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We know that if $A$ is symmetric, then we can determine the sign of $x^T A x$ for $x \ne 0$ by looking at its eigenvalues. But what if $A$ is not symmetric, then we cannot say that it is positive definite, etc. So is it still possible to judge the sign of $x^T A x$?

What if $A$ is a Metler matrix?

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You can decompose each matrix onto two matrices: symmetric and skew-symmetric. It can be done the same way: $A = \frac{1}{2} (A + A^T) + \frac{1}{2}(A - A^T)$. It's easy to figure out that the first matrix is symmetric, while the second one is skew-symmetric. Also, we know that for skew-symmetric matrices $x^TCx = 0 \ \forall x$. As a result $x^TAx$ is equal to $x^TSx$, where $A = S + C$, ($S = \frac{1}{2}(A + A^T)$, $C = \frac{1}{2}(A - A^T)$). And in symmetric case you already know, what to do!

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