The rank of the characteristic polynomial of a $\lambda$-matrix

Question

Assume $F(\lambda, y) = \mathrm{det}(yI - V)$, where $V = (V_{ij}(\lambda))_{4\times 4}$ and each $V_{ij}(\lambda)$ is a polynomial of $\lambda$.

If $F(\lambda, y) = 0$ always has four different roots $y_1, y_2, y_3, y_4$ for all $\lambda\in\mathbb{C}$ except a finite number $\lambda_1, \dots, \lambda_g$.

Prove $$\mathrm{rank}(yI - V) = 3$$ under constraint $F(\lambda, y) = 0$ and $\lambda\neq \lambda_j, j=1,\dots,g$.

Can anyone give me some hints?

Answer 1

The hint is to discard the other possible ranks by using the Rank-Nullity Theorem.