# What do real eigenvalues imply for a matrix

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.

• 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. – Qiaochu Yuan Feb 18 '14 at 21:52
• @QiaochuYuan good point. – ivan Feb 18 '14 at 22:12

• Ok, so $A$ must be a stretching and/or reflection. So basically, realness of eigenvalues just rules out rotations. – ivan Feb 18 '14 at 22:10