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Suppose I have a matrix:

$A \in \mathbb{R}^{n \times n}$, where $A$ not necessarily symmetric, but $x^TAx \geq 0$, or $\leq 0$ for $x \in \mathbb{R}^n$.

Will such matrix have imaginary eigenvalues?

It seems that symmetry is a sufficient but not necessary condition for having real imaginary values. Will positive + negative (semi)definite plus being real ensure that $A$ will not have imaginary eigenvalues?

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  • $\begingroup$ real imaginary values? real eigenvalues? $\endgroup$ – mathreadler Jan 14 '17 at 7:05
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They can have imaginary eigenvalues. Consider, $$A = \left[ \begin{matrix} 1 & -1 \\ 1 & 1\end{matrix} \right].$$ For $x = \binom u v \in \mathbb R^2$, we see $$x^tAx = u^2 + v^2 \ge 0 $$ with equality iff $x = 0$. Thus $A$ is positive definite. However, the eigenvalue of $A$ are given by $$(\lambda -1)^2 + 1 = 0 \,\,\,\,\, \implies \,\,\,\,\, \lambda_{1,2} = 1 \pm i.$$ The real part of the eigenvalues of such a matrix will always be positive (resp. non-negative, negative or non-positive if the matrix is positive semi-definite, negative definite or negative semi-definite).

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To expand on User8128s answer, $\det(A-\lambda I)$ is a polynomial of real coefficients. Such a polynomial always has roots which are either real or come in complex conjugate pairs.

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