Calculating determinant using eigenvalues (real and complex)

Question

I was trying to find the eigenvalues for the following characteristic polynomial:

$$p_M(\lambda)=(\lambda^{4}+1)(\lambda^2-4)$$

For this I solved the equation $p_M=0$ and I found the following two real eigenvalues:

$$\lambda=2\ \vee \ \lambda=-2$$

and I also got the four complex eigenvalues from

$$\exp({i(\pi/4\ + k\pi/2}))$$

Multiplying all $6$ eigenvalues I got the answer that the determinant of the corresponding matrix must be $-4$. This is the product of the two real eigenvalues.

Is this coincidence? Or is the determinant always the product of the real eigenvalues?

Answer 1

No, the determinant is not always the product of the real eigenvalues. Take the diagonal matrix (the Jordan form of a real matrix, if you want only real matrices) $$ A=\begin{pmatrix} 1 & 0 & 0 \cr 0 & 2i & 0 \cr 0 & 0 & -2i \end{pmatrix}. $$ Here the characteristic polynomial is $$ \chi_A(t)=(t^2 + 4)(t - 1). $$