Prove or disprove: if $\lambda$ is an eigenvalue of square matrix T then so is $\bar\lambda$

Question

I think this is true. I know that complex eigenvalues come out as conjugate pairs, but can't think of a formal proof. any lead?

Answer 1

The roots of the characteristic equation of a (real) square matrix $A$ is given by $\det(A-\lambda I)$ is a polynomial with real coefficients in $\lambda$ whose roots are the eigenvalues of $A$. A result that follows from the Fundamental Theorem of Algebra, is that the roots in a complex polynomial comes in conjugate pairs, from which your result follows.