Eigenvalues of nonsymmetric positive definite matrices I consider a non-symmetric matrix $A\in\Re^{n\times n}$ and try to estimate the region of the eigenvalues

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*Is it true, that a nonsymmetric matric $A$ is positive definite (in the sense of $x^T\cdot A\cdot x>0$), when the symmetric part of the matrix $A_\text{sym}=\frac12(A^T+A)$ is positive definite?

*I know for a symmetric matric, that $A_\text{sym} \text{ is pos. def} \Leftrightarrow \text{all Eigenvalues } \lambda_i > 0$. What can I say about the eigenvalues of $A$? I read that the real part of these Eigenvalues must be $\Re(\lambda_i)>0$.

Thanks in advance.
 A: Both claims are true, though be wary that your definition of a matrix being positive-definite in the sense that $x^T A x > 0$ for all non-zero real vectors $x$, is not common and usually restricted to Hermitian matrices. To be clear, let us denote $A \succ 0$ to mean that $A$ is positive-definite in the sense you have defined, for any real matrix.

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*We have
$$ x^T A x = \frac{1}{2} \left( x^T (A^T+A) x + x^T (A^T-A)x \right) = x^T A_{sym} x > 0$$
where $x^T (A^T-A)x = 0$ because $$x^T (A^T-A)x = (x^T (A^T-A)x)^T = x^T (A-A^T)x = -x^T (A^T-A)x \,.$$

*Take any eigenvalue $\lambda = a+ib$ where $a,b \in \mathbb{R}$ of a positive-definite matrix $A$, with eigenvector $x$. Then $Ax = \lambda x$ implies $x^* A x = a+ib$ (where $^*$ denotes the conjugate transpose) and so
$$ a = Re(x^* A x) = \frac{1}{2} \left( x^* A x + x^* A^* x \right) = x^* A_{sym} x > 0 $$
by using part 1 above and the fact that $A^* = A^T$ since $A$ is real.

A: I think you misunderstood what you wrote in point 1. Any positive operator is Hermitian, so if your $A$ is nonsymmetric but positive definite, well... it fulfills $A=A^\dagger$. What I mean with this is that $A$ has at least two off-diagonal complex entries (otherwise it would be symmetric) such that these elements fulfill $A_{ij}=A^*_{ji}$.
The eigenvalues of $H$ positive definite, hence Hermitian, are positive (nonegative).
