Properties of minimal polynomial with respect to the field elements

Question

Let F be a field and V be a vector space on F. Let $m_\alpha$ be the minimal polynomial of linear operator $\alpha$ on $V$, and $deg\ (m_\alpha)=n$.

1. Show that for each $\lambda_i\in F$ such that $m_\alpha(\lambda_i)\neq0_F$, $\exists !\ p_{\lambda_{i}}(x)\in F[x]$, $deg(p_{\lambda_{i}}) \leq n-1$ such that
   $$(\alpha-\lambda_i\ id_V)p_{\lambda_{i}}(\alpha)=id_V$$
2. For $n$ distinct such $\lambda_i$, show that $\exists !\ \{k_i\}_{i=1}^n\in F$ such that $$\sum_{i=1}^n k_i p_{\lambda_{i}}(\alpha)=id _V$$

I know the first part is by the division algorithm. However, I do not know where to proceed with the second part. Any hints will be appreciated.

Answer 1

Here is a way to prove part $2$.

Let $\lambda_1, ..., \lambda_n$ be $n$ distinct elements of $F$ such that $m_{\alpha}(\lambda_i) \neq 0$ for any $1 \leq i \leq n$. Let's prove that the family $(p_{\lambda_i}(\alpha))_{ 1 \leq i \leq n}$ is linearly independant. We suppose that there exists $k_1, ..., k_n \in F$ such that $$\sum_{i=1}^n k_i p_{\lambda_i}(\alpha) = 0$$

Then, multiplying by $\displaystyle{\prod_{j=1}^n (\alpha - \lambda_j\mathrm{Id}_V) }$, you get that $$\sum_{i=1}^n k_i \prod_{j=1 \\ j \neq i}^n (\alpha - \lambda_j \mathrm{Id}_V) = 0$$

In particular, the polynomial $$Q(X)= \sum_{i=1}^n k_i \prod_{j=1 \\ j \neq i}^n (X - \lambda_j \mathrm{Id}_V) $$ has degree at most $n-1$ and satisfies $Q(\alpha)=0$, so because $n$ is the degree of the minimal polynomial of $\alpha$, one must have $Q=0$. But for every $1 \leq i \leq n$, one has $Q(\lambda_i)=k_i \displaystyle{\prod_{j=1\\ j\neq i}^n (\lambda_i - \lambda_j)}$, so because the $\lambda_i$'s are distinct, you get $k_i = 0$.

This proves that the family $(p_{\lambda_i}(\alpha))_{ 1 \leq i \leq n}$ is linearly independant.

Now consider $A = \mathrm{Span} \left(\alpha ^k \right)_{k \geq 0}$. Then since $m_{\alpha}$ has degree $n$, one has $\mathrm{dim}(A)=n$. Since the family $(p_{\lambda_i}(\alpha))_{ 1 \leq i \leq n}$ is linearly independant and has $n$ elements, then it is a basis of $A$ : since $\mathrm{Id}_V \in A$, then there exists a unique family $(k_i)_{1 \leq i \leq n}$ such that $$\sum_{i=1}^n k_i p_{\lambda_{i}}(\alpha)=\mathrm{Id}_V$$