Positive definiteness of $\frac{1}{2}(A+A^T)$ and positive eigenvalues of $A$

Question

Given a (non-symmetric) matrix $A \in \mathbb{R}^{n\times n}$, is it possible to relate the sign of its eigenvalues (only the real part) to the definiteness of the corresponding symmetric matrix

$$\frac{1}{2}(A+A^T)?$$

I am very confused due to the non-consistent definitions of positive definiteness of matrices across the literature:

• in engineering or applied mathematics textbooks, a general matrix $M$ is positive definite $\iff$ all eigenvalues have positive real parts.

• in classical matrix analysis textbooks, however, the matrix $M$ needs to be Hermitian.

I was wondering if this is kind of justified by the statement above relating the signs of eigenvalues.

Answer 1

Counterexample (if I understand the question correctly):

$$A:=\begin{pmatrix}2&2\\-2&-1\end{pmatrix},\qquad \tfrac{1}{2}(A+A^\top)=\begin{pmatrix}2&0\\0&-1\end{pmatrix}$$

The eigenvalues of $A$ are $\frac{1}{2}\pm \frac{1}{2}i\sqrt7$, yet its symmetric part $\tfrac{1}{2}(A+A^\top)$ is not positive definite.