A real square matrix $A$ has eigenvalues lying on the imaginary axis. If $A$ is not normal is it possible to say anything about the eigenvalues of $A+A^{T}$ or maybe the matrix $A+A^T$ as a whole?

I found that when $A$ is normal and unitarily diagonalizable, the eigenvalues of $A+A^T$ will be all zeros but when it is not I can still decompose $A+A^T=Q\Lambda{Q^T}$, since $A+A^T$ is a real symmetric matrix with $Q$ being an orthogonal matrix containing the eigenvectors of $A$. But I don't see any way forward. Any suggestion is much appreciated.


In a nutshell: not much, but you might find the analysis below interesting.

If $A$ is a $2 \times 2$ matrix with real entries and imaginary eigenvalues, then it can be unitarily upper triangularized to the form $$ UAU^* = B = \pmatrix{\lambda i & \alpha\\0 & -\lambda i} $$ where $\alpha \in \Bbb C$ is arbitrary and will be zero if and only if $A$ is normal (in general, this is called a Schur decomposition). With that, we note that $$ U(A + A^*)U^* = B + B^* = \pmatrix{0 & \alpha\\ \bar \alpha & 0} $$ In other words: with a suitable choice of a real matrix $A$ with purely imaginary eigenvalues, we can make $A + A^T$ an arbitrary trace-zero symmetric matrix.

I haven't applied this argument carefully, but I believe that it generalizes to the following: a matrix $M$ can be written in the form $M = A + A^T$ for a real matrix $A$ with purely imaginary eigenvalues if and only if it is symmetric with a trace of zero.


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