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Consider the invertible matrix $A$ with all eigenvalues $\lambda_{i}$ with positive real part. Then consider its symmetric part: $$ A_{s}=\frac{A+A^{T}}{2}, $$ with real eigenvalues $\eta_{i}$. Are these eigenvalues $\eta_{i}$ all positive?

P.S. What I already know is that $\text{Trace}[A_{s}]=\text{Trace}[A]$.

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2 Answers 2

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No. E.g. the eigenvalues of $\dfrac12\left(\pmatrix{1&4\\ 0&1}+\pmatrix{1&0\\ 4&1}\right)=\pmatrix{1&2\\ 2&1}$ are $3$ and $-1$.

The converse is true, however. That is, if $A$ is a real square matrix such that $\frac{A+A^T}{2}$ is positive definite, then all eigenvalues of $A$ have positive real parts. To prove this, note that if $u$ is a unit eigenvector of $A$ corresponding to an eigenvalue $\lambda$, then $$ \Re(\lambda) =\frac{u^\ast(Au)+(u^\ast A^\ast)u}{2} =u^\ast\frac{A+A^\ast}{2}u =u^\ast\frac{A+A^T}{2}u>0. $$

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The matrix $A = \begin{bmatrix}1 & 2 \\ 0&1\end{bmatrix}$ has only one eigenvalue: $1$.

But $A+A^T = \begin{bmatrix}2 & 2 \\ 2&2\end{bmatrix}$ has the eigenvalue $0$.

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