# 矩陣對角化

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

$\newcommand{\trans}{^\top}
\newcommand{\adj}{^{\rm adj}}
\newcommand{\cof}{^{\rm cof}}
\newcommand{\inp}[2]{\left\langle#1,#2\right\rangle}
\newcommand{\dunion}{\mathbin{\dot\cup}}
\newcommand{\bzero}{\mathbf{0}}
\newcommand{\bone}{\mathbf{1}}
\newcommand{\ba}{\mathbf{a}}
\newcommand{\bb}{\mathbf{b}}
\newcommand{\bc}{\mathbf{c}}
\newcommand{\bd}{\mathbf{d}}
\newcommand{\be}{\mathbf{e}}
\newcommand{\bh}{\mathbf{h}}
\newcommand{\bp}{\mathbf{p}}
\newcommand{\bq}{\mathbf{q}}
\newcommand{\br}{\mathbf{r}}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\by}{\mathbf{y}}
\newcommand{\bz}{\mathbf{z}}
\newcommand{\bu}{\mathbf{u}}
\newcommand{\bv}{\mathbf{v}}
\newcommand{\bw}{\mathbf{w}}
\newcommand{\tr}{\operatorname{tr}}
\newcommand{\nul}{\operatorname{null}}
\newcommand{\rank}{\operatorname{rank}}
%\newcommand{\ker}{\operatorname{ker}}
\newcommand{\range}{\operatorname{range}}
\newcommand{\Col}{\operatorname{Col}}
\newcommand{\Row}{\operatorname{Row}}
\newcommand{\spec}{\operatorname{spec}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\Vol}{\operatorname{Vol}}
\newcommand{\sgn}{\operatorname{sgn}}
\newcommand{\idmap}{\operatorname{id}}$

from lingeo import random_int_list, random_good_matrix

## Main idea

We know that an $n\times n$ matrix is diagonalizable if and only if there is a basis of $\mathbb{R}^n$ composed of eigenvectors.
We also know how to find the eigenvalues through the characteristic polynomial.
Indeed, that is all we need to perform the diagonalization of a matrix.

Let $A$ be an $n\times n$ matrix.
Here are the steps for its diagonalization:
1. Calculate the characteristic polynomial $p_A(x) = \det(A - xI)$ and solve the roots for $\spec(A)$.
2. For each $\lambda\in\spec(A)$, find a basis $\beta_\lambda$ for $\ker(A - \lambda I)$.
3. Let $\beta$ be the union of all such $\beta_\lambda$.
4. If $\beta$ is a basis of $\mathbb{R}^n$, then $A$ can be diagonalized through $\beta$.
That is, $[f_A]_\beta^\beta = D$ is a diagonal matrix.
Equivalently, $D = Q^{-1}AQ$, where $Q$ is the matrix whose columns are the vectors in $\beta$.

##### Remark

If $\beta = \{\bv_1, \ldots, \bv_n\}$ is a basis of $\mathbb{R}^n$ such $\bv_i$ is an eigenvector of $A$ with respect to $\lambda_i$ for all $i$.
Then the equalities $A\bv_i = \lambda_i\bv_i$ is equivalent to
$$
AQ =
A\begin{bmatrix}
| & ~ & | \\
{\bf v}_1 & \cdots & {\bf v}_n \\
| & ~ & |
\end{bmatrix}
=
\begin{bmatrix}
| & ~ & | \\
\lambda_1{\bf v}_1 & \cdots & \lambda_n{\bf v}_n \\
| & ~ & |
\end{bmatrix}
=
\begin{bmatrix}
| & ~ & | \\
{\bf v}_1 & \cdots & {\bf v}_n \\
| & ~ & |
\end{bmatrix}
\begin{bmatrix}
\lambda_1 & ~ & ~ \\
~ & \ddots & ~ \\
~ & ~ & \lambda_n
\end{bmatrix}
= QD.
$$

##### Remark

The above process might not able to finish for a few different reasons:

- The root of $p_A(x)$ might not be real. If we instite the basis should be using real vectors, then it cannot be done.
- It is possible that $\beta_\lambda$ does not provide enough independent eigenvectors, so $\beta$ is not a bsis of $\mathbb{R}^n$.

Here is an example of diagonalizing
$$
A = \begin{bmatrix}
2 & 3 \\
3 & 2
\end{bmatrix}.
$$

_Solve for the eigenvalues_:

The characteristic polynomial of $A$ is
$$
p_A(x) =
\det\begin{bmatrix}
2 - x & 3 \\
3 & 2 - x
\end{bmatrix}
= x^2 - 4x - 5,
$$
which as roots $-1, 5$.

_Solve for the eigenvectors_:

For $\lambda = 5$, calculate the basis of $\ker(A - 5I)$ and get
$$
\ker(A - 5I) = \ker \begin{bmatrix} -3 & 3 \\ 3 & -3 \end{bmatrix} = \vspan\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}.
$$

For $\lambda = -1$, calculate the basis of $\ker(A + I)$ and get
$$
\ker(A + I) = \ker \begin{bmatrix} 3 & 3 \\ 3 & 3 \end{bmatrix} = \vspan\left\{\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}.
$$

By setting
$$
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}
\text{ and }
Q = \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix},
$$
we get
$$
[f_A]_\beta^\beta = Q^{-1}AQ =
\begin{bmatrix}
5 & 0 \\
0 & -1
\end{bmatrix}.
$$

## Side stories

- Jordan block
- discrete Fourier transform

## Experiments

##### Exercise 1

執行以下程式碼。

### 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)
pA = (-1)^n * A.charpoly()
print("characteristic polyomial =", pA)
print("                         =", factor(pA))

if print_ans:
    print("eigenvalues are:" + ", ".join("%s"%val for val in spec))
    print("corresponding eigenvectors are:")
    for i in range(n):
        pretty_print(LatexExpr(r"\lambda ="), spec[i], ", eigenvector =", Q[:,i])
    pretty_print(LatexExpr("Q ="), Q)

藉由 `seed = 2`得到
$$A=\begin{bmatrix}
-1 & 2 & -3\\
6 & 0 & 6\\
4 & -2 & 6
\end{bmatrix}.
$$
##### Exercise 1(a)

求出 $A$ 的所有特徵值。
答:
特徵多項式為
$$
p_A(x)=\det (A-xI)= \det \begin{bmatrix}
-1-x & 2 & -3\\
6 & 0-x & 6\\
4 & -2 & 6-x
\end{bmatrix}=6x^2-18x=-x^3+5x^2-6x=-x(x-3)(x-2).
$$

特徵值為 $0$、$2$、$3$。

##### Exercise 1(b)

對每個特徵值，求出對應的特徵向量。

:::warning

- [x] 沒有寫清楚答案，比如：所以對應的特徵向量分別為 ...
:::

答:

$$\lambda_1=2，
\ker (A-2I)= \ker \begin{bmatrix}
-3 & 2 & -3\\
6 & -2 & 6\\
4 & -2 & 4
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}
\right\}，\bv_1=(1,0,-1).
$$
$$\lambda_2=0，
\ker (A-0I)= \ker \begin{bmatrix}
-1 & 2 & -3\\
6 & 0 & 6\\
4 & -2 & 6
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}
\right\}，\bv_2=(-1,1,1).
$$
$$\lambda_3=3，
\ker (A-3I)= \ker \begin{bmatrix}
-4 & 2 & -3\\
6 & -3 & 6\\
4 & -2 & 3
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}
\right\}，\bv_3=(1,2,0).
$$

##### Exercise 1(c)

求出 $Q$ 使得 $D = Q^{-1}AQ$ 是一個對角矩陣。

:::warning

- [x] 要在上一題說明什麼是 $\bv_1,\bv_2,\bv_3$
- [x] 標點
:::

答:
得到
$$
Q= \begin{bmatrix}
| & | & | \\
\bv_1& \bv_2 & \bv_3 \\
| & | & | \\
\end{bmatrix}=
\begin{bmatrix}
1 & -1 & 1\\
0 & 1 & 2\\
-1 & 1 & 0
\end{bmatrix}，
$$
$$
D= \begin{bmatrix}
\lambda_1& 0 & 0 \\
0& \lambda_2& 0 \\
0 & 0 & \lambda_3 \\
\end{bmatrix}=\begin{bmatrix}
2 & 0 & 0 \\
0& 0 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}.
$$
使得 $D = Q^{-1}AQ$。

## Exercises

##### Exercise 2

將以下矩陣 $A$ 對角化。
（求出可逆矩陣 $Q$ 和對角矩陣 $D$ 使得 $D = Q^{-1}AQ$。）

##### Exercise 2(a)

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

:::warning

- [x] 特徵值為 2、3。 <-- 把 $2$ 和 $3$ 丟進數學模式
- [x] 跟第一題一樣，要說明 $\bv_1, \bv_2$ 是什麼
:::

答:
特徵多項式為
$$
p_A(x)=\det (A-xI)= \det \begin{bmatrix}
0-x & 1\\
-6 & 5-x\\
\end{bmatrix}=x^2-5x+6=(x-3)(x-2).
$$
特徵值為 $2$、$3$。
$$\lambda_1=2，
\ker (A-2I)= \ker \begin{bmatrix}
-2 & 1\\
-6 & 3\\
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} 1 \\ 2 \end{bmatrix}
\right\}，\bv_1=(1,2)，
$$
$$\lambda_2=3，
\ker (A-3I)= \ker \begin{bmatrix}
-3 & 1\\
-6 & 2\\
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} 1 \\ 3 \end{bmatrix}
\right\}，\bv_2=(1,3).
$$
得到
$$
Q= \begin{bmatrix}
| & | \\
\bv_1& \bv_2 \\
| & | \\
\end{bmatrix}=
\begin{bmatrix}
1 & 1\\
2 & 3\\
\end{bmatrix}、
$$
$$
D= \begin{bmatrix}
\lambda_1& 0\\
0& \lambda_2 \\
\end{bmatrix}=\begin{bmatrix}
2 & 0 \\
0 & 3 \\
\end{bmatrix}，
$$
使得 $D = Q^{-1}AQ$。

##### Exercise 2(b)

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

:::warning

- [x] 同上題
- [x] 呃... 特徵值有錯喔，所以特徵向量也有算錯
:::

答:
特徵多項式為
$$
p_A(x)=\det (A-xI)= \det \begin{bmatrix}
0-x & -1\\
1 & 0-x\\
\end{bmatrix}=x^2+1.
$$
特徵值為 $i,-i$。
$$
\lambda_1=i,\ \ker (A-iI)= \ker \begin{bmatrix}
-i & -1\\
1 & -i\\
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} i \\ 1 \end{bmatrix}
\right\}，\bv_1=(i,1).
$$
$$
\lambda_2=-i,\ \ker (A+iI)= \ker \begin{bmatrix}
i & -1\\
1 & i\\
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} -i \\ 1 \end{bmatrix}
\right\}，\bv_2=(-i,1).
$$
得到
$$
Q= \begin{bmatrix}
| & | \\
\bv_1& \bv_2 \\
| & | \\
\end{bmatrix}=
\begin{bmatrix}
i & -i\\
1 & 1\\
\end{bmatrix}、
$$
$$
D= \begin{bmatrix}
\lambda_1& 0\\
0& \lambda_2 \\
\end{bmatrix}=\begin{bmatrix}
i & 0 \\
0 & -i \\
\end{bmatrix}，
$$
使得 $D = Q^{-1}AQ$。

##### Exercise 3

將以下矩陣 $A$ 對角化。
（求出可逆矩陣 $Q$ 和對角矩陣 $D$ 使得 $D = Q^{-1}AQ$。）

##### Exercise 3(a)

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

:::warning

- [x] $P_A$ --> $p_A$；後面幾題也改一下
:::

sol:

The characteristic polynomial of $A$ is
$$\begin{split}
p_A(x)
& = \det(A - xI) \\
& = \det \left[\begin{array}{ccc}
4 - x & 0 & -1 \\
0 & 4 - x & -1 \\
-1 & -1 & 5 - x
\end{array}\right] \\
& = -x^3 + 13 x^2 - 54x + 72 \\
& = -(x - 3)(x - 4)(x - 6).
\end{split}$$

If characteristic polynomial $p_A(x) = 0$, then $-(x - 3)(x - 4)(x - 6) = 0$. We can get eigenvalues $x = 3, 4, 6$.

For $x = 3$, calculate the basis for $\ker(A - 3I)$ and get
$$\begin{split}
\ker(A - 3I)
= \ker \left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1 \\
-1 & -1 & 2
\end{array}\right]
= \operatorname{span} \left\lbrace \left[\begin{array}{c}
1 \\
1 \\
1
\end{array}\right] \right\rbrace.
\end{split}$$

For $x = 4$, calculate the basis for $\ker(A - 4I)$ and get
$$\begin{split}
\ker(A - 4I)
= \ker \left[\begin{array}{ccc}
0 & 0 & -1 \\
0 & 0 & -1 \\
-1 & -1 & 1
\end{array}\right]
= \operatorname{span} \left\lbrace \left[\begin{array}{c}
1 \\
-1 \\
0
\end{array}\right] \right\rbrace.
\end{split}$$

For $x = 6$, calculate the basis for $\ker(A - 6I)$ and get
$$\begin{split}
\ker(A - 6I)
= \ker \left[\begin{array}{ccc}
-2 & 0 & -1 \\
0 & -2 & -1 \\
-1 & -1 & -1
\end{array}\right]
= \operatorname{span} \left\lbrace \left[\begin{array}{c}
1 \\
1 \\
-2
\end{array}\right] \right\rbrace.
\end{split}$$
By setting
$$\beta = \left\lbrace
\left[\begin{array}{c}
1 \\
1 \\
1
\end{array}\right],
\left[\begin{array}{c}
1 \\
-1 \\
0
\end{array}\right],
\left[\begin{array}{c}
1 \\
1 \\
-2
\end{array}\right]
\right\rbrace
\text{, and }
Q = \left[\begin{array}{ccc}
1 & 1 & 1\\
1 & -1 & 1 \\
1 & 0 & -2
\end{array}\right]$$
such that
$$[f_A]^\beta_\beta
= Q^{-1}AQ
= \left[\begin{array}{ccc}
3 & 0 & 0\\
0 & 4 & 0 \\
0 & 0 & 6
\end{array}\right].
$$

:::info
寫得很完整 :smiley: :dog: :rabbit: :cat: :rage:
:::

##### Exercise 3(b)

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

sol:

The characteristic polynomial of $A$ is
$$\begin{split}
p_A(x)
& = \det(A - xI) \\
& = \det \left[\begin{array}{ccc}
- x & 1 & 0 \\
0 & - x & 1 \\
6 & -11 & 6 - x
\end{array}\right] \\
& = -x^3 + 6x^2 - 11x + 6 \\
& = -(x - 1)(x - 2)(x - 3).
\end{split}$$

If characteristic polynomial $p_A(x) = 0$, then $-(x - 1)(x - 2)(x - 3) = 0$. We can get eigenvalues $x = 1, 2, 3$.

For $x = 1$, calculate the basis for $\ker(A - I)$ and get
$$\begin{split}
\ker(A - I)
= \ker \left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
6 & -11 & 5
\end{array}\right]
= \operatorname{span} \left\lbrace \left[\begin{array}{c}
1 \\
1 \\
1
\end{array}\right] \right\rbrace.
\end{split}$$

For $x = 2$, calculate the basis for $\ker(A - 2I)$ and get
$$\begin{split}
\ker(A - 2I)
= \ker \left[\begin{array}{ccc}
-2 & 1 & 0 \\
0 & -2 & 1 \\
6 & -11 & 4
\end{array}\right]
= \operatorname{span} \left\lbrace \left[\begin{array}{c}
1 \\
2 \\
4
\end{array}\right] \right\rbrace.
\end{split}$$

For $x = 3$, calculate the basis for $\ker(A - 3I)$ and get
$$\begin{split}
\ker(A - 3I)
= \ker \left[\begin{array}{ccc}
-3 & 1 & 0 \\
0 & -3 & 1 \\
6 & -11 & 3
\end{array}\right]
= \operatorname{span} \left\lbrace \left[\begin{array}{c}
1 \\
3 \\
9
\end{array}\right] \right\rbrace.
\end{split}$$

By setting
$$\beta = \left\lbrace
\left[\begin{array}{c}
1 \\
1 \\
1
\end{array}\right],
\left[\begin{array}{c}
1 \\
2 \\
4
\end{array}\right],
\left[\begin{array}{c}
1 \\
3 \\
9
\end{array}\right]
\right\rbrace
\text{, and }
Q = \left[\begin{array}{ccc}
1 & 1 & 1\\
1 & 2 & 3\\
1 & 4 & 9
\end{array}\right]$$
such that
$$[f_A]^\beta_\beta
= Q^{-1}AQ
= \left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 2 & 0 \\
0 & 0 & 3
\end{array}\right].
$$

##### Exercise 3(c)

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

**[以下答案說明 $A$ 在實數中不可對角化]**
The characteristic polynomial of $A$ is
$$\begin{split}
p_A(x)
& = \det(A - xI) \\
& = \det \left[\begin{array}{ccc}
- x & 1 & 0 \\
0 & - x & 1 \\
1 & 0 & - x
\end{array}\right] \\
& = -x^3 + 1 \\
& = -(x - 1)(x^2 + x + 1).
\end{split}
$$

Since $(x^2 + x + 1)$ can not be spilt in $\mathbb{R}$, thus $A$ can't diagonalizable.

:::warning

- [ ] 可以用 $\omega = e^{\frac{2\pi}{3}i}$ 改寫，讓答案變乾淨一些
:::

討論 $A$ 在複數空間中的特徵值為 $(1, \frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2})$。
$$
\lambda_1=1,\ker(A-1I)=
\ker\begin{bmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
1 & 0 & -1 \\
\end{bmatrix}=\vspan\left\{\begin{bmatrix} 1\\1\\1\end{bmatrix}\right\}
$$
$$
\lambda_2=\frac{-1+\sqrt{3}i}{2},\ker(A-\frac{-1+\sqrt{3}i}{2}I)=
\ker\begin{bmatrix}
\frac{1-\sqrt{3}i}{2} & 1 & 0 \\
0 & \frac{1-\sqrt{3}i}{2} & 1 \\
1 & 0 & \frac{1-\sqrt{3}i}{2} \\
\end{bmatrix}
$$
$$
\lambda_3=\frac{-1-\sqrt{3}i}{2},\ker(A-\frac{-1-\sqrt{3}i}{2}I)=
\ker\begin{bmatrix}
\frac{1+\sqrt{3}i}{2} & 1 & 0 \\
0 & \frac{1+\sqrt{3}i}{2} & 1 \\
1 & 0 & \frac{1+\sqrt{3}i}{2} \\
\end{bmatrix}
$$

:::info
答案可接受，不用更動
但沒特別說明的情況下應該對複數對角化，像 2(b)
可以在後面再加一個 $A$ 在複數中可對角化的答案
:::

##### Exercise 4

將以下矩陣 $A$ 對角化，
並說明 $f_A$ 的作用。

##### Exercise 4(a)

$$
A = \begin{bmatrix}
\frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\
-\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}
\end{bmatrix}.
$$

:::warning

- [x] 雖然根沒有解錯，但 $p_A(x)$ 應該有少一些係數
- [x] $\ker(...)$ 應該是等於 $\vspan\left\{...\right\}$, 不會等於一兩個向量
- [x] $\{ \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},\
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\}$ --> $\left\{ \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},\
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\right\}$

:::

**Ans:**

先求 $A$ 的特徵多項式，依定義 $p_A(x) = \det (A-xI)$ 。
即
$$
\frac{1}{27}\det \begin{bmatrix}
2-3x & -1 & -1 \\
-1 & 2-3x & -1 \\
-1 & -1 & 2-3x
\end{bmatrix}.
$$
經計算後得 $p_A(x) = -x{(x-1)}^2$ ，根為 $0,1,1$ 。

當 $x=0$ ，
$$
\ker(A-xI) = \vspan\left\{\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}\right\}.
$$

當 $x=1$ ，
$$
\ker(A-xI) = \vspan\left\{ \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},\
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\right\}.
$$

則可得
$$
Q = \begin{bmatrix}
1 & -1 & -1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}.
$$
且將 $Q$ 做到 $rref$ 後得到 $I$ ，故 $Q^{-1}$ 存在。

依照定義，將 $A$ 對角化後得到的矩陣 $D = Q^{-1}AQ$ 。
故
$$
D = \begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}.
$$
可知 $f_A$ 是一個將 $3$ 維向量投影到 $2$ 維平面上的函數。

##### Exercise 4(b)

$$
A = \begin{bmatrix}
\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} \\
-\frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\
-\frac{2}{3} & -\frac{2}{3} & \frac{1}{3}
\end{bmatrix}.
$$

:::warning

- [x] 跟上一題類似
:::

**Ans:**

先求 $A$ 的特徵多項式，依定義 $p_A(x) = \det (A-xI)$ 。
即
$$
\frac{1}{27}\det \begin{bmatrix}
1-3x & -2 & -2 \\
-2 & 1-3x & -2 \\
-2 & 2 & 1-3x
\end{bmatrix}.
$$

經計算後得 $p_A(x) = -(x+1){(x-1)}^2$ ，根為 $-1,1,1$ 。

當 $x=-1$ ，
$$
\ker(A-xI) = \vspan\left\{\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}\right\}.
$$

當 $x=1$ ，
$$
\ker(A-xI) =\vspan\left\{ \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},\
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\right\}.
$$

則可得
$$
Q = \begin{bmatrix}
1 & -1 & -1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}.
$$
且將 $Q$ 做到 $rref$ 後得到 $I$ ，故 $Q^{-1}$ 存在。

依照定義，將 $A$ 對角化後得到的矩陣 $D = Q^{-1}AQ$ 。
故
$$
D = \begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}.
$$
可知 $f_A$ 是一個鏡射的函數。

##### Exercise 5

令
$$
A = \begin{bmatrix}
3 & 1 & 0 \\
0 & 3 & 1 \\
0 & 0 & 3
\end{bmatrix}.
$$
說明 $A$ 矩陣只有一個特徵值 $3$（重根三次），
但 $\ker(A - 3I)$ 的維度只有 $1$，
找不到足夠的向量將 $A$ 對角化。

答:
特徵多項式為
$$
P_A(x)=\det (A-xI)= \det \begin{bmatrix}
3-x & 1 & 0 \\
0 & 3-x & 1 \\
0 & 0 & 3-x
\end{bmatrix}=-x^3+9x^2-27x+27=-(x-3)^3.
$$
得到特徵值
$$
\lambda_1=\lambda_2=\lambda_3=3.
$$
使得
$$\lambda=3，
\ker (A-3I)= \ker \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}= \operatorname{span}\left\{
\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}
\right\}.
$$
得到 $\lambda=3$ 時，只有一個特徵向量 $\bv=(1,0,0)$，使得 $\ker(A - 3I)$ 的維度只有 $1$，所以找不到足夠的向量將 $A$ 對角化。
##### Exercise 6

令
$$
A = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
並令 $\zeta = e^{\frac{2\pi}{5}i}$ 為 $1$ 的五次方根。

這個練習告訴我們，如果運氣很好有找到所有的特徵向量，則不見得要解特徵多項式也可以找得到 $\spec(A)$。

##### Exercise 6(a)

令
$$
Q = \begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^6 & \zeta^8 \\
1 & \zeta^3 & \zeta^6 & \zeta^9 & \zeta^{12} \\
1 & \zeta^4 & \zeta^8 & \zeta^{12} & \zeta^{16} \\
\end{bmatrix}.
$$
驗證 $Q^*Q = 5I_5$，
因此 $Q$ 可逆且其行向量集合是一組 $\mathbb{C}^5$ 的基底。

（這裡 $Q^*$ 的意思是將 $Q$ 轉置後再逐項取共軛。）

答:
因為尤拉公式(Euler's formula)，對於任意實數 $x$，$$e^{ix}=\cos(x)+i\sin(x)$$ 恆成立。
所以可以得到 $\zeta = e^{\frac{2\pi}{5}i} = \cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5}).$
可以推出 $\zeta^n = e^{\frac{2n\pi}{5}i} = \cos(\frac{2n\pi}{5})+i\sin(\frac{2n\pi}{5})$，且 $\zeta^1$ 和 $\zeta^4$ 共軛
，$\zeta^2$ 和 $\zeta^3$ 共軛。
計算 $Q$ 和 $Q^*:$
$$Q = \begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^6 & \zeta^8 \\
1 & \zeta^3 & \zeta^6 & \zeta^9 & \zeta^{12} \\
1 & \zeta^4 & \zeta^8 & \zeta^{12} & \zeta^{16} \\
\end{bmatrix}=
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
\end{bmatrix}.
$$
$$
Q^* = \begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
\end{bmatrix}.
$$
計算 $Q^*Q:$
運用 $x^5 = 1$，計算
$$x^5-1=
(x-1)(x^4+x^3+x^2+x+1)=0.$$
可以得到 $\zeta^4+\zeta^3+\zeta^2+\zeta^1+1=0.$
因此
$$Q^*Q =
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
\end{bmatrix}=
\begin{bmatrix}
5 & 0 & 0 & 0 & 0 \\
0 & 5 & 0 & 0 & 0 \\
0 & 0 & 5 & 0 & 0 \\
0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & 5 \\
\end{bmatrix}.
$$
因此 $Q$ 存在反矩陣，$Q$ 可逆且其行向量集合是一組 $\mathbb{C}^5$ 的基底。

##### Exercise 6(b)

說明 $Q$ 的每個行向量都是 $A$ 的特徵向量，並找出對應的特徵值。
利用這點將 $A$ 對角化。

:::warning

- [x] 向量用粗體
:::

答:
計算 $AQ=QD:$
$$\begin{aligned}AQ&=
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^6 & \zeta^8 \\
1 & \zeta^3 & \zeta^6 & \zeta^9 & \zeta^{12} \\
1 & \zeta^4 & \zeta^8 & \zeta^{12} & \zeta^{16} \\
\end{bmatrix}= \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix}=
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & \zeta^1 & \zeta^2 & \zeta^3 & \zeta^4 \\
1 & \zeta^2 & \zeta^4 & \zeta^1 & \zeta^3 \\
1 & \zeta^3 & \zeta^1 & \zeta^4 & \zeta^2 \\
1 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta^1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & \zeta^1 & 0 & 0 & 0 \\
0 & 0 & \zeta^2 & 0 & 0 \\
0 & 0 & 0 & \zeta^3 & 0 \\
0 & 0 & 0 & 0 & \zeta^4
\end{bmatrix}=QD.
\end{aligned}$$
令 $Q$ 的各行向量為 $\{\bv_1,\cdots,\bv_5\}$，則可找出特徵值依序為 $\{1,\zeta^1, \ldots,\zeta^4\}$ 與各行向量對應，組成對角矩陣 $D$，使得 $A$ 對角化為 $D$。

:::success
寫得很棒 :thumbsup: :u5272:
:::

:::info
目前分數 = 6.5 &times; 檢討 = 6.5
:::