Suppose we have a matrix $A \in \mathbb{R}^{n \times n}$ with $\textrm{eig}(A)=\{ \lambda_1, \lambda_2, \ldots, \lambda_n\}$ such that $\lambda_i \in \mathbb{R}$.

Does the realness of the eigenvalues imply some structure in $A$?

I know of that $A=A^{\sf T}$ (real symmetric) implies all the eigenvalues will be real, but I'm wondering about implications in the other direction. Is there some additional condition + realness of eigenvalues which would imply the matrix is symmetric?

Context: I'm trying to classify special types of matrices according to the properties of their eigenvalues. e.g. $\lambda_i \geq 0$, then $A$ is positive semidefinite, etc. If anyone can point me to a list of such correspondences between eigenvalue constraints and matrix properties it would be much appreciated.

  • $\begingroup$ Positive semidefinite is a term that should really only be applied to self-adjoint matrices. It's not really a condition on a linear operator but a condition on a symmetric bilinear form. $\endgroup$ – Qiaochu Yuan Feb 18 '14 at 21:52
  • $\begingroup$ @QiaochuYuan good point. $\endgroup$ – ivan Feb 18 '14 at 22:12

If all eigenvalues are real then your matrix is equivalent to stretching and/or basic reflection with respect to some basis. If your matrix is real, then the only other possibility is that there are complex eigenvalues that come in conjugate pairs, in which case your matrix also involves some sort of rotations.

  • $\begingroup$ Ok, so $A$ must be a stretching and/or reflection. So basically, realness of eigenvalues just rules out rotations. $\endgroup$ – ivan Feb 18 '14 at 22:10
  • $\begingroup$ @Ivan Right. :) $\endgroup$ – user2566092 Feb 18 '14 at 22:56

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