linear algebra - Cayley-Hamilton Theorem: characteristic polynomial and distinct eigenvalues

I do not know where to start for solving this exercise although I have the "official" solution. In the solution, I see that some variables are exchanged but I cannot connect the steps to a coherent "story". Some direct help is highly appreciated. I do know how to get the poles from a characteristic equation.

Exercise

Consider a matrix $A \in \mathbb{R}^{n \times n}$ with the characteristic polynomial
$$\det(sI-A)=a(s)=s^n+a_{n-1}s^{n-1}+ \cdots + a_1 s+a_0$$

a) Show that if $A$ has distinct eigenvalues $(λ_1, λ_2, \ldots , λ_n)$, the following relationship holds:
$$\Lambda^n + a_{n-1}\Lambda^{n-1}+ \cdots + a_0 I = 0$$
with $\Lambda = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}$

b) Now show that
$$A^n + a_{n-1}A^{n-1}+ \cdots + a_0 I = 0$$

(This proves the Cayley-Hamilton Theorem for distinct eigenvalues.)
Hint: Use the fact that a matrix A with distinct eigenvalues can be written as
$A = T ΛT^{−1}$; where $Λ$ is diagonal.

Solution:

a) The characteristic equation is true for all eigenvalues of $A$, $λ_1 . . . λ_n$
$$\lambda^n + a_{n-1}\lambda^{n-1}+\cdots + a_0 = 0$$
$$\Lambda = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}$$
so $\Lambda^n + a_{n-1} \Lambda^{n-1}+ \cdots + a_0 I = 0$

This is the matrix characteristic equation.

b) With distinct eigenvalues and diagonal $Λ$ we have
$$A=T\Lambda T^{-1} \\ A^2 = T \Lambda T^{-1} T \Lambda T^{-1} = T \Lambda^2 T^{-1} \\ \vdots \\ A^m=T \Lambda^m T^{-1}$$

Multiply the matrix characteristic equation by $T$ (left) and $T^{-1}$ (right) to obtain $$T \Lambda^n T^{-1} + a_{n-1}T \Lambda^{n-1} T^{-1}+\cdots + a_0 TT^{-1}= 0$$
$$A^n + a_{n-1}A^{n-1}+\cdots+a_0 I = 0$$

Answer

It seems that you have done. Let
$\det(sI-A)=a(s)=s^n+a_{n-1}s^{n-1}+ \cdots + a_1 s+a_0$ be the characteristic polynomial. We know by assumption that it has exactly $n$ distinct roots, namely: $\lambda_1,\dots,\lambda_n$.

Let $\Lambda = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}\text{ }$ be the diagonal matrix whose entries are the roots of $a(s)$.

We want to show that $\Lambda^n + a_{n-1}\Lambda^{n-1}+ \cdots + a_0 I = 0.$ Since

$$\Lambda^k = \begin{pmatrix} \lambda_1^k & 0 & \cdots & 0 \\ 0 & \lambda_2^k & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n^k \end{pmatrix},$$

thus $$\Lambda^n + a_{n-1}\Lambda^{n-1}+ \cdots + a_0 I= \begin{pmatrix} a(\lambda_1) & 0 & \cdots & 0 \\ 0 & a(\lambda_2) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & a(\lambda_n) \end{pmatrix}= \begin{pmatrix} 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 0 \end{pmatrix}$$
because $\lambda_i$ are roots of $a(s)$.

The second part ask to prove the theorem for a matrix $A$ similar to $\Lambda$, i.e.

$$A^n + a_{n-1}A^{n-1}+ \cdots + a_0 I = 0 .$$

Thus $A = T ΛT^{−1}$; where $Λ$ is diagonal, by definition of similarity; and

$$A^k = T\Lambda^k T^{-1} \quad \forall k\in \Bbb N.$$

Finally: $$A^n + a_{n-1}A^{n-1}+\cdots+a_0 I = 0\Longleftrightarrow T \Lambda^n T^{-1} + a_{n-1}T \Lambda^{n-1} T^{-1}+\cdots + a_0 TT^{-1}= 0$$
$$\Longleftrightarrow T( \Lambda^n + a_{n-1} \Lambda^{n-1} +\cdots + a_0I)T^{-1}= 0 \Longleftrightarrow \Lambda^n + a_{n-1} \Lambda^{n-1} +\cdots + a_0I =T^{-1}0T=0.$$