對一個特徵向量化簡

# 對一個特徵向量化簡 Reduction ![Creative Commons License](https://i.creativecommons.org/l/by/4.0/88x31.png) This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/). $\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$ ```python from lingeo import random_int_vec ``` ## Main idea We start with some basic ideas on complex matrices. Let $A$ be a complex matrix. Then the **conjugate transpose** of $A$ is the matirx obtained from $A\trans$ by taking conjugate entrywisely. Recall that if $\bx$ and $\by$ are complex column vectors, then their inner product is $\inp{\bx}{\by} = \by^* \bx$. If a complex matrix $A$ satisfies $A^* A = AA^* = I$, then it is called a **unitary** matrix. In comparison, if a real matrix satisfies $A\trans A = AA\trans = I$, then it is an orthogonal matrix. Let $A$ be an $n\times n$ complex matrix. Then the following are equivalent: - $A$ is a unitary matrix. - $A^{-1} = A^*$. - The columns of $A$ form an orthonormal basis of $\mathbb{C}^n$. - The rows of $A$ form an orthonormal basis of $\mathbb{C}^n$. Let $\bv\in\mathbb{C}^n$ be a nonzero vector. Then one may expand $\bv$ into a basis $\beta$ of $\mathbb{C}^n$ whose first vector is $\bv$. Let $Q$ be the matrix whose columns are the vectors in $\beta$. Then $Q$ is an invertible matrix whose first column is $\bv$. If necessary, one may apply the Gram–Schimdt process to obtain an orthonormal basis of $\mathbb{C}^n$ whose first vector is $\frac{\bv}{\|\bv\|}$. Thus, there is a unitary matrix $Q$ whose first column is $\frac{\bv}{\|\bv\|}$. ##### Reduction lemma Let $A$ be a complex matrix. Suppose $\bv$ is an eigenvector of $A$ with respect to the eigenvalue $\lambda$. Let $Q$ be an invertible matrix whose first column is $\bv$. Then $Q^{-1}AQ$ has the form $$ \begin{bmatrix} \lambda & * \\ \bzero & A_2 \end{bmatrix}. $$ Moreover, $Q$ can be chosen as a unitary matrix whose first column is $\frac{\bv}{\|\bv\|}$. ##### Remark Note that the eigenvalues of a real matrix are not necessarily all real. Suppose $A$ is a real matrix and $\lambda$ is a real eigenvalue of $A$. Then the eigenvector $\bv\in\ker(A - \lambda I)$ can be chosen to be real. Also, the $Q$ matrix in the reduction lemma can be chosen to be orthogonal. However, $A_2$ can still possibly have a non-real eigenvalue. ## Side stories - all-ones vector - cases of real matrices - properties of unitary/orthogonal matrices - discrete Fourier transform matrix ## Experiments ##### Exercise 1 執行以下程式碼。令 $\beta = \{\bu_1,\ldots,\bu_n\}$ 為 $Q$ 的行向量集合。  Run the code below. Let $\beta = \{\bu_1,\ldots,\bu_n\}$ be the columns of $Q$.  ```python ### code set_random_seed(0) print_ans = False n = 4 Q = identity_matrix(n) Q[1:,0] = random_int_vec(n-1, 3) D = matrix(n, random_int_vec(n**2,3)) D[1:,0] = vector([0] * (n-1)) A = Q * D * Q.inverse() print("n =", n) pretty_print(LatexExpr("A ="), A) pretty_print(LatexExpr("Q ="), Q) if print_ans: print("The representation of f_A(u1) with respect to beta is") pretty_print(D[:,0]) pretty_print(LatexExpr("Q^{-1} ="), Q.inverse()) pretty_print(LatexExpr("Q^{-1} A Q ="), Q.inverse() * A * Q) ``` When `seed = 0`, we can get matrices $$ A = \begin{bmatrix} 3 & -3 & -3 & -1\\ -1 & 10 & 7 & 4\\ 23 & -8 & -12 & -5\\ 16 & -2 & -6 & -2\\ \end{bmatrix}, Q = \begin{bmatrix} 1 & 0 & 0 & 0\\ -3 & 1 & 0 & 0\\ 3 & 0 & 1 & 0\\ 1 & 0 & 0 & 1\\ \end{bmatrix}. $$ ##### Exercise 1(a) 求 $[f_A(\bu_1)]_\beta$。  Find $[f_A(\bu_1)]_\beta$.  Ans: $f_A(\bu_1)=A\cdot\bu_1= \begin{bmatrix} 3 & -3 & -3 & -1\\ -1 & 10 & 7 & 4\\ 23 & -8 & -12 & -5\\ 16 & -2 & -6 & -2\\ \end{bmatrix}\cdot\begin{bmatrix} 1\\ -3\\ 3\\ 1\\ \end{bmatrix}$ $=\begin{bmatrix} 2\\ -6\\ 6\\ 2\\ \end{bmatrix}$. And we can know that $[f_A(\bu_1)]_\beta$ is equal to $2\cdot\bu_1+0\cdot\bu_2+0\cdot\bu_3+0\cdot\bu_4$, so $[f_A(\bu_1)]_\beta = \begin{bmatrix} 2\\ 0\\ 0\\ 0\\ \end{bmatrix}$. --- ##### Exercise 1(b) 求 $Q^{-1}$。  Find $Q^{-1}$.  Ans: By calculating the reduced echelon form of$$ \left[\begin{array}{rrrr|rrrrr} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ -3 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{array}\right]$$ we can get $Q^{-1}$= $\begin{bmatrix} 1 & 0 & 0 & 0\\ 3 & 1 & 0 & 0\\ -3 & 0 & 1 & 0\\ -1 & 0 & 0 & 1\\ \end{bmatrix}$ . --- ##### Exercise 1(c) 求 $[f_A]_\beta^\beta$.  Find $[f_A]_\beta^\beta$.  Ans: We can get $Q$ and $Q^{-1}$ respectively from the code and Exercise 1(b) above. $[f_A]_\beta^\beta = Q^{-1}AQ=$ $\begin{bmatrix} 1 & 0 & 0 & 0\\ 3 & 1 & 0 & 0\\ -3 & 0 & 1 & 0\\ -1 & 0 & 0 & 1\\ \end{bmatrix}\begin{bmatrix} 3 & -3 & -3 & -1\\ -1 & 10 & 7 & 4\\ 23 & -8 & -12 & -5\\ 16 & -2 & -6 & -2\\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0\\ -3 & 1 & 0 & 0\\ 3 & 0 & 1 & 0\\ 1 & 0 & 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 2 & -3 & -3 & -1\\ 0 & 1 & -2 & 1\\ 0 & 1 & -3 & -2\\ 0 & 1 & -3 & -1\\ \end{bmatrix}.$ :::info What do the experiments try to tell you? (open answer) ... ::: --- ## Exercises ##### Exercise 2 令 $\bone$ 為全一向量。（其長度將由文意決定。）已知以下矩陣 $A$ 皆有 $\bone$ 這個特徵向量。求出 $A$ 的所有特徵值。  Let $\bone$ be the all-ones vector (whose dimension will be clear by the context). It is known that each of the following matrices has $\bone$ as an eigenvector. Find all eigenvalues of $A$.  ##### Exercise 2(a) $$ A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}. $$ Ans: We can get the eigenvalues of $A$ by solving the characteristic equation: $det(A - λI) = 0$. $$ A-λI = \begin{bmatrix} -λ & 1 & 1 \\ 1 & -λ & 1 \\ 1 & 1 & -λ \end{bmatrix}, $$ so the characteristic polynomial of $A$ is ${-λ^3 + 3λ^2 + 2 =(λ+1)^2(-λ+2) = 0}$, and the eigenvalues are $\{ -1,-1,2\}$. --- ##### Exercise 2(b) $$ A = \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{bmatrix}. $$ Ans: Let $\bv_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{bmatrix}$, Expand $\bv_1$ into the basis of $\mathbb{R}^{4}$ $\beta = \{\bv_1,\bv_2,\bv_3,\bv_4\}$, and let it as the basis of $Q$ row vector, we can get $$ Q = \begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} , $$ so $$ [f_A]_\beta^\beta = Q^{-1}AQ = \begin{bmatrix} 1 & 0 & 0 & 0\\ -1 & 1 & 0 & 0\\ -1 & 0 & 1 & 0\\ -1 & 0 & 0 & 1\\ \end{bmatrix}\begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 3 & 1 & 1 & 1\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{bmatrix}. $$ We can know that there is an eigenvalue from the above formula $\lambda_1 = 3$. And let $A_2 = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\ 0 & 0 & -1 \end{bmatrix}$, so $\spec(A_2) = \{-1,-1,-1\}$. We can know $\spec(A) =\{3\}\cup\{-1,-1,-1\} = \{-1,-1,-1,3\}$. --- ##### Exercise 2(c) $$ A = \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix}. $$ Ans: let $\bv_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, Expand $\bv_1$ into $\mathbb{R}^{3}$ basis $\beta = \{\bv_1,\bv_2,\bv_3\}$, and let it as a row vector of $Q$, we can get $$ Q = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}, $$ so $[f_{A}]_\beta^\beta = Q^{-1}AQ = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & -1 & 0\\ 0 & 3 & -1 \\ 0 & 0 & 1 \end{bmatrix}.$ We can know that there is an eigenvalue from the above formula $\lambda = 0$. Let matrix $A_2 = \begin{bmatrix} 3 & -1 \\ 0 & 1 \\ \end{bmatrix}$，so $\spec(A_2) = \{1,3\}$. We can know $\spec(A) =\{0\}\cup\{1,3\} = \{0,1,3\}$. --- ##### Exercise 2(d) $$ A = \begin{bmatrix} 0.2 & 0.8 & 0 \\ 0.4 & 0.2 & 0.4 \\ 0 & 0.8 & 0.2 \end{bmatrix}. $$ Ans: Let $\bv_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, Expand $\bv_1$ to the basis of $\mathbb{R}^{3}$ $\beta = \{\bv_1,\bv_2,\bv_3\}$ , and let it as a row vector of $Q$, we can get $$Q = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} , $$ so $[f_{A}]_\beta^\beta = Q^{-1}AQ = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} 0.2 & 0.8 & 0 \\ 0.4 & 0.2 & 0.4 \\ 0 & 0.8 & 0.2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0.8 & 0\\ 0 & -0.6 & 0.4 \\ 0 & 0 & 0.2 \end{bmatrix}.$ We can know that there is an eigenvalue from the above formula $\lambda = 1$. Let matrix $A_2 = \begin{bmatrix} -0.6 & 0.4 \\ 0 & 0.2 \\ \end{bmatrix}$，so $\spec(A_2) = \{-0.6,0.2\}$. We can know $\spec(A) =\{1\}\cup\{-0.6,0.2\} = \{1,-0.6,0.2\}$. --- ##### Exercise 3 令 $$ A = \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 2 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 2 \\ \end{bmatrix}. $$ 已知 $\bone$ 為 $A$ 的一特徵值。求 $A$ 的所有特徵值。提示：將 $A$ 對 $\bone$ 化簡後，再對 $A_2$ 化簡一次。  Let $$ A = \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 2 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 2 \\ \end{bmatrix}. $$ It is known that $\bone$ is an eigenvector of $A$. Find all eigenvalues of $A$. Hint: Apply the reduction lemma to $A$ and $\bone$ to get $A_2$. Then apply the lemma again to $A_2$.  $Ans:$ Let $\bv_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{bmatrix}.$ $\beta = \{\bv_1,\bv_2,\bv_3,\bv_4\}$, $$ Q = \begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} $$ Then $[f_A]_\beta^\beta = Q^{-1}AQ = \begin{bmatrix} 1 & 0 & 0 & 0\\ -1 & 1 & 0 & 0\\ -1 & 0 & 1 & 0\\ -1 & 0 & 0 & 1\\ \end{bmatrix}\begin{bmatrix} 0 & 1 & 1 & 1\\ 1 & 2 & 0 & 0\\ 1 & 0 & 2 & 0\\ 1 & 0 & 0 & 2\\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 3 & 1 & 1 & 1\\ 0 & 1 & -1 & -1\\ 0 & -1 & 1 & -1\\ 0 & -1 & -1 & 1\\ \end{bmatrix}.$ Since we can get an eigenvalue $\lambda_1 = 3$ 。 Let $A_2 = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 1 & -1\\ -1 & -1 & 1 \\ \end{bmatrix}$, $\bv_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}。$ $\beta = \{\bv_1,\bv_2,\bv_3\}$, $$ Q_2 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} $$ Then $[f_{A_2}]_\beta^\beta = Q_2^{-1}A_2Q_2 = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} -1 & -1 & -1\\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}.$ We can get $\lambda_2 = -1$ 。 Let matrix $A_3 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ \end{bmatrix}$, then $\spec(A_3) = \{2,2\}$。 Thus, $\spec(A) =\{3\}\cup\{-1\}\cup\{2,2\} = \{-1,2,2,3\}$。 ##### Exercise 4 令 $$ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}. $$  Let $$ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}. $$  ##### Exercise 4(a) 令 $\omega = e^{\frac{2\pi}{3}i}$ 且 $$ \bv = \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix}. $$ 求出 $\bv$ 所對應的特徵值 $\lambda$，並說明如何找到一個么正矩陣 $Q$ 使得 $$ Q^* AQ = \begin{bmatrix} \lambda & * \\ \bzero & A_2 \end{bmatrix}. $$  Let $\omega = e^{\frac{2\pi}{3}i}$ and $$ \bv = \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix}. $$ Find the eigenvalue $\lambda$ corresponding to $\bv$. Then explain how to find a unitary matrix $Q$ such that $$ Q^* AQ = \begin{bmatrix} \lambda & * \\ \bzero & A_2 \end{bmatrix}. $$  ##### Exercise 4(b) 令 $$ \bv = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}. $$ 求出 $\bv$ 所對應的特徵值 $\lambda$，並說明如何找到一個實垂直矩陣 $Q$ 使得 $$ Q\trans AQ = \begin{bmatrix} \lambda & * \\ \bzero & A_2 \end{bmatrix}. $$  Let $$ \bv = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}. $$ Find the eigenvalue $\lambda$ corresponding to $\bv$. Then explain how to find a unitary matrix $Q$ such that $$ Q^* AQ = \begin{bmatrix} \lambda & * \\ \bzero & A_2 \end{bmatrix}. $$  ##### Exercise 4(c) 令 $\omega = e^{\frac{2\pi}{3}i}$ 且 $$ \bv = \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix}. $$ 已知 $\bv$ 所對應的特徵值為 $\omega = a + bi$。令 $\bv = \bx + \by i$，也就是 $\bx$ 和 $\by$ 分別為 $\bv$ 的實部和虛部向量。驗證 $$ \begin{aligned} A \bx &= a\bx - b\by, \\ A \by &= b\bx + a\by. \end{aligned} $$ 並說明如何找到一個可逆矩陣 $Q$ 使得 $$ Q^{-1} AQ = \begin{bmatrix} a & b & * \\ -b & a & * \\ 0 & 0 & A_2 \end{bmatrix}. $$  Let $\omega = e^{\frac{2\pi}{3}i}$ 且 $$ \bv = \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix}. $$ It is known that $\omega = a + bi$ is the eigenvalue corresponding to $\bv$. Let $\bv = \bx + \by i$ such that $\bx$ and $\by$ are both real vectors. Verify that $$ \begin{aligned} A \bx &= a\bx - b\by, \\ A \by &= b\bx + a\by. \end{aligned} $$ Then find the eigenvalue $\lambda$ corresponding to $\bv$. Then explain how to find a unitary matrix $Q$ such that $$ Q^{-1} AQ = \begin{bmatrix} a & b & * \\ -b & a & * \\ 0 & 0 & A_2 \end{bmatrix}. $$  ##### Exercise 5 令 $Q$ 為一 $n\times n$ 么正矩陣，而 $\bx,\by\in\mathbb{C}^n$。證明 $\inp{\bx}{\by} = \inp{Q\bx}{Q\by}$。（上述性質在當 $Q$ 是實垂直矩陣而 $\bx$ 和 $\by$ 為實向量時也對。）這表示 $\bv\mapsto Q\bv$ 這個動作不會改變 $\bv$ 的長度，因此么正矩陣和實垂直矩陣常被視為高維度的鏡射和旋轉。（我們沒有說清楚高維度的鏡射和旋轉是什麼意思。）  Let $Q$ be an $n\times n$ unitary matrix and $\bx,\by\in\mathbb{C}^n$. Show that $\inp{\bx}{\by} = \inp{Q\bx}{Q\by}$. (The same statement also holds when $Q$ is a real orthogonal matrix and $\bx$ and $\by$ are real vectors.) Therefore, the mapping $\bv\mapsto Q\bv$ preserves the length of any vector $\bv$, so unitary matrices and real orthogonal matrices are usually viewed as reflections or rotations in higher dimensions. (However, we did not clarify the meaning of reflections and rotations.)  Ans: Since the inverse of $Q$ equals it conjugate transpose, which means that $Q^*Q=I$. We can know that $\inp{\bx}{\by} = \inp{Q^*Q\bx}{Q\by} = \inp{Q\bx}{Q\by}$. --- ##### Exercise 6 固定一個正整數 $n$。令 $\zeta = e^{\frac{2\pi}{n}i}$，並令 $Q$ 為一 $n\times n$ 矩陣，其第 $a,b$-項為 $\zeta^{a-1}{b-1}$。證明 $Q$ 是一個么正矩陣。（這個矩陣稱為**離散傅立葉變換矩陣** 。）  Fix a positive integer $n$. Let $\zeta = e^{\frac{2\pi}{n}i}$ and $Q$ the $n\times n$ matrix whose $a,b$-entry is $\zeta^{a-1}{b-1}$. Show that $Q$ is a unitary matrix. (The matrix $Q$ is known as the **discrete Fourier transform matrix** .)  :::info collaboration: 2 3 problems: 3 - 2ab, 3 extra: 1.5 - 2cd, 5 moderator: 1 qc: 1 :::

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