The eigenvalue problem can be rewritten as $$ \left( \mathbf{A}-\lambda^{(\nu)} \mathbf{I} \right) \mathbf{x}^{(\nu)} = 0, $$ with \( \mathbf{I} \) being the unity matrix. This equation provides a solution to the problem if and only if the determinant is zero, namely $$ \left| \mathbf{A}-\lambda^{(\nu)}\mathbf{I}\right| = 0, $$ which in turn means that the determinant is a polynomial of degree \( n \) in \( \lambda \) and in general we will have \( n \) distinct zeros.