Cayley–Hamilton定理的证明的疑问

hbghlyj · 发表于 2022-10-29 00:59

本帖最后由 hbghlyj 于 2022-10-29 12:16 编辑 我对Cayley–Hamilton_定理的这个证明有疑问:

记 $B (x) = x I - A$, 并视之为 $R[x]$ 上的矩阵.
记 $\widetilde{B} (x)$ 为 $B (x)$ 的伴随矩阵, 从而
\[
      B (x) \, \widetilde{B} (x) =
      \widetilde{B} (x) \, B (x) =
      p (x) \, I_n ,
\]
其中 $I_n$ 为 $n \times n$ 单位矩阵.
对任何列向量 $u (x) = u_0 + x u_1 + \cdots + x^m u_m \in R[x]^n$,
其中每个 $u_i \in R^n$, 我们记
\[
      u (x) \, |_{x = A} =
      u_0 + A u_1 + \cdots + A^m u_m.
\]
则对任何列向量 $v \in R^n$, 有
\begin{align*}
      p (A) \, v
      & = p (x) \, v \, |_{x = A} \\
      & = \widetilde{B} (x) \, B (x) \, v \, |_{x = A} \\
      & = \widetilde{B} (x) \, (x v - A v) \, |_{x = A} \\
      & = 0.
\end{align*}
因此, $p (A)$ 必为零矩阵.

这里的恒等式$p(x)I=\widetilde{B} (x) B (x)$是假定了乘法交换的,但是代入一个矩阵就不一定成立了.(见A proof using polynomials with matrix coefficients)
令$A=\pmatrix{a&b\\c&d}$,
把$A$换成其它的矩阵$M=\pmatrix{u&v\\s&t}$,具体验证一下这个等式:
\begin{align*}
\tilde B(x)B (x) v|_{x=M}&=(M-A)(M-\tilde A)\\
&=
\left(
\begin{array}{cc}
a d-a u-b c-b s+c v-d u+s v+u^2 & -2 a v-b t+b u+t v+u v \\
c t-c u-2 d s+s t+s u & a d-a t-b c+b s-c v-d t+s v+t^2 \\
\end{array}
\right)
\end{align*}
另一方面
\begin{align*}p(M)&=(ad-b c ) I- (a + d) M + M^2\\
&=\pmatrix{
a d-a u-b c-d u+s v+u^2 & -a v-d v+t v+u v \\
-a s-d s+s t+s u & a d-a t-b c-d t+s v+t^2}
\end{align*}

hbghlyj · 发表于 2022-10-29 19:17

在讨论页举了具体例子

Czhang271828 · 发表于 2022-10-31 18:47

证明里一直在玩 $A$ 的多项式吧, 其实没有出现第二"种"矩阵.
一种便于理解的方式是将整个证明放在有理多项式域 $\mathbb F(x)$ 上的商空间 $\mathbb F(x)/ (m_A(x))$ 上. 然后把 $x$ 看成 $A$ 就可以了.

hbghlyj · 发表于 2022-11-1 02:20

LaTeX to HTML with lwarp
MA20216

3.5 The Cayley–Hamilton theorem

Theorem 3.12 (Cayley–Hamilton Theorem). Let $\phi \in L(V)$ be a linear operator on a finite-dimensional vector space over a field $\F $.

Then $\Delta _{\phi }(\phi )=0$.

Equivalently, for any $A\in M_n(\F )$, $\Delta _A(A)=0$.

Proof of Theorem 3.12. We will prove the matrix version. So let $A\in M_n(\F )$ and write

\begin{equation*} \Delta _{A}=a_0+\dots +a_nx^n. \end{equation*}

Thus, our mission is to show that

\begin{equation*} a_0I_n+a_1A+\dots +a_nA^n=0. \end{equation*}

The key is the adjugate formula:

\begin{equation} \label {eq:9} \mathrm {adj}(A-xI_{n})(A-xI_{n})=\det (A-xI_{n})I_{n}. \end{equation}

Each entry of $\mathrm {adj}(A-xI_n)$ is a polynomial in $x$ of degree at most $n-1$ so we write

\begin{equation*} \mathrm {adj}(A-xI_n)=B_0+B_1x+\dots +B_{n-1}x^{n-1}, \end{equation*}

with each $B_k\in M_n(\F )$. Substitute this into \eqref{eq:9} to get

\begin{equation*} (B_0+B_1x+\dots +B_{n-1}x^{n-1})(A-xI)=(a_0+\dots +a_nx^n)I_n \end{equation*}

and compare coefficients of $x^{k}$ to get

\begin{equation} \label {eq:10} B_kA-B_{k-1}=a_kI_{n}, \end{equation}

for $0\leq k\leq n$, where we have set $B_{-1}=B_n=0\in M_n(\F )$.

Multiply \eqref{eq:10} by $A^k$ on the right to get

\begin{equation*} B_kA^{k+1}-B_{k-1}A^k=a_kA^k \end{equation*}

and sum:

\begin{equation*} \Delta _{A}(A)=\sum _{k=0}^na_kA^k=\sum _{k=0}^n(B_kA^{k+1}-B_{k-1}A^k)=B_nA^{n+1}-B_{-1}=0 \end{equation*}

because nearly all terms in the penultimate sum cancel. □

Corollary 3.13. Let $\phi \in L(V)$ be a linear operator on a finite-dimensional vector space over a field $\F $.
- (1) $m_{\phi }$ divides $\Delta _{\phi }$. Equivalently, $m_A$ divides $\Delta _A$, for any $A\in M_n(\F )$.
- (2) The roots of $m_{\phi }$ are exactly the eigenvalues of $\phi $.

Proof. By Theorem 3.12, $\Delta _{\phi }(\phi )=0$ so $m_{\phi }$ divides $\Delta _{\phi }$ by Proposition 3.7. As a result, any root of $m_{\phi }$ is a root of $\Delta _{\phi }$ and so an eigenvalue. Conversely, any eigenvalue is a root of $m_{\phi }$ by Corollary 3.11. □

Let us summarise the situation when $\F =\C $ so that any polynomial is a product of linear factors. So let $\phi \in L(V)$ be a linear operator on a finite-dimensional complex vector space with distinct eigenvalues $\lst \lambda 1k$. Then

\begin{align*} \Delta _{\phi }&=\pm \prod _{i=1}^k(x-\lambda _i)^{r_i}\\ m_{\phi }&=\prod _{i=1}^k(x-\lambda _i)^{s_i}, \end{align*} where $r_i=\am (\lambda _i)$ and $\bw 1{s_i}{r_i}$, for $\bw 1ik$.

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