Finding the minimal polynomial of a matrix A

Question

$$ Let \qquad A= \begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ The characteristic polynomial of $A$ is: $p_A(x)=\det(xI-A)$
Now $ \begin{bmatrix} x & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & x \\ \end{bmatrix} $- $ \begin{bmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \\ \end{bmatrix} $ = $ \begin{bmatrix} x-3 & 0 & 0 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \\ \end{bmatrix} $
So, $$\det \begin{pmatrix} x-3 & 0 & 0 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \\ \end{pmatrix}= (x-3)(x-1)^2=p_A(x)$$ Since the minimal polynomial must be the divisor of the characteristic polynomial, the minimal polynomial is either $(x-3)(x-1)$ or $(x-3)(x-1)^2$

Until here these are what I know. I saw a guy on youtube when finding the minimal polynomial did this computation below: $$ \begin{bmatrix} 1 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} -\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \end{bmatrix} \begin{bmatrix} 3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} -\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \end{bmatrix}=\begin{pmatrix} -2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$ , so the minimal polynomial of $A$ is:$(x-1)(x-3)$
What is the computation above ? Is it $m_A(x)$ or $m_A(A)$ ?

Answer 1

The minimal polynomial of a diagonalizable matrix has distinct roots and every eigenvalue must be a root.

For the particular case, you can dispense with the theorem above, as it's easy to verify that $(A-I)(A-3I)=0$, but neither $(A-I)^2=0$ nor $A-3I=0$. So the minimal polynomial must be $(x-1)(x-3)$.