Eigenvalues and rank connection

Question

Let $A$ be a $4 \times 4$ matrix and let $\dim \mathcal N (A) = 2$. What can you tell about the eigenvalues of $A$?

Answer 1

As $\dim(N(A))=2$, $0$ is an eigenvalue with multiplicity 2 in $A$'s characteristic polinomial $\chi_A$ then $\chi_A=x^2p(x)$.

Now, if $A\in \mathbb{R}^{4\times 4}$ $\chi_A$'s coefficients are real so if $p(x)=(x-\lambda)(x-\mu)$ then either $\lambda,\mu\in \mathbb{R}$ or if $\lambda\in \mathbb{C}\setminus \mathbb{R} \implies \mu =\overline{\lambda}$. In the former case you have two eigenvalues which may be equal or not. In the latter $0$ is the only real eigenvalue of the matrix.

if $A\in \mathbb{C}^{4\times 4}$ then $\chi_A=x^2(x-\lambda)(x-\mu)$ with any $\lambda,\mu \in \mathbb{C}$. Then you have two more eigenvalues which may be equal.