For $A\in \mathbb R^{n\times n}$, when is $A+A^T$ positive semi-definite?

Question

Let $A\in \mathbb R^{n\times n}$. What are conditions on $A$ that ensure that $A+A^T$ is p.s.d. ?

Since $x^TA^Tx= ( x^TA^Tx)^T =x^TAx$, $A+A^T$ is positive semi-definite if and only if $\forall x\in \mathbb R^n, x^TAx\geq 0$.

Is there a caracterization of non-symmetric matrices that verify $\forall x\in \mathbb R^n, x^TAx\geq 0$ (in terms of eigenvalues e.g.) ?

Answer 1

Here is an example which shows that there is no characterization "in terms of eigenvalues".

Let $A = \begin{pmatrix} 1 & 0\\ 0 & 2\end{pmatrix}$ and let $B = \begin{pmatrix} 1 & 3\\ 0 & 2\end{pmatrix}$. Then both $A$ and $B$ have eigenvalues $1, 2$.

But $A + A^T$ is positive definite, while $B + B^T$ is not positive semidefinite.

Answer 2

This seems equivalent to the condition of $A$ being an "accretive operator" (from this paper)

Looking at the action of $B$ from the example above, you can see that for quite a few vectors it takes them into opposite direction, so it's not accretive.

From algebraic viewpoint, this comes down to asking whether numerical range of $A$ contains the origin. Called "zero inclusion question" in this paper.

The reason for this is the connection between numeric range of $A$ and symmetric part of $A$, see Mark Embers slides.

Matrices $A$ and $B$ in WhatsUp example above have same eigenvalues, but different numeric ranges

We can visualize numeric range for 3 representative examples

1. $X+X^T$ positive definite

2. $X+X^T$ positive semidefinite but not positive definite

3. $X+X^T$ not positive semidefinite

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