 Solution

The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.