Minimal Polynomial

Minimal Polynomial

A complex polynomial $p$ annihilates $A$ when $p(A) = 0$.

There is a unique monic polynomial of least degree $m$ which annihilates $A$. This is the minimal polynomial $q_A$.

Proof Let $q, q'$ be monic polynomials of degree $m$ that annihilate $A$. Then $g = q - q'$ has degree $< m$ and $g(A) = q(A) - q'(A) = 0$. If $g \neq 0$, scaling it gives a monic annihilating polynomial of degree $< m$, so $g = 0$ and $q = q'$.

For any $p$ annihilating $A$, $q_A \mid p$.

Proof Divide, $p(t) = d(t) q_A(t) + r(t)$ with $\deg r < \deg q_A$. Then $0 = p(A) = d(A) q_A(A) + r(A) = r(A)$, so $r$ is the zero polynomial.

By Cayley-Hamilton, $q_A \mid p_A$.

Roots of the Minimal Polynomial

Every root of $p_A$ is a root of $q_A$. I.e. if $p_A(t) = \prod_{i=1}^{r} (t - \lambda_i)^{n_i}$ with distinct $\lambda_i$, then

$$q_A(t) = \prod_{i=1}^{r} (t - \lambda_i)^{m_i}, \qquad 1 \le m_i \le n_i$$

Proof Let $x$ be an eigenvector of $\lambda$ and $q_A(t) = \sum_{i=0}^{m} c_i t^i$. Then

$$\vec{0} = q_A(A)x = \sum_{i=0}^{m} c_i A^i x = \sum_{i=0}^{m} c_i \lambda^i x = q_A(\lambda) x$$

so $q_A(\lambda) = 0$.

Minimal Polynomial and Jordan Blocks

$m_i$ is the size of the largest Jordan block associated with $\lambda_i$.

Proof Let $A = SJS^{-1}$ and $q_A(t) = \prod_{i=1}^{r} (t - \lambda_i)^{b_i}$. Then

$$q_A(A) = (A - \lambda_1 I)^{b_1} \cdots (A - \lambda_r I)^{b_r} = S\big[(J - \lambda_1 I)^{b_1} \cdots (J - \lambda_r I)^{b_r}\big]S^{-1}$$

In $(J - \lambda_i I)^{b_i}$ the blocks of $\lambda_i$ become $J_\cdot(0)^{b_i}$ and the others stay invertible. $J_k(0)^{b} = 0$ exactly when $b \ge k$, so the product is 0 iff each $b_i$ is at least the size of the largest block of $\lambda_i$.

In the Ferrers diagram of the Jordan blocks of $\lambda_i$, the number of blocks is the geometric multiplicity, the largest block is $m_i$, and the total size is the algebraic multiplicity.

Diagonalizable and Nonderogatory Matrices

$A \in M_n$ is diagonalizable $\iff$ $q_A$ is a product of distinct linear factors, since then the largest Jordan blocks have size 1 and $J$ is diagonal.

$A \in M_n$ is nonderogatory when every $\lambda \in \sigma(A)$ has one Jordan block, i.e.

• $\mathrm{gemu}(\lambda_i) = 1$ for all $i$,
• $m_i = n_i$ for all $i$,
• $p_A = q_A$.

$A$ is diagonalizable and nonderogatory $\iff$ the eigenvalues of $A$ are distinct.

Companion Matrix

Let $g(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_1 t + a_0$. The companion matrix of $g$ is

$$C_g = \begin{bmatrix} 0 & 1 & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1 \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{bmatrix} \in M_n$$

The first $n - 1$ rows of $C_g x = \lambda x$ force $x_{i+1} = \lambda x_i$, so eigenvectors have the form

$$x = \alpha \begin{bmatrix} 1 \\ \lambda \\ \lambda^2 \\ \vdots \\ \lambda^{n-1} \end{bmatrix}$$

The last equation gives $-\alpha \sum_{i=0}^{n-1} a_i \lambda^i = \alpha \lambda^n \iff g(\lambda) = 0$. So $\lambda$ is an eigenvalue $\iff$ $\lambda$ is a root of $g$, and its eigenvectors span a 1-dimensional space. Each eigenvalue has $\mathrm{gemu} = 1$, so $C_g$ is nonderogatory and

$$p_{C_g} = q_{C_g} = g$$

Commuting Matrices

$A \in M_n$ is nonderogatory $\iff$

$$\{B \in M_n : AB = BA\} = \{p(A) : p \text{ is a complex polynomial}\}$$

$(\supseteq)$ is trivial, since any $p(A)$ commutes with $A$.