行列式值的定義

# 行列式值的定義 Definition of the determinant ![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_list ``` ## Main idea For $n\times n$ matrices $A$, the **determinant** $\det(A)$ is defined through the following rules. - $\det(I_n) = 1$. - If $E$ is the elementary matrix of $\rho_i\leftrightarrow\rho_j$, then $\det(EA) = -\det(A)$ and we define $\det(E) = -1$. - If $E$ is the elementary matrix of $\rho_i:\times k$, then $\det(EA) = k\det(A)$ and we define $\det(E) = k$. (Note that this statement still holds even when $k = 0$.) - If $E$ is the elementary matrix of $\rho_i:+k\rho_j$, then $\det(EA) = \det(A)$ and we define $\det(E) = 1$. Note that the determinants for $2\times 2$ and $3\times 3$ matrices agree with this definition. As a consequence, if a matrix $A$ is invertible and can be written as the product a sequence of elementary matrices $F_1\cdots F_k$, then $\det(A) = \det(F_1)\cdots\det(F_k)\det(I_n) = \det(F_1)\cdots\det(F_k)$. In contrast, if $A$ is not invertible, then $\det(A) = 0$. In particular, this happens when - $A$ has a zero row, or - $A$ has repeated rows. By definitions, $\det(A) = \Vol_C(A) = \Vol_R(A)$ for any matrix $A$. ##### Remark Thanks to row operations, the definition of $\det(A)$ assigns at least a value to $\det(A)$. However, _maybe_ the rules assigns more than one values to it. That is, the function might not be _well-defined_. A matrix $A$ can be written as the product of different sequences of elementary matrices. For example, one may write $$ A = F_1 \cdots F_k = E_1 \cdots E_h $$ for elementary matrices $F_1,\ldots, F_k$ and $E_1,\ldots, E_h$. However, it is not yet clear by the definition that $\det(F_1) \cdots \det(F_k) = \det(E_1) \cdots \det(E_h)$. We will deal with this issue at the end of this chapter. ## Side stories - well-defined - permutation matrices ## Experiments ##### Exercise 1 執行以下程式碼。  Run the code below.  ```python ### code set_random_seed(0) print_ans = False n = 4 while True: A = matrix(n, random_int_list(n^2, 3)) if A.det() != 0: break print("A =") pretty_print(A) if print_ans: print("determinant of A =", A.det()) ``` ##### Exercise 1(a) 將 $A$ 消成最簡階梯形式、並記錄下每一步的列運算。  Find the reduced echelon form of $A$ and record each step of the row operation.  ##### Exercise 1(a) - answer here Let `seed = 15`, $$\begin{aligned} A = \left[\begin{array}{cccc} 2 & 2 & 2 & 0 \\ -2 & -1 & 2 & -2\\ 2 & 0 & -2 & 1\\ 2 & 0 & -2 & -2 \end{array}\right] \xrightarrow[\rho_2:\rho_1+\rho_2]{\rho_3:\rho_3-\rho_1}\left[\begin{array}{cccc} 2 & 2 & 2 & 0\\ 0 & 1 & 4 & -2\\ 0 & -2 & -4 & 1\\ 2 & 0 & -2 & -2 \end{array}\right]\\ \xrightarrow[\rho_3:\rho_3+2\rho_2]{\rho_4:\rho_4-\rho_1}\left[\begin{array}{cccc} 2 & 2 & 2 & 0\\ 0 & 1 & 4 & -2\\ 0 & 0 & 4 & -3\\ 0 & -2 & -4 & -2 \end{array}\right]\\ \xrightarrow[\rho_1:\frac{1}{2}\rho_1]{\rho_4:\rho_4+2\rho_2}\left[\begin{array}{cccc} 1 & 1 & 1 & 0\\ 0 & 1 & 4 & -2\\ 0 & 0 & 4 & -3\\ 0 & 0 & 4 & -6 \end{array}\right]\\ \xrightarrow[\rho_4:\rho_4-\rho_3]{\rho_2:\rho_2-\rho_3}\left[\begin{array}{cccc} 1 & 1 & 1 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 4 & -3\\ 0 & 0 & 0 & -3 \end{array}\right]\\ \xrightarrow[\rho_1:\rho_1-\rho_2]{\rho_4:-\frac{1}{3}\rho_4}\left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 4 & -3\\ 0 & 0 & 0 & 1 \end{array}\right]\\ \xrightarrow[\rho_2:\rho_2-\rho_4]{\rho_3:\frac{1}{4}\rho_3}\left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & -\frac{3}{4}\\ 0 & 0 & 0 & 1 \end{array}\right]\\ \xrightarrow[\rho_3:\rho_3+\frac{3}{4}\rho_4]{\rho_1:\rho_1+\rho_3}\left[\begin{array}{cccc} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\\ \xrightarrow[]{\rho_1:\rho_1-\rho_3}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\\ \end{aligned} $$ :::success Nice work! ::: ##### Exercise 1(b) 求出 $\det(A)$。  Find $\det(A)$.  ##### Exercise 1(b) -- answer here $\det(A)=1\cdot 4\cdot (-3)\cdot 2= -24.$ :::info What do the experiments try to tell you? (open answer) > We can know that after rref, we can find $\det(A)$ more easily . ::: :::success Good. - [x] No space before a comma, put a space after a comma. ::: ## Exercises ##### Exercise 2 對以下矩陣 $A$，求出 $\det(A)$。  For the following matrices $A$, find $\det(A)$.  ##### Exercise 2(a) $$ A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix}. $$ ##### Exercise 2(a) - answer here We have to turn the last row to the first row, so we times $-1$ three times. Because we swap the rows three times, then we can get the matrix after swapping. $$ A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}. $$ Then we can calculate $\det(A)=-1\cdot -1\cdot -1\cdot 1\cdot 1\cdot 1\cdot 1= -1.$ :::warning The idea is correct. However, $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\cdot(-1)\cdot(-1)\cdot(-1) = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix} $$ is not related to $A$. ::: ##### Exercise 2(b) $$ A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{bmatrix}. $$ ##### Exercise 2(b) - answer here First swap the third row to the first row and last row to the second row. So we have to times -1 four times, then we can get the matrix after swapping. $$ A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}. $$ Then we can calculate $\det(A)=-1\cdot -1\cdot -1\cdot -1\cdot 1\cdot 1\cdot 1\cdot 1 = 1.$ :::warning Same as 2(b). ::: ##### Exercise 2(c) $$ A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix} $$ ##### Exercise 2(c) - answer here First swap the third row to the first row and last row to the second row. So we have to times -1 two times, then we can get the matrix after swapping. $$ A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}. $$ $\det(A)= (-1)^2\cdot 1\cdot 1\cdot 1\cdot 1=1$. :::warning Same as 2(b). ::: ##### Exercise 2(d) $$ A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 0 \\ 0 & 3 & 0 & 0 \\ 4 & 0 & 0 & 0 \\ \end{bmatrix}. $$ ##### Exercise 2(d) - answer here $$ A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 0 \\ 0 & 3 & 0 & 0 \\ 4 & 0 & 0 & 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}. $$ By observing 2(c), we can know that $\det(A)=1\cdot 2\cdot 3\cdot 4 = 24.$ ##### Exercise 2(e) $$ A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 3 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ \end{bmatrix}. $$ ##### Exercise 2(e) - answer here First swap the third row to the first row and last row to the second row. So we have to times -1 four times, then we can get the matrix after swapping. $$ A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 3 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 3 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}. $$ Then we can calculate $\det(A)$. $\det(A)=-1\cdot -1\cdot -1\cdot -1\cdot 3\cdot 4\cdot 1\cdot 2 = 24.$ :::warning Same as 2(b). ::: ##### Exercise 3 對以下矩陣 $A$，求出 $\det(A)$。  For the following matrices $A$, find $\det(A)$.  ##### Exercise 3(a) $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{bmatrix}. $$ ##### Exercise 3(a) -- answer here Subtracting the other rows by first row to get, $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 \\ 0 & 1 & 2 & 3 \end{bmatrix}, $$ and use second row to kill rows beneath it, $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 2 \end{bmatrix}, $$ $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}, $$ $$ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. $$ Since row subtraction doesn't change determinant, $\det(A)=1.$ :::success Good. ::: --- ##### Exercise 3(b) $$ A = \begin{bmatrix} 1 & 1 & \cdots & 1 \\ 1 & 2 & \cdots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & \cdots & n \end{bmatrix}. $$ **[由 :cloud: 提供]** ##### Exercise 3(b) -- answer here By applying the same way of **Exercise 3(a)**, we found that $\det(A)=1.$ ##### Exercise 4 對以下矩陣 $A$，求出 $\det(A)$。  For the following matrices $A$, find $\det(A)$.  ##### Exercise 4(a) $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \\ 1 & 4 & 16 & 64 \end{bmatrix}. $$ ##### Exercise 4(a) - answer here Subtracting the other rows by first row to get, $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 \\ 0 & 2 & 8 & 26 \\ 0 & 3 & 15 & 63 \end{bmatrix}, $$ and do the row operation by second row to obtain $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 \\ 0 & 0 & 2 & 12 \\ 0 & 0 & 6 & 42 \end{bmatrix}. $$ Now we can do the matrix deflation to calculate the determinant of A. $\det(A)=1\cdot 1\cdot (2\cdot 42 - 12\cdot 6) = 12.$ :::success Good. ::: ##### Exercise 4(b) $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 4 & 8 & 16 \\ 1 & 3 & 9 & 27 & 81 \\ 1 & 4 & 16 & 64 & 256 \\ 1 & 5 & 25 & 125 & 625 \end{bmatrix}. $$ ##### Exercise 4(b) - answer here We do the row operation to matrix A to get $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1\\ 0 & 1 & 3 & 7 & 15 \\ 0 & 0 & 2 & 12 & 50 \\ 0 & 0 & 0 & 6 & 60 \\ 0 & 0 & 0 & 0 & 24 \end{bmatrix}, $$ and we do the matrix deflation to calculate the determinant of A. $\det(A)=1\cdot 1\cdot 2\cdot (6\cdot 24 - 60\cdot 0) = 288.$ :::success Good. ::: ##### Exercise 5 對以下矩陣 $A$，求出 $\det(A)$。提示：把所有列加到第一列。  For the following matrices $A$, find $\det(A)$. Hint: Add each row to the first row.  ##### Exercise 5(a) $$ A = \begin{bmatrix} 4 & -1 & -1 & -1 \\ -1 & 4 & -1 & -1 \\ -1 & -1 & 4 & -1 \\ -1 & -1 & -1 & 4 \end{bmatrix}. $$ **[由 :cloud: 提供]** After adding each row to the first row, the matrix $A$ becomes $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ -1 & 4 & -1 & -1 \\ -1 & -1 & 4 & -1 \\ -1 & -1 & -1 & 4 \end{bmatrix}. $$ Next, we subtract the the first row from other rows, eliminating the remaining $(-1)$'s. Then we divide these rows by $5$ to make an upper triangular matrix as follows. $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{bmatrix}, $$ $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, $$ Thus, $\det(A)=1\cdot 5\cdot 5\cdot 5=125.$ ##### Exercise 5(b) 令 $A$ 為一 $n\times n$ 矩陣 $$ \begin{bmatrix} n & -1 & \cdots & -1 \\ -1 & n & \ddots & \vdots \\ \vdots & \ddots & \ddots & -1 \\ -1 & \cdots & -1 & n \end{bmatrix}. $$  Let $A$ be the $n\times n$ matrix $$ \begin{bmatrix} n & -1 & \cdots & -1 \\ -1 & n & \ddots & \vdots \\ \vdots & \ddots & \ddots & -1 \\ -1 & \cdots & -1 & n \end{bmatrix}. $$  **[由 :cloud: 提供]** ##### Exercise 5(b) -- answer here By applying the same way of **Exercise 5(a)**, it will become something like $$ \begin{bmatrix} 1 & 1 & \cdots & 1 \\ 0 & (n+1) & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & (n+1) \end{bmatrix}. $$ So $\det(A)=1\cdot (n+1)^{(n-1)}.$ ##### Exercise 6 利用 $\det$ 定義中的四條準則，說明以下性質。  Use the four rules in the definition of the determinant to explain the following properties.  ##### Exercise 6(a) 若 $A$ 中有一列為零向量，說明 $\det(A) = 0$。  If $A$ has a zero row, then $\det(A) = 0$.  **[由 :cloud: 提供]** ##### Exercise 6(a) -- answer here Suppose the $i$th row of $A$ are zero, after applied row operation : $$ A\xrightarrow{\rho_i:\times k}B. $$ By the definition, $\det(B)=k\cdot \det(A),$ and since the $i$th row are all $0$ times any number is still $0$, so actually $B=A.$ We can know from $B=A$ that $\det(B)=\det(A),$ to achieve this equality (with any $k\neq 1$), the only possibility is $\det(A)=0.$ ##### Exercise 6(b) 若 $A$ 中有兩列向量相等，說明 $\det(A) = 0$。  If $A$ has two rows in common, then $\det(A) = 0$.  **[由 :cloud: 提供]** ##### Exercise 6(b) -- answer here Suppose $A$ has two rows in common, we can eliminate one of them by subtracting each other. There will be a zero row in $A$, applying the same way in **Exercise 6(a)**, we can find $\det(A) = 0.$ ##### Exercise 6(c) 若 $A$ 中的列向量集合線性相依，說明 $\det(A) = 0$。  If some rows of $A$ form a linearly dependent set, then $\det(A) = 0$.  **[由 :cloud: 提供]** ##### Exercise 6(c) -- answer here Since some rows of $A$ form a linearly dependent set, there are coefficients $c_1,\ldots, c_n$ that is not all zero such that $$ c_1\br_1 + \cdots + c_n\br_n = \bzero, $$ where $\br_1,\ldots,\br_n$ are the rows of $A$. Without loss of generality, we may assume $c_1 \neq 0$. $$ \br_1 = -\frac{1}{c_1}(c_2\br_2 + \cdots + c_n\br_n). $$ Consequently, we may apply a sequence of row operations $\rho_1: +\frac{c_2}{c_1}\br_2$, $\ldots$, $\rho_1: +\frac{c_n}{c_1}\br_n$ to generate a zero row. Therefore, $\det(A)=0.$ :::info collaboration: 2 3 problems: 3 * 2a~c extra problems: 2.5 * 2d~e, 3a, 4a~b moderator: 1 qc: 1 :::

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