I've been struggling with proving the following statement:
Let $V$ be a vector space over general field $\mathbb F$. Let $P(x)=(x-c_1)^{m_1} \dots (x-c_k)^{m_k}$ be a polynomial in $\mathbb F[x]$. Let $A \in M_n(\mathbb F)$ be a square matrix such that $P(A)=0$. Prove: $$V = ker(A-c_1I_n)^{m_1} \oplus \dots \oplus ker(A-c_kI_n)^{m_k}$$.
I was looking to prove this by taking $u_1, \dots, u_k \in ker(A-c_1I_n)^{m_1}, \dots, ker(A-c_kI_n)^{m_k}$ such that $u_1+ \dots + u_k = 0$ and showing that they all equal the zero vector, but I'm really not sure how to approach this.
Consider the polynomials $$Q_j(x)={ P(x)\over (x-c_j)^{m_j}}$$ The polynomials $Q_1,Q_2,\ldots, Q_k$ are relatively prime. Therefore there exist polynomials $R_j$ such that $$R_1Q_1+R_2Q_2+\ldots +R_kQ_k=1$$ hence $$R_1(A)Q_1(A)+R_2(A)Q_2(A)+\ldots +R_k(A)Q_k(A)=I\quad (*)$$ Then for any vector $v$ we have $$R_1(A)Q_1(A)v+R_2(A)Q_2(A)v+\ldots +R_k(A)Q_k(A)v=v$$ The $j$th term belongs to $V_j=\ker(A-c_jI)^{m_j}.$ This gives the decomposition $V=V_1+\ldots + V_k.$ Let $v_j\in \ker(A-c_jI)^{m_j}$ and $$v_1+v_2+\ldots +v_k=0$$ Applying $Q_j(A)$ gives $$Q_j(A)v_j=0$$ On the other hand $(*)$ implies $$R_j(A)Q_j(A)v_j=v_j$$ hence $v_j=0.$ Summarizing we get $$V=V_1\oplus V_2+\oplus \ldots \oplus V_k$$