Eigenvalues of sum $A+A^{\top}$ where $A$ is not symmetric but has positive eigenvalues

Question

What can be said about the eigenvalues of $C=A+A^T$ where $A$ is not symmetric but has positive eigenvalues? Can we say $C$ is a positive definite matrix? (It is symmetric, but are all eigenvalues positive ?) Thanks a lot!

Answer 1

When $A$ has a positive spectrum, $A+A^T$ must have a positive trace. Therefore, it is guaranteed that $A+A^T$ has at least one positive eigenvalue.

However, $A+A^T$ may possess negative eigenvalues. For instance, consider the rank-one matrix $$ B=\pmatrix{1\\ -1}\pmatrix{2&-1}=\pmatrix{2&-1\\ -2&1}. $$ Its eigenvalues are $\lambda_1=\operatorname{tr}(B)=3$ and $\lambda_2=0$, which are nonnegative. Yet, $$ B+B^T=\pmatrix{4&-3\\ -3&2} $$ has a negative eigenvalue because $\det(B+B^T)=-1<0$. It follows that when $t>0$ is sufficiently small, $A=B+tI$ has a positive spectrum but $A+A^T=(B+B^T)+2tI$ has a negative eigenvalue.

In general, if an $n\times n$ real square matrix $A$ has a positive spectrum, $A+A^T$ can have up to $n-1$ negative eigenvalues. See my other answer for a concrete example.

Answer 2

We have that $A+A^T$ can be any symmetric matrix. To see this, let $M=M^T$ and $S=-S^T$ be any symmetric and skew symmetric matrices. If $A=(M+S)/2$, then $A+A^T=M$ and $A-A^T=S$.

However, this doesn't rule out that the eigenvalues of $A$ might be somehow related to the eigenvalues of $A+A^T$.

Answer 3

Consider the $2 \times 2$ matrix $A = \begin{bmatrix}a&b\\0&a\end{bmatrix}$ where $a > 0$ and $b > 2a$.

The eigenvalues of $A$ are $a$ with multiplicity $2$, which are both positive.

The eigenvalues of $A+A^T = \begin{bmatrix}2a&b\\b&2a\end{bmatrix}$ are $2a+b$ and $2a-b$. The eigenvalue $2a-b$ is negative.

So the eigenvalues of $A$ being positive is not enough to guarantee that $A+A^T$ is positive definite.