求不依赖基与矩阵分块行列式的不变子空间限制算子特征多项式整除性证明

Hey there, let's tackle this cleanly without relying on bases or matrix determinant tricks—just using core linear algebra concepts and polynomial properties.

First, let's recall key definitions to set the stage:

• For a linear operator $S \in L(X,X)$ on a finite-dimensional space $X$, its characteristic polynomial $p_S(t)$ is the unique monic polynomial of degree $\dim X$ such that $p_S(S) = 0$ (this follows from the Cayley-Hamilton theorem, a foundational result we can take as given here).
• A subspace $W \preccurlyeq V$ is $T$-invariant if $T(W) \subseteq W$, so the restriction $T_W = T|_W$ is a well-defined operator on $W$.

Step 1: $p_T(T_W) = 0$ (the characteristic polynomial of $T$ annihilates $T_W$) #

By Cayley-Hamilton, $p_T(T) = 0$—meaning applying the polynomial $p_T$ to the operator $T$ gives the zero operator on $V$. Now take any $w \in W$:
$$p_T(T_W)(w) = p_T(T)(w) = 0$$
This holds because every term in $p_T(T)$ is a power of $T$, and since $T(W) \subseteq W$, applying any power of $T$ to $w$ keeps it in $W$. So $p_T(T_W)$ maps every element of $W$ to 0, which means $p_T(T_W) = 0 \in L(W,W)$.

Step 2: If a polynomial annihilates an operator, its characteristic polynomial divides that polynomial #

Now we need to show that if $q(t)$ is a polynomial with $q(S) = 0$ for some operator $S$, then $p_S(t) \mid q(t)$. Here's how to do this without bases:

1. Algebraic closed field case: Suppose $F$ is algebraically closed. Then $S$ has eigenvalues $\lambda_1, \lambda_2, ..., \lambda_k$ (with multiplicities matching the dimension of their generalized eigenspaces). For each eigenvalue $\lambda_i$, there exists a generalized eigenvector $v_i$ such that $(S - \lambda_i I)^m v_i = 0$ for some $m > 0$. Since $q(S) = 0$, applying $q(S)$ to $v_i$ gives $q(\lambda_i)v_i = 0$ (polynomials commute with operators, so we can factor $q(S)$ and evaluate directly on generalized eigenvectors). Since $v_i \neq 0$, $q(\lambda_i) = 0$, so $(t - \lambda_i)$ divides $q(t)$. Since $p_S(t)$ is the product of $(t - \lambda_i)$ raised to their multiplicities, $p_S(t)$ divides $q(t)$.
2. General field case: Take the algebraic closure $\overline{F}$ of $F$, and extend $S$ to an operator $\overline{S}$ on $\overline{F} \otimes_F X$ (the space of $\overline{F}$-linear combinations of elements of $X$). The characteristic polynomial $p_{\overline{S}}(t) = p_S(t)$ (it's the same monic polynomial over $F$, hence over $\overline{F}$), and $q(\overline{S}) = 0$ (since $q(S)=0$ extends naturally). From the algebraic closed field case, $p_{\overline{S}}(t) \mid q(t)$ in $\overline{F}[t]$, but since both $p_S(t)$ and $q(t)$ are in $F[t]$, the divisibility must hold in $F[t]$ too.

Step 3: Putting it all together #

We have $q(t) = p_T(t)$ annihilates $S = T_W$, so by Step 2, $p_{T_W}(t) \mid p_T(t)$. That's exactly what we needed to prove!

Alternatively, if you prefer a module-theoretic perspective (still basis-free):

• Treat $V$ as a module over the polynomial ring $F[t]$, where the action of $t$ is given by $t \cdot v = T(v)$. A $T$-invariant subspace is exactly an $F[t]$-submodule.
• For a finite-dimensional $F[t]$-module $X$ corresponding to operator $S$, the order of $X$ (the monic generator of the annihilator ideal ${ q(t) \in F[t] \mid q \cdot X = 0 }$) is exactly the characteristic polynomial $p_S(t)$.
• For a submodule $W \subseteq V$, we have the order of $V$ is the product of the order of $W$ and the order of the quotient module $V/W$ (a standard result for torsion modules over a PID like $F[t]$). Translating back to operators, this means $p_T(t) = p_{T_W}(t) \cdot p_{\overline{T}}(t)$, where $\overline{T}$ is the quotient operator on $V/W$, so divisibility follows immediately.

内容的提问来源于stack exchange，提问作者Benjamin Bray