Minimal polynomial and algebraic multiplicy 
Let $A\in \mathbb{R}^{4 \times 4}$ and the Minimal polynomial $m_A(x)=x^3-2x^2$
What are the eigenvalues and the multiplicy?

$m_A(x)=x^3-2x^2=x^2(x-2)$ so $\lambda_1=0$ and $ \lambda_2=2$ so we now look at the option of the characteristic polynomial we know that it must be formed by $x$ and $(x-2)$ with at a degree that at most is 4. So it can be $x^2(x-2)^2$ so the multiplicity of $\lambda_1=0$ is 2 and $\lambda_2=2$ is 2
Why can the multiplicity be $\lambda_1=0$ is 3 and $\lambda_2=2$ is 1?
 A: Because $x^{2}(x-2)$ is the minimal polynomial, then the Jordan canonical form can be built from the following blocks along the diagonal:
$$
                 \left[2\right], \left[0\right], \left[\begin{array}{cc} 0 & 1 \\ \cdot & 0\end{array}\right]
$$
You must have the first and third blocks, which, together, occupy 3 of the 4 diagonal entries. So there are two possibilities:
$$
                \left[\begin{array}{cccc}0 & 1 & 0 & 0 \\
                                         0 & 0 & 0 & 0 \\
                                         0 & 0 & 2 & 0 \\
                                         0 & 0 & 0 & 2 \end{array}\right],
                \left[\begin{array}{cccc}0 & 1 & 0 & 0 \\
                                         0 & 0 & 0 & 0 \\
                                         0 & 0 & 0 & 0 \\
                                         0 & 0 & 0 & 2 \end{array}\right].
$$
A: As you said, the characteristic polynomial is of degree $4$ and is a multiple of $x^2(x-2)$, hence it is either $x^2(x-2)^2$ (which you considered) or $x^3(x-2)$. The second case is indeed possible (think of an example!)
