什么是 minimal polynomial？

Example 3.13 (Minimal polynomials found by hand)

The minimal polynomial is computed by testing successive degrees. For the all-ones matrix $J \in \mathcal{M}_3(\R)$: $J \neq \lambda I$ (degree $1$ is out), and $J^2 = 3J$, so

degree $2$, split, simple roots — $J$ is diagonalizable with spectrum $\{0, 3\}$ (Corollary 3.17 below), confirming Example 2.19 without a single determinant. For the swap matrix $A$ of Example 3.15: $A \neq \pm I$ and $A^2 = I$ give $\mu_A = X^2 - 1$. In both cases the pattern is the same: guess a low-degree identity from the structure (rank one forces $J^2 = (\operatorname{tr}J)\,J$; an involution forces $A^2 = I$), then check that no proper divisor annihilates. Minimal polynomials are usually found, not computed from $\chi$.

Example 3.15 (The kernel lemma with explicit projectors)

Let $A = \left(\begin{smallmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{smallmatrix}\right)$ (swap the first two coordinates). Then $A^2 = I$: the polynomial $X^2 - 1 = (X - 1)(X + 1)$ annihilates $A$, its factors are coprime, and Bézout is explicit:

Following the proof of Theorem 3.14, the projections onto $\ker(A - I)$ and $\ker(A + I)$ are the polynomials in $A$

Check: $\pi_+ + \pi_- = I$, $\pi_+\pi_- = 0$, $\pi_\pm^2 = \pi_\pm$, and the images are the plane $\{x = y\}$ (symmetric vectors, eigenvalue $1$) and the line $\R(1, -1, 0)$ (antisymmetric, eigenvalue $-1$). The kernel lemma is not an existence statement: Bézout coefficients are the projector formulas.

Example 3.20 (Minimal versus characteristic)

For $D = \operatorname{diag}(2, 2, 3)$: $\chi_D = (X - 2)^2(X - 3)$ but $\mu_D = (X - 2)(X - 3)$, since $(D - 2I)(D - 3I) = 0$ (check on the canonical basis) while neither factor alone kills $D$. For the shift block $N = \left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right) \oplus (3)$, i.e. $N' = \left(\begin{smallmatrix}0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 3\end{smallmatrix}\right)$: $\chi_{N'} = X^2(X - 3)$ and $\mu_{N'} = X^2(X - 3)$ — the double root is genuinely needed because $N'$ is not diagonalizable on $\ker$-side ($N'e_2 = e_1 \neq 0$). Rule of thumb: $\mu$ and $\chi$ share their roots (Proposition 3.12); the multiplicity in $\mu$ measures the size of the largest nilpotent block, the one in $\chi$ the total dimension of the characteristic subspace.