Eigenvalue spectrum of a real marix

Question

Is it true that for any real matrix, the complex eigenvalues occur in complex conjugate pairs? If yes, how does one prove that?

Answer 1

Eigenvalues are roots of the characteristic polynomials $$P(\lambda)=\det(A-\lambda I)$$

The coefficients of the characteristic polynomial are real numbers.

If $P(\lambda)=0$, then upon conjugation we get $$ P( \lambda)=0 \implies p(\overline{\lambda})=0$$

Answer 2

Guide:$$Ax = \lambda x$$

Try to take conjugate of the equation above and see what do you get.

Answer 3

The characteristic polynomial $p(\lambda)$ of a real square matrix has real coefficients.

So, it holds for any complex solution $\lambda_0$ $$0 =\overline{p(\lambda_0)} = p(\overline{\lambda_0})$$.

Answer 4

If $\det(\lambda I -A) = 0$ then $\overline{\det(\lambda I -A)} = \det(\overline{\lambda}I -A) = 0$.