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Do positive-definite matrices always have real eigenvalues?

I tried looking for examples of matrices without real eigenvalues (they would have even dimensions). But the examples I tend to see all have zero diagonal entries. So they are not positive definite.

Would anyone have an example of positive-definite matrix without any real eigenvalue? Or it is a property of positive-definite matrices that they always have some real eigenvalues?

EDIT:

A matrix $A$ is positive definite iff $\forall x, x^TAx>0$.

No symmetry is implied here.

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  • $\begingroup$ Try the matrix $\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}$. $\endgroup$ – Yuval Filmus May 28 '15 at 23:04
  • $\begingroup$ Nice. Would you like to post the example to your answer for me to close the question? $\endgroup$ – Argyll May 30 '15 at 2:44
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The matrix $\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}$ is a counterexample. It is positive-definite since for $(x,y) \neq (0,0)$ we have $$ \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 2x^2 - 2xy + y^2 = x^2 + (x-y)^2 > 0. $$ On the other hand, its characteristic polynomial is $$(t-2)(t-1)+3 = t^2-3t+5,$$ whose discriminant $9-20 = -11$ is negative, so both of its roots are complex.

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Depends on your definition of positive-definite matrices. Usually they're assumed to be symmetric, and so all their eigenvalues are real. In fact, since they're positive-definite, all their eigenvalues are positive: if $v$ is an eigenvector with eigenvalue $\lambda$ of the positive-definite matrix $A$ then $0 < v'Av = \lambda \|v\|^2$.

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  • $\begingroup$ Symmetry is not assumed $\endgroup$ – Argyll May 28 '15 at 20:58

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