# Minimal and characteristic polynomials of $\begin{bmatrix} 3 & 0 \\1 & 3 \end{bmatrix}$

I have to determine the characteristic polynomial and the minimal polynomial of: $$\begin{bmatrix} 3 & 0 \\1 & 3 \end{bmatrix}$$

After computing $\det(A-\lambda I_2)$, I get that characteristic polynomial is $P_A(\lambda) = (3-\lambda)^2$ and $\lambda=3$ (if $\det(A-\lambda I_2)=0$).

Now, I am trying to determine the minimal polynomial:

$$\text{Null}(A-\lambda I_2) = \text{Null}( \begin{bmatrix} 0 & 0 \\1 & 0 \end{bmatrix})$$

And then I solve the linear system of equations:

$$0\cdot x_1 + 0\cdot x_2 = 0 \\ and \\ 1\cdot x_1 + 0\cdot x_2 = 0 \\ \Rightarrow x_1 = 0\ \text{and}\ x_2 \in \mathbb{R}$$

Now, $\text{Null}( \begin{bmatrix} 0 & 0 \\1 & 0 \end{bmatrix}) = x_2\begin{bmatrix} 0 \\1 \end{bmatrix}, x_2\in\mathbb{R}$ (because $x_1=0\cdot x_2$ and $x_2 = 1\cdot x_2$)

So, $\text{Null}( \begin{bmatrix} 0 & 0 \\1 & 0 \end{bmatrix}) = \text{span}\{ \begin{bmatrix} 0 \\1 \end{bmatrix}\}$ and the dimension of the null is 1 so the minimal polynomial is: $m_A(\lambda) = (3-\lambda)$

But the answer should be $m_A(\lambda) = (3-\lambda)^2$, as I saw here (exercise 2.1 - for answer scroll a bit to bottom).

Why? Please explain me. Thank you!

• $m_A(\lambda) = 1$ (clearly not the case)
• $m_A(\lambda) = (3-\lambda)$ (again, not true since $3\cdot I - A\neq 0$
• $m_A(\lambda) = (3-\lambda)^2$.
• Thank you! The power of minimal polynomial could be found by checking at which power $A-\lambda I_n = O_n$. I was really wrong, I was searching the eigenspace. – MM PP Aug 22 '16 at 10:37