# What properties should a matrix have if all its eigenvalues are real?

Recently, I’m trying to prove all the eigenvalues of a class of matrices are real. The matrices are complex and not hermitian. The problem for me is I don't know any properties for a matrix with all the eigenvalues real. So would you please tell me any sufficient or necessary conditions for such matrices? Thank you!

• Read up on normal matrices, i.e., $AA^H=A^HA$. – LutzL Feb 9 '14 at 12:38
• Well for a symmetric matrix all eigenvalues are real. – randomname Feb 9 '14 at 12:50
• Are we to assume the matrix has a complete basis of eigenvectors? As in the real matrix case, there can be a deficiency of the geometric multiplicity versus algebraic multiplicity of an eigenvalue. – hardmath Feb 9 '14 at 12:59
• The Gershgorin circle theorem might be handy (en.wikipedia.org/wiki/Gershgorin_circle_theorem). – Calle Feb 9 '14 at 13:25

Further, if $A$ is a complex matrix with real eigenvalues, so will be $PAP^{-1}$ for any invertible matrix $P$, by similarity. Any eigenvalue of $A$, say $Av = \lambda v$, will give the same eigenvalue for $PAP^{-1}$:
$$PAP^{-1} u = \lambda u \; \text{ where } \; u=Pv$$
Indeed this can be framed as a necessary and sufficient condition, however difficult to apply in practice. Complex matrix $B$ has only real eigenvalues if and only if $B$ is similar to an upper triangular matrix $A$ with real diagonal entries. If we wish to impose a condition of a complete basis of eigenvectors for $B$, then in addition we require $A$ to be a real diagonal matrix (and to avoid hermitian examples, impose a condition that $P$ is not unitary).