Let $A$ be an $n \times n$ matrix with characteristic polynomial $$f=(x-c_1)^{d_1}\cdots (x - c_k)^{d_k}$$ Show that $$c_ld_1+\cdots + c_kd_k=\text{trace} (A)$$
My attempt: Characteristic polynomial function of $A$ is $f:F\to F$ such that $f(x)=\det (xI_n-A)$, $\forall x\in F$. By exercise 10 section 5.4, $$f(x)=\prod_{i=1}^n\left(x-A(i,i)\right) +\sum_{\sigma \in S_n \setminus \{\text{id}_{J_n}\}}(\text{sgn}\sigma) \prod_{i=1}^nx\delta_{i\sigma(i)}-A(i,\sigma (i))$$ and $\deg \left(\prod_{i=1}^nx\delta_{i\sigma(i)}-A(i,\sigma (i))\right)\leq n-2$, $\forall \sigma \in S_n\setminus \{\text{id}_{J_n}\}$. So $f(x)=x^n+(-1)\text{tr} (A)x^{n-1}+\cdots +\det (A)$. By hypothesis, $$f(x)=(x-c_1)^{d_1}\dots (x-c_k)^{d_k}=\underbrace{(x-c_1)\cdots (x-c_1)}_{d_1 \text{times}}\cdots \underbrace{(x-c_k)\cdots (x-c_k)}_{d_k \text{times}}.$$ By distributive law, coefficient of $x^{n-1}$ is $$\underbrace{(-c_1)+…+(-c_1)}_{d_1 \text{times}}+\dots +\underbrace{(-c_k)+\cdots +(-c_k)}_{d_k \text{times}}=(-c_1d_1)+\cdots +(-c_kd_k)=(-1)(c_1d_1+\cdots +c_kd_k).$$ Thus $\text{trace}(A)= c_1d_1+\cdots c_kd_k$. Is my proof correct?
the result and method of proof is correct but with your definition ( $ f(\lambda) = det(\lambda I_n - A)$ ). The coefficient of $tr(A)$ shouldn't be $(-1)^n$, it's just $-1$. $$ \underbrace{(-c_1)+\dots+(-c_1)}_{d_1 times} \quad \text{is just} \quad -d_1c_1 $$ and $$ \prod_{i=1}^n(x-A(i,i)) = x^n + (-A(1,1))x^{n-1} + \dots + (-A(2,2))x^{n-1} + \dots = x^n +\underline{(-1)tr(A)} + \dots $$