
A: 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 $M.$
A: 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.
