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.