$A = \operatorname{diag}(2, 3, 2, 5)$. Eigenvalues: $2, 3, 5$.

$p_A(\lambda) = (\lambda - 2)^2(\lambda - 3)(\lambda - 5)$.

$m_A(\lambda) = (\lambda - 2)(\lambda - 3)(\lambda - 5)$ (each eigenvalue appears once, since diagonal matrices are diagonalizable).

Verify: $(A - 2I)(A - 3I)(A - 5I)$. The $(1,1)$ entry: $(2-2)(2-3)(2-5) = 0$. The $(2,2)$ entry: $(3-2)(3-3)(3-5) = 0$. All diagonal entries are $0$ ✓.

$A = cI$: $m_A(\lambda) = \lambda - c$.

$p_A(\lambda) = (\lambda - c)^n$, but $m_A$ has degree just $1$.

This is the extreme case where $m_A$ is much smaller than $p_A$.

The minimal polynomial of a Jordan form $J = \operatorname{diag}(J_{k_1}(\lambda_1), \ldots, J_{k_r}(\lambda_r))$ is:

$m_J(\lambda) = \prod_i (\lambda - \lambda_i)^{s_i}$

where $s_i$ is the size of the largest Jordan block for eigenvalue $\lambda_i$.

For $J = \operatorname{diag}(J_3(2), J_1(2), J_2(5))$: $m_J(\lambda) = (\lambda - 2)^3(\lambda - 5)^2$. For $J = \operatorname{diag}(J_2(2), J_2(2), J_1(5))$: $m_J(\lambda) = (\lambda - 2)^2(\lambda - 5)$.

The companion matrix of $p(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_0$ has $m_A = p_A = p$. For companion matrices, the minimal and characteristic polynomials coincide.

$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$: $p_A(\lambda) = (\lambda - 2)^2$. Is $(A - 2I) = 0$? No: $A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \neq 0$.

So $m_A(\lambda) = (\lambda - 2)^2$ (has a repeated root). Conclusion: $A$ is not diagonalizable.

$B = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2I$: $m_B(\lambda) = \lambda - 2$ (no repeated roots). Conclusion: $B$ is diagonalizable.

$A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}$: $p_A = (\lambda - 3)^3$. Check: $(A - 3I) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \neq 0$. $(A-3I)^2 = 0$. So $m_A = (\lambda - 3)^2$ (repeated root). Not diagonalizable.

$B = \operatorname{diag}(3, 3, 3) = 3I$: $m_B = \lambda - 3$. Diagonalizable.

$C = \operatorname{diag}(3, 3, 5)$: $m_C = (\lambda - 3)(\lambda - 5)$ (distinct linear factors). Diagonalizable.

$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$.

$p_A(\lambda) = (\lambda - 1)^2(\lambda - 2)$.

The candidates for $m_A$ (monic divisors of $p_A$ sharing the same roots) are:

• $(\lambda - 1)(\lambda - 2)$ (degree $2$),
• $(\lambda - 1)^2(\lambda - 2)$ (degree $3$, same as $p_A$).

Test $(\lambda - 1)(\lambda - 2)$: $(A - I)(A - 2I) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \neq 0$.

So $m_A \neq (\lambda - 1)(\lambda - 2)$, hence $m_A = (\lambda - 1)^2(\lambda - 2) = p_A$.

$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$.

$p_A(\lambda) = (\lambda - 1)^2(\lambda - 2)$.

Test $(\lambda - 1)(\lambda - 2)$: $(A - I)(A - 2I) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0$.

So $m_A = (\lambda - 1)(\lambda - 2)$, which is strictly smaller than $p_A$. The matrix is diagonalizable (it already is diagonal).

Jordan form $\operatorname{diag}(J_2(1), J_1(1), J_3(2))$:

• Largest block for $\lambda = 1$: size $2$.
• Largest block for $\lambda = 2$: size $3$.
• $m_A = (\lambda - 1)^2(\lambda - 2)^3$.
• $p_A = (\lambda - 1)^3(\lambda - 2)^3$ (sum of block sizes for each eigenvalue).

Jordan form $\operatorname{diag}(J_2(0), J_2(0))$:

• $p_A = \lambda^4$, $m_A = \lambda^2$.
• The matrix is nilpotent with index $2$: $A^2 = 0$ but $A \neq 0$.

Jordan form $\operatorname{diag}(J_3(0), J_2(0), J_1(0))$:

• $p_A = \lambda^6$, $m_A = \lambda^3$.
• Nilpotent index $3$.

$A = \operatorname{diag}(J_2(0), J_1(0))$ and $B = \operatorname{diag}(J_1(0), J_1(0), J_1(0))$:

Both have $p(\lambda) = \lambda^3$. But $m_A = \lambda^2$ while $m_B = \lambda$.

The minimal polynomial distinguishes the two similarity classes that the characteristic polynomial cannot.

$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$: $m_A = \lambda^2$, so $\dim F[A] = 2$.

$F[A] = \operatorname{span}\{I, A\} = \left\{\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} : a, b \in F\right\}$.

This is a $2$-dimensional commutative subalgebra of $M_2(F)$.

$A = \operatorname{diag}(1, 2, 3)$: $m_A = (\lambda - 1)(\lambda - 2)(\lambda - 3)$, $\deg m_A = 3$.

$F[A] = \{p(A) : p \in F[\lambda]\}$ has dimension $3$, spanned by $I, A, A^2$.

$F[A] = \{\operatorname{diag}(a, b, c) : a, b, c \in F\}$ (all diagonal matrices), which has dimension $3$.

$A = 5I$ ($n \times n$): $m_A = \lambda - 5$, $\dim F[A] = 1$. $F[A] = \{cI : c \in F\}$, the scalar matrices.

If $A$ is nilpotent, then $m_A = \lambda^k$ where $k$ is the smallest positive integer with $A^k = 0$ (the nilpotent index). By Cayley--Hamilton, $k \leq n$.

For the $n \times n$ shift matrix: $m_A = \lambda^n$ (the maximum possible). For $A = 0$: $m_A = \lambda$ (the minimum for a nilpotent matrix).

The number of Jordan blocks of size $\geq j$ equals $\operatorname{rank}(A^{j-1}) - \operatorname{rank}(A^j)$.