In "Groupes et algebres de Lie" in chapter VII of "Elements de mathematique" by Bourbaki in Proposition 3 there are the following arguments:
Let $ P(X) = \Pi(X-\lambda_i)^{q_i} (i=1,...,n)$, $Q(X)= (X-\lambda_0)^{q_0} )$, where all $\lambda$ are different. Suppose that for vector space $V$ there exists linear map $A$ and vector $v$ such that $P(A)(v)=0$ and $Q(A)(v)=0$. Because polynomials $P, Q$ are coprime then from Bezout's identity it follows that $v=0$.
I don't understand what they mean under "Bezout's identity" and how it applies here.
