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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.

Rodrigo de Azevedo
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Trb2
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  • Positive (or negative) definiteness (semi- or not) is almost exclusively used about symmetric real, or Hermitian complex matrices (at least as far as I have seen). And for such matrices, all eigenvalues are real, and you can ask about whether they are all positive or all negative. – Arthur Dec 02 '20 at 08:39
  • @Arthur And can I relate the sign of the eigenvalues of a non-symmetric $A$ to the definiteness of $\frac{1}{2}(A+A^T)$? E.g. if $\frac{1}{2}(A+A^T)$ pos. def, $\Rightarrow Re(\lambda(A))>0$, or viceversa? – Trb2 Dec 02 '20 at 08:51
  • Take a look at this. – Rodrigo de Azevedo Dec 02 '20 at 17:54
  • If we consider the reals. $x^T A x = x^T A^T x.$ Hence $\frac{1}{2}x^T (A +A^T)x = x^T A x.$ – William Zheng Nov 07 '23 at 11:14

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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.

Chrystomath
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