This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
Cayley–Hamilton theorem
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
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from lingeo import random_int_list
from linspace import vtop
Let $A$ be an $n\times n$ matrix and
$$ p(x) = a_0x^d + a_1x^{d-1} + \cdots + a_d $$a polynomial.
Then define
Let $A$ be an $n\times n$ matrix and
$$ p_A(x) = s_0(-x)^n + s_1(-x)^{n-1} + \cdots + s_n $$its characteristic polyonimal.
Then
Although $p_A(x) = \det(A - xI)$, it is not correct that $p_A(A) = \det(A - A\cdot I)$ by definition.
In particular, $p_A(A)$ is a matrix, while $\det(A - AI)$ is a number.
Let $A_x = A - xI$.
Then we know $\det(A_x) = p_A(x)$ and the adjucate $A_x\adj$ of $A_x$ has the property
On the other hand, each entry of $A_x\adj$ is a polynomial of degree at most $n-1$, so we may write
$$ A_x\adj = C_1(-x)^{n-1} + C_2(-x)^{n-2} + \cdots + C_n $$for some constant matrices $C_1,\ldots, C_n$.
Therefore,
By comparing the coefficients of the two expressions of $A_x\adj A_x$ we get
$$ \begin{array}{rclcl} s_0 I &= & & &C_1 \\ s_1 I &= &C_1A &+ &C_2\\ \vdots &&&& \\ s_{n-1}I &= &C_{n-1}A &+ &C_n\\ s_nI &= &C_nA. & & \end{array} $$By post-multiplying $(-A)^n-k$ to both sides of the $k$-th equation for $k = 0,\ldots, n$, we get
$$ \begin{aligned} p_A(A) &= s_0(-A)^n + s_1(-A)^{n-1} + \cdots + s_nI \\ &= C_1(-A)^n + C_1A(-A)^{n-1} + C_2(-A)^{n-1} + C_2A(-A)^{n-2} + \cdots + C_n(-A) + C_nA\\ &= O. \end{aligned} $$Notice that $s_0,\ldots,s_n$ can be written as polynomials in entries of $A$.
Also, each entry of $A^0,\ldots,A^n$ can be written as a polynomial in entries of $A$.
This mean each entry of $p_A(A)$ is a polynomial in entries of $A$, and it is zero polynomial!
How come these terms cancel with each other can be viewed from graph theory, but it is beyond the scope.
See the book A Combinatorial Approach to Matrix Theory and Its Applications by Richard A. Brualdi and Dragoš Cvetković for more details.
### code
set_random_seed(0)
print_ans = False
n = 2
A = matrix(n, random_int_list(n^2))
cs = random_int_list(n + 1)
p = vtop(cs)
print("n =", n)
pretty_print(LatexExpr("A ="), A)
pretty_print(LatexExpr("p ="), p)
if print_ans:
for i in range(n + 1):
pretty_print(LatexExpr("A^{%s} ="%i), A^i)
pofA = sum([cs[i] * A^i for i in range(n + 1)], identity_matrix(n))
pretty_print(LatexExpr("p(A) ="), pofA)
令
$$ p(x) = x^2 - 5x + 6 = (x - 2)(x - 3). $$說明
$$ p(A) = A^2 - 5A + 6 = (A - 2I)(A - 3I). $$因此 $p(A)$ 不會因為 $p$ 的表達式不同而影響結果。
Let
$$ p(x) = x^2 - 5x + 6 = (x - 2)(x - 3). $$Verify
$$ p(A) = A^2 - 5A + 6 = (A - 2I)(A - 3I). $$and explain why $p(A)$ does not depend on the different expressions of $p(A)$.
對以下 $n\times n$ 矩陣 $A$，
求出其特徵多項式 $p_A(x)$ 並計算 $A^0, \ldots, A^n$，
最後驗證 $p_A(A) = O$。
For the following matrices $A$, find the characteristic polynomial $p_A(x)$ and calculate $A^0, \ldots, A^n$, and then verify $p_A(A) = O$.
令
$$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}. $$計算 $B_1 = A - I$、$B_2 = A - 2I$、及 $B_3 = A - 3I$。
說明 $B_1,B_2,B_3$ 彼此兩兩可交換，並驗證 $B_1B_2B_3 = O$。
求出特徵多項式 $p_A(x)$ 並說明為什麼 $p_A(A) = O$。
Let
$$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}. $$Calculate $B_1 = A - I$, $B_2 = A - 2I$, and $B_3 = A - 3I$. Explain why $B_1, B_2, B_3$ are mutually commutative and verify $B_1B_2B_3 = O$. Find the characteristic polynomial $p_A(x)$ and explain why $p_A(A) = O$.
令
$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}. $$計算 $A^0, A^1, \ldots, A^5$。
求出特徵多項式 $p_A(x)$ 並說明為什麼 $p_A(A) = O$。
Let
$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}. $$Calculate $A^0, A^1, \ldots, A^5$. Then find the characteristic polynomial $p_A(x)$ and explain why $p_A(A) = O$.
令
$$ A = \begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} $$計算 $B_1 = (A - I)^2$、$B_2 = (A - 2I)^3$、及 $B_3 = (A - 3I)^4$。
說明 $B_1,B_2,B_3$ 彼此兩兩可交換，並驗證 $B_1B_2B_3 = O$。
求出特徵多項式 $p_A(x)$ 並說明為什麼 $p_A(A) = O$。
Let
$$ A = \begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} $$Calculate $B_1 = A - I$, $B_2 = A - 2I$, and $B_3 = A - 3I$. Explain why $B_1, B_2, B_3$ are mutually commutative and verify $B_1B_2B_3 = O$. Find the characteristic polynomial $p_A(x)$ and explain why $p_A(A) = O$.
令
$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} $$而 $A_x = A - xI$。
計算 $A_x\adj$ 並將其寫成
Let
$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} $$and $A_x = A - xI$. Compute $A_x\adj$ and write it into the form
$$ A_x\adj = C_1(-x)^{2} + C_2(-x)^{1} + C_3. $$當 $z = a + bi$ 為一複數時，定義
$$ A(z) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}. $$驗證對於任兩複數 $z_1,z_2$ 都有
所以當 $p$ 是一個多項式滿足 $p(z) = 0$ 時，則有 $p(A(z)) = O$。
For any complex number $z = a + bi$, define
$$ A(z) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}. $$Verify that
for any two complex numbers $z_1,z_2$. Therefore, if $p$ is a polynomial with $p(z) = 0$, then we also have $p(A(z)) = O$.