# 特徵多項式係數

Coefficients of the characteristic polynomial

This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/).

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from lingeo import random_int_list, random_good_matrix

## Main idea

Let $A$ be an $n\times n$ matrix.
Let $\alpha \subseteq [n]$ be a subset of indices.
We let $A[\alpha]$ be the submatrix of $A$ induced on rows and columns in $\alpha$.
Such a matrix is called a **principal submatrix** of $A$, and its determinant is called a **principal minor**.

Let $s_k = s_k(A)$ be the sum of all $k\times k$ principal minors.
Vacuously, we define $s_0 = 1$.
Then

$$
\det(A - xI) = s_0(-x)^n + s_1(-x)^{n-1} + \cdots + s_n.
$$

In particular,
$s_1 = \tr(A)$ is the sum of all diagonal entries of $A$, and
$s_n = \det(A)$.

This identity follows from the expansion of the characteristic polynomial by the distributive law on each column.
Here is an example.

$$
\begin{aligned}
\det\begin{bmatrix}
1 - x & 2 & 3 \\
4 & 5 - x & 6 \\
7 & 8 & 9 - x
\end{bmatrix}
&=
\det\begin{bmatrix}
-x & 0 & 0 \\
0 & -x & 0 \\
0 & 0 & -x
\end{bmatrix}
+
\det\begin{bmatrix}
-x & 0 & 3 \\
0 & -x & 6 \\
0 & 0 & 9
\end{bmatrix}
+
\det\begin{bmatrix}
-x & 2 & 0 \\
0 & 5 & 0 \\
0 & 8 & -x
\end{bmatrix}
+
\det\begin{bmatrix}
1 & 0 & 0 \\
4 & -x & 0 \\
7 & 0 & -x
\end{bmatrix}
+ \\
&\mathrel{\phantom{=}}
\det\begin{bmatrix}
-x & 2 & 3 \\
0 & 5 & 6 \\
0 & 8 & 9
\end{bmatrix}
+
\det\begin{bmatrix}
1 & 0 & 3 \\
4 & -x & 6 \\
7 & 0 & 9
\end{bmatrix}
+
\det\begin{bmatrix}
1 & 2 & 0 \\
4 & 5 & 0 \\
7 & 8 & -x
\end{bmatrix}
+
\det\begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{bmatrix}.
\end{aligned}
$$

## Side stories

- Vieta's formulas
- Cauchy–Binet formula

## Experiments

##### Exercise 1

執行以下程式碼。

<!-- eng start -->
Run the code below.
<!-- eng end -->

### code
set_random_seed(0)
print_ans = False

n = 3
spec = random_int_list(n, 3)
D = diagonal_matrix(spec)
Q = random_good_matrix(n,n,n,2)
A = Q * D * Q.inverse()

pretty_print(LatexExpr("A ="), A)

if print_ans:
    for k in [3,2,1]:
        print("k =", k)
        kmtx = []
        kmnr = []
        for alpha in Combinations(list(range(3)), k):
            kmtx.append(A[alpha, alpha])
            kmnr.append(A[alpha, alpha].det())
        print("all k x k principal submatricies:")
        pretty_print(*kmtx)
        print("all k x k principal minors:")
        print(kmnr)
        print("sk =", sum(kmnr))
        print("---")
        
    pA = (-1)^n * A.charpoly()
    print("characteristic polyomial =", pA)
    print("spectrum is the set { " + ", ".join("%s"%val for val in spec) + " }")
    

We get
$$
A = \begin{bmatrix}
-63 & -60 &24 \\
74 & 71 & -28 \\
20 & 20 & -7
\end{bmatrix}.
$$

##### Exercise 1(a)

列出所有的 $3\times 3$ 主子矩陣，並計算 $s_3$。

<!-- eng start -->
List all the $3\times 3$ principal submatrices and find $s_3$.
<!-- eng end -->

##### Exercise 1(a) - answer
When $|\alpha| = 3$ , $\alpha = \{1, 2, 3\}$, so the $3\times 3$ principal submatrices $$
A[\alpha] = \begin{bmatrix}
-63 & -60 &24 \\
74 & 71 & -28 \\
20 & 20 & -7
\end{bmatrix},
$$ and $s_3 = \det (A[\alpha]) = -9$

##### Exercise 1(b)

列出所有的 $2\times 2$ 主子矩陣，並計算 $s_2$。

<!-- eng start -->
List all the $2\times 2$ principal submatrices and find $s_2$.
<!-- eng end -->

##### Exercise 1(b) - answer
When $|\alpha| = 2$ , $\alpha = \{1, 2\}, \{1, 3\}, \{2, 3\}$, so the $2\times 2$ principal submatrices $$
A[\alpha] = \begin{bmatrix}
-63 & -60 \\
74 & 71
\end{bmatrix},
\begin{bmatrix}
-63 & 24 \\
20 & -7
\end{bmatrix},
\begin{bmatrix}
71 & -28 \\
20 & -7
\end{bmatrix}
$$ and $s_2 = (-33) + (-39) + 63 = -9$

##### Exercise 1(c)

列出所有的 $1\times 1$ 主子矩陣，並計算 $s_1$。

<!-- eng start -->
List all the $1\times 1$ principal submatrices and find $s_1$.
<!-- eng end -->

##### Exercise 1(c) - answer
When $|\alpha| = 1$ , $\alpha = \{1\}, \{2\}, \{3\}$, so the $1\times 1$ principal submatrices $$
A[\alpha] = \begin{bmatrix}
-63
\end{bmatrix},
\begin{bmatrix}
71
\end{bmatrix},
\begin{bmatrix}
-7
\end{bmatrix}
$$ and $s_1 = (-63) + 71 + (-7) = 1$

##### Exercise 1(d)

計算 $A$ 的特徵多項式及 $\spec(A)$。

<!-- eng start -->
Find the characteristic polynomial of $A$ and $\spec(A)$.
<!-- eng end -->

##### Exercise 1(d) - answer
From the above exercises, we can find
$$
\begin{aligned}
p_A(x) &= s_0(-x)^3 + s_1(-x)^2 + s_2(-x) + s_3\\
&= 1(-x)^3 + 1(-x)^2 + (-9)(-x) + (-9)\\
&= -(x-1)(x-3)(x+3)
\end{aligned}
$$
When $x = 1, 3 ,-3$ , $p_A = 0$.
So $\spec(A) = \{1, 3, -3\}$

:::info
What do the experiments try to tell you? (open answer)

...
:::

:::info
The experiments try to tell us, we can find the characteristic polynomial easier by its principal submatrices.
:::

## Exercises

##### Exercise 2

利用 $s_k$ 計算以下矩陣 $A$ 的特徵多項式以及 $\spec(A)$。

<!-- eng start -->
Find the characteristic polynomial of $A$ and $\spec(A)$ through $s_k$.
<!-- eng end -->

##### Exercise 2(a)

$$
A = \begin{bmatrix}
5 & -1 \\
-1 & 5
\end{bmatrix}.
$$

##### Exercise 2(a) answer:
$s_0 = 1$\
$s_1 = \tr(A) = 5+5 = 10$\
$s_2 = \det(A) = (5 \times 5) - (-1 \times -1) = 24$

The Characteristic Polynomial of A:

$$
\begin{aligned}
P_A(x) &= s_0 (-x)^2 + s_1 (-x) + s_2 \\
&= x^2 - 10x +24 \\
&= (x-4)(x-6)
\end{aligned}
$$

$\spec(A) = \{4,6\}$

##### Exercise 2(b)

$$
A = \begin{bmatrix}
0 & 1 \\
-6 & 5
\end{bmatrix}.
$$

##### Exercise 2(b) - answer
$s_0 = 1$\
$s_1 = \tr(A) = 0 + 5 = 5$\
$s_2 = \det(A) = 0 \times 5 - (-6) \times 1 = 6$
$$
\begin{aligned}
P_A(x) &= s_0 (-x)^2 + s_1 (-x) + s_2 \\
&= x^2 - 5x +6 \\
&= (x-2)(x-3)
\end{aligned}
$$
$\spec(A) = \{2,3\}$

##### Exercise 2(c)

$$
A = \begin{bmatrix}
0 & 1 \\
-4 & 0
\end{bmatrix}.
$$

##### Exercise 2(c) - answer
$s_0 = 1$\
$s_1 = \tr(A) = 0 + 0 = 0$\
$s_2 = \det(A) = 0 \times 0 - (-4) \times 1 = 4$
$$
\begin{aligned}
P_A(x) &= s_0 (-x)^2 + s_1 (-x) + s_2 \\
&= x^2 - 0x +4 \\
&= (x-\sqrt{-4})^2
\end{aligned}
$$
$\spec(A) = \{\sqrt{-4}, \sqrt{-4}\}$

##### Exercise 2(d)

$$
A = \begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}.
$$

##### Exercise 2(d) - answer
$s_0 = 1$\
$s_1 = \tr(A) = 0 + 0 = 0$\
$s_2 = \det(A) = 0 \times 0 - (-1) \times (-1) = -1$
$$
\begin{aligned}
P_A(x) &= s_0 (-x)^2 + s_1 (-x) + s_2 \\
&= x^2 - 0x -1 \\
&= (x-1)(x+1)
\end{aligned}
$$
$\spec(A) = \{1, -1\}$

##### Exercise 3

利用 $s_k$ 計算以下矩陣 $A$ 的特徵多項式以及 $\spec(A)$。

<!-- eng start -->
Find the characteristic polynomial of $A$ and $\spec(A)$ through $s_k$.
<!-- eng end -->

##### Exercise 3(a)

$$
A = \begin{bmatrix}
4 & 0 & -1 \\
0 & 4 & -1 \\
-1 & -1 & 5
\end{bmatrix}.
$$

##### Exercise 3(a) - Ans by Kevin
$s_0 = 1$

$s_1 = 4+4+5 = 13$

$s_2 = 16+19+19 = 54$

$s_3 = det(A) = 4*(4*5-1)+(-1)(4) = 72$

We now then apply the calculated $s$ to the formula and get the characteristic polynomial:
$$
P_A(x) = -x^3 + 13x^2 - 54x +72
$$
Then by factorizing $P_A(x)$ and solving its roots, we get:
$$
P_A(x) = (x-3)(x-4)(x-6)
$$
$\spec(A)$ = $3$ or $4$ or $6$

##### Exercise 3(b)

$$
A = \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
6 & -11 & 6
\end{bmatrix}.
$$

##### Exercise 3(b) - answer
$s_0 = 1$
$s_1 = 0+ 0+ 6 = 6$
$s_2 = (0\times0) - (0\times1) + (0\times6) - (6\times0) + (0\times6) - ((-11)\times1) = 11$
$s_3 = (0\times0\times6) + (1\times1\times6) + (0\times(-11)\times0) - (6\times0\times0) - (0\times1\times6) - (0\times1\times(-11))=6$
$$
\begin{aligned}
P_A(x) &= s_0 (-x)^3 + s_1 (-x)^2 + s_2 (-x) + s_3 \\
&= -x^3 + 6x^2 -11x + 6 \\
&= -(x-1)(x-2)(x-3)
\end{aligned}
$$
$\spec(A) = \{1, 2, 3\}$

##### Exercise 3(c)

$$
A = \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{bmatrix}.
$$

##### Exercise 3(c) - answer
$s_0 = 1$
$s_1 = 0+ 0+ 0 = 0$
$s_2 = (0\times0) - (0\times1) + (0\times0) - (1\times0) + (0\times0) - (0\times1) = 0$
$s_3 = (0\times0\times0) + (1\times1\times1) + (0\times0\times0) - (1\times0\times0) - (0\times1\times0) - (0\times1\times0)=1$
$$
\begin{aligned}
P_A(x) &= s_0 (-x)^3 + s_1 (-x)^2 + s_2 (-x) + s_3 \\
&= -x^3 + 0x^2 -0x + 1 \\
&= -(x-1)(x-\cfrac{-1+\sqrt{-3}}{2})(x-\cfrac{-1-\sqrt{-3}}{2})
\end{aligned}
$$
$\spec(A) = \{1, \cfrac{-1+\sqrt{-3}}{2}, \cfrac{-1-\sqrt{-3}}{2}\}$

##### Exercise 4

令 $J_n$ 為 $n\times n$ 的全 $1$ 矩陣。
對每一個 $k = 0,\ldots, n$，計算 $s_k$，
並藉此求 $J_n$ 的特徵多項式及 $\spec(J_n)$。

<!-- eng start -->
Let $J_n$ be the $n\times n$ all-ones matrix. For each of $k = 0,\ldots, n$, find $s_k$. Then use them to find the characteristic polynomial of $J_n$ and $\spec(J_n)$.
<!-- eng end -->
##### Exercise 5

令 $J_{m,n}$ 為 $m\times n$ 的全 $1$ 矩陣，而

$$
A = \begin{bmatrix}
O_{n,n} & J_{n,m} \\
J_{m,n} & O_{m,m}
\end{bmatrix}.
$$

對每一個 $k = 0,\ldots, n$，計算 $s_k$，
並藉此求 $A$ 的特徵多項式及 $\spec(A)$。

<!-- eng start -->
Let $J_{m,n}$ be the $m\times n$ all-ones matrix and

$$
A = \begin{bmatrix}
O_{n,n} & J_{n,m} \\
J_{m,n} & O_{m,m}
\end{bmatrix}.
$$

For each of $k = 0,\ldots, n$, find $s_k$. Then use them to find the characteristic polynomial of $J_n$ and $\spec(J_n)$.
<!-- eng end -->

##### Exercise 6

令 $A$ 為一 $n\times n$ 矩陣。
若特徵多項式 $p_A(x) = s_0(-x)^n + s_1(-x)^{n-1} + \cdots + s_n$ 的根為 $\lambda_1,\ldots,\lambda_n$，則

$$
p_A(x) = (\lambda_1 - x) \cdots (\lambda_n - x).
$$

如此一來我們就有根與係數的關係

$$
\begin{aligned}
s_0 &= 1, \\
s_1 &= \lambda_1 + \cdots + \lambda_n, \\
s_2 &= \sum_{i<j}\lambda_i\lambda_j, \\
\vdots & \\
s_n &= \lambda_1\cdots\lambda_n.
\end{aligned}
$$

<!-- eng start -->
Let $A$ be an $n\times n$ matrix. Let $\lambda_1,\ldots,\lambda_n$ be the roots of the characteristic polynomial $p(A) = s_0(-x)^n + s_1(-x)^{n-1} + \cdots + s_n$. Then

$$
p_A(x) = (\lambda_1 - x) \cdots (\lambda_n - x).
$$

Thus, Vieta's formulas describe the relations between the coefficients of a polynomial and its roots by

$$
\begin{aligned}
s_0 &= 1, \\
s_1 &= \lambda_1 + \cdots + \lambda_n, \\
s_2 &= \sum_{i<j}\lambda_i\lambda_j, \\
\vdots & \\
s_n &= \lambda_1\cdots\lambda_n.
\end{aligned}
$$
<!-- eng end -->

##### Exercise 6(a)

令 $J_n$ 為 $n\times n$ 的全 $1$ 矩陣。
若已知 $\spec(J_n)$ 中有 $n-1$ 個零，
求最後一個特徵值。

（未來會發現這雖然是求 $\spec(J_n)$，
但也可以反推特徵多項式，
所以請不要用先前題目的結果來計算這題。）

<!-- eng start -->
Let $J_n$ be the $n\times n$ all-ones matrix. It is known that $\spec(J_n)$ contains $n-1$ copies of $0$. Find the last eigenvalues.

(In the future we will see that $\spec(J_n)$ can be used to find the characteristic polynomial, so please do not use the previous results on the characteristic polynomial to solve this problem.)
<!-- eng end -->

##### Exercise 6(b)

令 $J_{m,n}$ 為 $m\times n$ 的全 $1$ 矩陣，而

$$
A = \begin{bmatrix}
O_{n,n} & J_{n,m} \\
J_{m,n} & O_{m,m}
\end{bmatrix}.
$$

若已知 $\spec(A)$ 中有 $n-2$ 個零，
求最後兩個特徵值。

（未來會發現這雖然是求 $\spec(A)$，
但也可以反推特徵多項式，
所以請不要用先前提目的結果來計算這題。）

<!-- eng start -->
Let $J_{m,n}$ be the $m\times n$ all-ones matrix and

$$
A = \begin{bmatrix}
O_{n,n} & J_{n,m} \\
J_{m,n} & O_{m,m}
\end{bmatrix}.
$$

It is known that $\spec(J_n)$ contains $n-2$ copies of $0$. Find the last two eigenvalues.

(In the future we will see that $\spec(J_n)$ can be used to find the characteristic polynomial, so please do not use the previous results on the characteristic polynomial to solve this problem.)
<!-- eng end -->

##### Exercise 7

證明若 $A$ 和 $B$ 相似
（也就是存在可逆的 $Q$ 使得 $B = Q^{-1}AQ$），
則對每一個 $k = 0,\ldots, n$ 都有 $s_k(B) = s_k(A)$。

因此我們也可以把任一線性函數 $f:V\rightarrow V$ 的 $s_k(f)$ 定義成 $s_k([f]_\beta^\beta)$，
其中 $\beta$ 可以是 $V$ 的任意基底。

<!-- eng start -->
Show that if $A$ and $B$ are similar (meaning $B = Q^{-1}AQ$ for some invertible matrix $Q$), then $s_k(B) = s_k(A)$ for each $k = 0,\ldots, n$.

Therefore, for any linear function $f:V\rightarrow V$, we may define $s_k(f)$ as $s_k([f]_\beta^\beta)$, where $\beta$ could be any basis.
<!-- eng end -->

##### Exercise 8

令 $A$ 與 $B$ 分別為 $m\times n$ 與 $n\times n$ 矩陣，且 $m\leq n$。
若 $\alpha\subseteq [n]$ 是一些下標的集合，
定義 $A[:,\alpha]$ 是由 $A$ 中 $\alpha$ 裡的那些行所組成的 $m\times |\alpha|$ 矩陣，
而 $B[\alpha,:]$ 是由 $B$ 中 $\alpha$ 裡的那些列所組成的 $|\alpha|\times m$ 矩陣。

##### Theorem (Cauchy–Binet formula)

$$
\det(AB) = \sum_{\substack{\alpha\subseteq [n] \\ |\alpha| = m}} \det(A[:,\alpha])\det(B[\alpha,:]).
$$

<!-- eng start -->
Let $A$ and $B$ be $m\times n$ and $n\times n$ matrices with $m\leq n$. Let $\alpha\subseteq [n]$ be an index set. Define $A[:,\alpha]$ as the $m\times |\alpha|$ matrix obtained from $A$ by collecting those columns in $\alpha$. Similarly, define $B[\alpha,:]$ as the $|\alpha|\times m$ matrix obtained from $A$ by collecting those rows in $\alpha$.

##### Theorem (Cauchy–Binet formula)

$$
\det(AB) = \sum_{\substack{\alpha\subseteq [n] \\ |\alpha| = m}} \det(A[:,\alpha])\det(B[\alpha,:]).
$$
<!-- eng end -->

##### Exercise 8(a)

令

$$
A = \begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
\text{ and }
B = \begin{bmatrix}
7 & 8 \\
9 & 10 \\
11 & 12
\end{bmatrix}.
$$

對所有大小為 $2$ 的集合 $\alpha\subseteq [3]$，
求出所有的 $A[:,\alpha]$ 及 $B[\alpha,:]$，
並利用柯西比內公式計算 $AB$ 的行列式值。

<!-- eng start -->
Let

$$
A = \begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
\text{ and }
B = \begin{bmatrix}
7 & 8 \\
9 & 10 \\
11 & 12
\end{bmatrix}.
$$

For each $\alpha\subseteq [3]$ of size $2$, find $A[:,\alpha]$ and $B[\alpha,:]$. Then use the Cauchy–Binet formula to find the determinant of $AB$.
<!-- eng end -->

##### Exercise 8(b)

令

$$
M = \begin{bmatrix}
O_{n,n} & B \\
A & O_{m,m}
\end{bmatrix}.
$$

利用 506-6 求出 $p_M(x)$ 的 $(-x)^{n-m}$ 項係數。

<!-- eng start -->
Let

$$
M = \begin{bmatrix}
O_{n,n} & B \\
A & O_{m,m}
\end{bmatrix}.
$$

Use the results in 506-6 to find the coefficient of $(-x)^{n-m}$ in $p_M(x)$.
<!-- eng end -->

##### Exercise 8(b)

令

$$
M = \begin{bmatrix}
O_{n,n} & B \\
A & O_{m,m}
\end{bmatrix}.
$$

利用 $s_{2m}$ 的定義直接求出 $s_{2m}(M)$。
配合上一題證明柯西比內公式。

<!-- eng start -->
Let

$$
M = \begin{bmatrix}
O_{n,n} & B \\
A & O_{m,m}
\end{bmatrix}.
$$

Find $s_{2m}(M)$ by definition. Then use the previous problem to show the Cauchy–Binet formula.
<!-- eng end -->

##### Exercise 9

令 $A$ 為一 $n\times n$ 矩陣。
將 $p_A(x)$ 代入 $x = 0$ 可得

$$
s_n = p_A(0) = \det(A - 0I) = \det(A).
$$

類似地，我們可以利用 506-10 計算

$$
s_{n-1} = -\frac{dp_A(x)}{dx}\Big|_{x=0} = \sum_{i = 1}^n p_{A(i)}(x)\Big|_{x=0} = \sum_{i=1}^n\det(A(i)).
$$

利用數學歸納法證明

$$
\det(A - xI) = s_0(-x)^n + s_1(-x)^{n-1} + \cdots + s_n.
$$

<!-- eng start -->
Let $A$ be an $n\times n$ matrix. By substituting $x = 0$ in $p_A(x)$, we have

$$
s_n = p_A(0) = \det(A - 0I) = \det(A).
$$

Similarly, we may use the results in 506-10 to find

$$
s_{n-1} = -\frac{dp_A(x)}{dx}\Big|_{x=0} = \sum_{i = 1}^n p_{A(i)}(x)\Big|_{x=0} = \sum_{i=1}^n\det(A(i)).
$$

Prove that

$$
\det(A - xI) = s_0(-x)^n + S_1(-x)^{n-1} + \cdots + s_n
$$

by induction.
<!-- eng end -->

:::info
collaboration: 2
3 problems: 3

- 2abc

extra: 2

- 2d, 3abc

moderator: 1
qc: 1

:::