Eigenvalues that are complex numbers Have a square matrix problem that involves complex numbers and am at a loss.  $M$ is a square matrix with real entries.  $\lambda = a + ib$ is a complex eigenvalue of $M$, show that the complex conjugate $\bar{\lambda} = a - ib$ of $\lambda$ is also an eigenvalue of $M$.  Does solving this relate to a matrix having a characteristic polynomial of $(t^2-4)$? Could the community please explain? 

If you consider complex eigenvalues of a matrix $M$, then you are implicitly interpreting it as the matrix of a complex-linear operator $\phi$ acting on a complex vector space$~V$, even if $M$ has only real entries. The fact that $M$ has real entries then is basis dependent (it will probably not hold for the matrix of$~\phi$ with respect to any basis of$~V$); if $V_\Re$ denotes the real span of the chosen basis and $V_\Im$ the real span of the basis vectors multiplied by $\def\ii{\mathbf i}\ii$, then the fact that $M$ has real entries means that $\phi$ commutes, not only with multiplication by$~\ii$ as any complex-linear operator must, but also with "complex conjugation": the $\Bbb R$-linear operator $J:V\to V$ that acts by a scalar$~1$ on$~V_\Re$ and by a scalar$~{-}1$ on$~V_\Im$. Note that $J$ anti-commutes with multiplication by$~\ii$, so it is not complex-linear, but conjugate-linear: (it is $\Bbb R$-linear and) $J(\lambda v)=\overline\lambda J(v)$ for all $\lambda\in\Bbb C$ and $v\in V$.
Now if $v\in V$ is an eigenvector of$~\phi$ for an eigenvalue$~\lambda\in\Bbb C$, this means that $\phi(v)=\lambda v$. Now from the above $\phi(J(v))=J(\phi(v))=J(\lambda v)=\overline\lambda J(v)$, so $J(v)$ is an eigenvector for $\overline\lambda$.
