\(\mathbb{R}^n\) 中的向量表示法

This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.

from lingeo import random_int_list, random_good_matrix

Main idea

Recall that \(\mathcal{E}_n = \{ {\bf e}_1, \ldots, {\bf e}_n \}\) is the standard basis of \(\mathbb{R}^n\).
For any vector \({\bf v} = (c_1, \ldots, c_n)\in\mathbb{R}^n\), it can be written as
\[{\bf v} = c_1{\bf e}_1 + \cdots + c_n{\bf e}_n. \]

Similarly, let \(\beta = \{ {\bf u}_1, \ldots, {\bf u}_n \}\) be a basis of \(\mathbb{R}^n\).
Every vector \({\bf v}\in\mathbb{R}^n\) has a unique way to be written as a linear combination
\[{\bf v} = c_1{\bf u}_1 + \cdots + c_n{\bf u}_n. \]
We call the vector \((c_1,\ldots, c_n)\in\mathbb{R}^n\) the vector representation of \({\bf v}\) with respect to the basis \(\beta\), denoted as \([{\bf v}]_\beta\).
Since every vector in \(\mathbb{R}^n\) can be written as a linear combinatoin of \(\beta\) and the way of writing it is unique, it is a one-to-one correspondence between \({\bf v}\) and \([{\bf v}]_\beta\).

Let let \(\beta = \{ {\bf u}_1, \ldots, {\bf u}_n \}\) be a basis of \(\mathbb{R}^n\) and
\(A\) the \(n\times n\) matrix whose columns are vectors in \(\beta\).
Since \(\beta\) is a basis, \(A\) is invertible.
By definition, \[A[{\bf v}]_\beta = {\bf v} \text{ and } A^{-1}{\bf v} = [{\bf v}]_\beta. \]

When \(\beta\) is the standard basis of \(\mathbb{R}^n\), \(A = I_n\) nad \([{\bf v}]_\beta = {\bf v}\).
Therefore, our usual way of writting a vector is the vector representation with respect to the standard basis.

In the case when \(\beta\) is an orthonormal basis, \(A\) is an orthogonal matrix and \(A^{-1} = A^\top\).
Therefore, \[A[{\bf v}]_\beta = {\bf v} \text{ and } A^{-1}{\bf v} = [{\bf v}]_\beta. \]

Side stories

• vector representation algebra
• define new inner product

Experiments

Exercise 1

執行以下程式碼。
令 \(\beta = \{ {\bf u}_1, \ldots, {\bf u}_3 \}\) 為 \(A\) 的行向量且
已知其為 \(\mathbb{R}^3\) 的基底。

### code
set_random_seed(0)
print_ans = False
m,n,r = 3,3,3
A = random_good_matrix(m,n,r, bound=3)
x1 = vector(random_int_list(n, 3))
x2 = vector(random_int_list(n, 3))
v1,v2 = A*x1, A*x2
k = choice([3,4,5])

print("A =")
show(A)
print("v1 =", v1)
print("v2 =", v2)
print("k =", k)

if print_ans:
    Ainv = A.inverse()
    print("[v1]_beta =", Ainv * v1)
    print("[v2]_beta =", Ainv * v2)
    print("[v1 + v2]_beta =", Ainv * (v1 + v2))
    print("[v1]_beta + [v2]_beta =", Ainv * v1 + Ainv * v2)
    print("[k * v1]_beta =", Ainv * (k*v1))
    print("k * [v1]_beta =", k * Ainv * v1)
    print("< [v1]_beta, [v2]_beta > =", (Ainv * v1).inner_product(Ainv * v2))
    print("< v1, v2 > =", (v1).inner_product(v2))

執行程式碼得 \[A = \begin{bmatrix} 1 & 3 & 1 \\ -3 & -8 & -1 \\ 0 & -1 & 1 \end{bmatrix} \] 及 \({\bf v}_1 = (4,-15,1)\)、\({\bf v}_2 = (8,-20,-3)\)、\(k=3\)

Exercise 1(a)

求出 \([{\bf v}_1]_\beta\) 及 \([{\bf v}_2]_\beta\)。

答:
假設一個矩陣 \[ \left[\begin{array}{c|c} A & I \end{array}\right] = \left[\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \\ -3 & -8 & -1 & 0 & 1 & 0 \\ 0 & -1 & -1 & 0 & 0 & 1 \end{array}\right] \] 經過列運算得\[ \left[\begin{array}{c|c} R & B \end{array}\right] = \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 7 & 2 & 5 \\ 0 & 1 & 0 & -3 & -1 & -2 \\ 0 & 0 & 1 & 3 & 1 & 1 \end{array}\right] \] 因為 \({A}\) 的各行向量為 \(\mathbb{R}^3\) 的基底，所以 \({A}\) 可逆，可以得 \[B = \begin{bmatrix} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \end{bmatrix} \] 為 \({A}\) 的反矩陣使得 \({AB=I_3}\) ，\({B=A^{-1}}\)。
因此 \[[{\bf v}_1]_\beta=A^{-1}{\bf v}_1=\begin{bmatrix} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ -15 \\ 1\end{bmatrix}= \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}, \] \[[{\bf v}_2]_\beta=A^{-1}{\bf v}_2=\begin{bmatrix} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 8 \\ -20 \\ -3\end{bmatrix}= \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}. \]

Exercise 1(b)

判斷是否 \([{\bf v}_1 + {\bf v}_2]_\beta = [{\bf v}_1]_\beta + [{\bf v}_2]_\beta\)。

答:
計算 \({\bf v}_1 + {\bf v}_2 = (12,-35,-2)\) ， \[[{\bf v}_1+{\bf v}_2]_\beta=A^{-1}({\bf v}_1+{\bf v}_2)=\begin{bmatrix} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 12 \\ -35 \\ -2\end{bmatrix}= \begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix}\] \[[{\bf v}_1]_\beta + [{\bf v}_2]_\beta= \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}+\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=\begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix} \] 得 \([{\bf v}_1 + {\bf v}_2]_\beta = [{\bf v}_1]_\beta + [{\bf v}_2]_\beta\)。

Exercise 1©

判斷是否 \([k{\bf v}_1]_\beta = k[{\bf v}_1]_\beta\)。

答:
計算 \(k{\bf v}_1 = 3(4,-15,1) = (12,-45,3)\) ， \[[k{\bf v}_1]_\beta=A^{-1}(k{\bf v}_1)=\begin{bmatrix} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 12 \\ -45 \\ 3\end{bmatrix}= \begin{bmatrix} 9 \\ 3 \\ -6 \end{bmatrix}\] \[k[{\bf v}_1]_\beta = 3\begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}=\begin{bmatrix} 9 \\ 3 \\ -6 \end{bmatrix} \] 得 \([k{\bf v}_1]_\beta = k[{\bf v}_1]_\beta\)。

Exercise 1(d)

判斷是否 \(\langle {\bf v}_1, {\bf v}_2 \rangle = \langle [{\bf v}_1]_\beta, [{\bf v}_2]_\beta \rangle\)。

答:
計算 \[\langle {\bf v}_1, {\bf v}_2 \rangle = (4,-15,1)(8,-20,-3)= 32+300-3=329\] \[\langle [{\bf v}_1]_\beta, [{\bf v}_2]_\beta \rangle= (3,1,-2)(1,2,1)= 3+2-2=3 \] 得 \(\langle {\bf v}_1, {\bf v}_2 \rangle \neq \langle [{\bf v}_1]_\beta, [{\bf v}_2]_\beta \rangle\)。

Exercises

Exercise 2

已知 \[A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & -4 \\ 5 & -2 & 12 \end{bmatrix} \] 的反矩陣為 \[A^{-1} = \begin{bmatrix} 4 & -2 & -1 \\ -8 & 7 & 3 \\ -3 & 2 & 1 \end{bmatrix}. \]
令 \(\beta\) 為 \(A\) 的行向量集合。
令 \({\bf v}_1 = (1,1,1)\)、
\({\bf v}_2 = (1,2,3)\)。
求 \([{\bf v}_1]_\beta\) 和 \([{\bf v}_2]_\beta\)。

答：

\[[{\bf v}_1]_\beta =A^{-1}{\bf v}_1= \begin{bmatrix} 4 & -2 & -1 \\ -8 & 7 & 3 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\1\\1\end{bmatrix}= \begin{bmatrix} 1\\2\\0\end{bmatrix}, \] \[[{\bf v}_2]_\beta =A^{-1}{\bf v}_2= \begin{bmatrix} 4 & -2 & -1 \\ -8 & 7 & 3 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\2\\3\end{bmatrix}= \begin{bmatrix} -3\\15\\4\end{bmatrix}. \]

Exercise 3

已知 \[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \end{bmatrix} \] 為一垂直矩陣。
令 \(\beta\) 為 \(A\) 的行向量集合。
令 \({\bf v}_1 = (1,1,1)\)、
\({\bf v}_2 = (1,2,3)\)。
求 \([{\bf v}_1]_\beta\) 和 \([{\bf v}_2]_\beta\)。

答：

因為 \(A\) 為一垂直矩陣，所以 \(A\) 可逆且 \[A^{-1} = A^\top = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \end{bmatrix}. \] 則 \[[{\bf v}_1]_\beta=A^{-1}{\bf v}_1=\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}=\begin{bmatrix} {\sqrt{3}} \\ 0 \\ 0\end{bmatrix}, \] \[[{\bf v}_2]_\beta=A^{-1}{\bf v}_2= \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix}=\begin{bmatrix} 2{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} \\ -\frac{3}{\sqrt{6}}\end{bmatrix}. \]

Exercise 3

令 \(\beta\) 為 \(\mathbb{R}^n\) 中的一組基底。
定義
\[\begin{aligned} f : \mathbb{R}^n &\rightarrow \mathbb{R}^n \\ {\bf v} &\mapsto [{\bf v}]_\beta \\ \end{aligned} \]
為一函數。

Exercise 3(a)

驗證 \(f\) 為一線性函數。

答:
令 \(\beta = \{\bu_1,\ldots,\bu_n\}\)。
因為 \(\beta\) 是 \(\mathbb{R}^n\) 中的基底，
所以任何 \(\bv_1,\bv_2\in\mathbb{R}^n\) 都可以寫成
\[ \begin{aligned} \bv_1 &= c_1{\bf u}_1 + \cdots + c_n{\bf u}_n , c_i \in \mathbb{R} , i=\{1,\cdots,n\}, \\ \bv_2 &= d_1{\bf u}_1 + \cdots + d_n{\bf u}_n , d_i \in \mathbb{R} , i=\{1,\cdots,n\}, \end{aligned} \] 因此對任意實數 \(k\) 都有
\[ \begin{aligned} \bv_1 + \bv_2 &= (c_1 + d_1){\bf u}_1 + \cdots +(c_n + d_n){\bf u}_n, \\ k\bv_1 &= kc_1{\bf u}_1 + \cdots + kc_n{\bf u}_n,\\ k\bv_2 &= kd_1{\bf u}_1 + \cdots + kd_n{\bf u}_n. \end{aligned} \] 如此一來，
\[ \begin{aligned}\\ [\bv_1]_\beta &= (c_1 , \cdots , c_n ) , \\ [\bv_2]_\beta &= (d_1 , \cdots , d_n ) , \end{aligned} \] 因此 \[ \begin{aligned} \\ [\bv_1 + \bv_2]_\beta &= ((c_1+d_1) , \cdots , (c_n+d_n)), \\ [k\bv_1]_\beta &= (kc_1 , \cdots , kc_n),\\ [k\bv_2]_\beta &= (kd_1 , \cdots , kd_n). \end{aligned} \] 因為 \[f({\bf v}_1+{\bf v}_2)=[\bv_1 + \bv_2]_\beta=[\bv_1]_\beta+[\bv_2]_\beta=f({\bf v}_1)+f({\bf v}_2), \] \[kf({\bf v})=k[\bv]_\beta=[k\bv]_\beta=f(k{\bf v}), \] 所以綜上所述，\(f\) 為一線性函數得證。

Exercise 3(b)

判斷 \(f\) 是否是嵌射。

答:

嵌射的定義：對任意定義域中的 \(\bv_1,\bv_2\) 都有「若 \(f(\bv_1) = f(\bv_2)\)，則 \(\bv_1 = \bv_2\)」的性質。

令 \(\bv_1,\bv_2\in\mathbb{R}^n\)。
若 \(f({\bf v}_1) = f({\bf v}_2)= (c_1,\dots,c_n)\)，則 \(\bv_1 = c_1\bu_1 + \dots + c_n\bu_n = \bv_2\)。

符合嵌射的定義，因此得證 \(f\) 是嵌射。

Exercise 3©

判斷 \(f\) 是否是映射。

答:
令 \(\beta= \{ {\bf u}_1, \ldots, {\bf u}_n \}\)，
則任何 \([{\bf v}]_\beta=(c_1, \cdots ,c_n)\) 都有 \({\bf v}=c_1{\bf u}_1+ \cdots +c_n{\bf u}_n\)，
使得 \(f({\bf v})=[{\bf v}]_\beta\).
根據定義, 得證 \(f\) 是映射。

Exercise 3(d)

求出 \(f\) 的矩陣表示法 \([f]\)。

答:

依照 \([f]\) 的定義可知
\[ [f] = \begin{bmatrix} | & ~ & | \\ [\be_1]_\beta & \cdots & [\be_n]_\beta \\ | & ~ & | \\ \end{bmatrix}. \] 實際上，若 \(A\) 為由 \(\beta\) 中向量當作行向量的矩陣，
則 \(A[\bv]_\beta = \bv\) 且 \(A^{-1}\bv = [\bv]_\beta\)。
因此，我們也有 \([f] = A^{-1}\)。

Exercise 4

回顧一個 內積 \(\langle \cdot, \cdot \rangle\) 必須符合以下的條件：

1. \(\langle{\bf x}_1 + {\bf x}_2,{\bf y}\rangle = \langle{\bf x}_1,{\bf y}\rangle + \langle{\bf x}_2,{\bf y}\rangle\).
2. \(\langle k{\bf x},{\bf y}\rangle = k\langle{\bf x},{\bf y}\rangle\).
3. \(\langle {\bf x},{\bf y}\rangle = \langle {\bf y},{\bf x}\rangle\).
4. \(\langle {\bf x}, {\bf x} \rangle \geq 0\), and the equality holds if and only if \({\bf x} = {\bf 0}\).

Exercise 4(a)

令 \(\beta\) 為 \(\mathbb{R}^n\) 中的一組基底。
定義一個新的雙變數函數 \(\langle {\bf x}, {\bf y} \rangle_\beta = \langle [{\bf x}]_\beta, [{\bf y}]_\beta \rangle\)﹐
其中 \(\langle [{\bf x}]_\beta, [{\bf y}]_\beta \rangle\) 指的是 \(\mathbb{R}^n\) 中的標準內積。
驗證 \(\langle \cdot, \cdot \rangle_\beta\) 也是 \(\mathbb{R}^n\) 上的另一種內積。

答:

1. 可驗證 \[\begin{aligned} \inp{\bx_1 + \bx_2}{\by}_\beta &= \inp{[\bx_1 + \bx_2]_\beta}{[\by]_\beta} \\ &= \inp{[\bx_1]_\beta + [\bx_2]_\beta}{[\by]_\beta} \\ &= \inp{[\bx_1]_\beta}{[\by]_\beta}+\inp{[\bx_2]_\beta}{[\by]_\beta} \\ &= \inp{\bx_1}{\by}_\beta + \inp{\bx_2}{\by}_\beta. \end{aligned} \]
2. 可驗證 \[\begin{aligned} \inp{k\bx}{\by}_\beta &= \inp{[k\bx]_\beta}{[\by]_\beta} \\ &= \inp{k[\bx]_\beta}{[\by]_\beta} \\ &= k\inp{[\bx]_\beta}{[\by]_\beta} \\ &= k\inp{\bx}{\by}_\beta. \end{aligned} \]
3. 可驗證 \[\begin{aligned} \inp{\bx}{\by}_\beta &= \inp{[\bx]_\beta}{[\by]_\beta}\\ &= \inp{[\by]_\beta}{[\bx]_\beta} \\ &= \inp{\by}{\bx}_\beta. \end{aligned} \]
4. 若 \(\beta = \{\bu_1,\ldots,\bu_n\}\) 且 \(\bx = c_1\bu_1 + \cdots + c_n\bu_n\)。
   則 \([\bx]_\beta = (c_1,\ldots,c_n)\)。
   則 \(\inp{\bx}{\bx}_\beta = \inp{[\bx]_\beta}{[\bx]_\beta} = c_1^2 + \cdots + c_n^2 \geq 0\)，
   且等式成立時只有在 \(c_1=\cdots=c_n=0\) 時。
   此時 \(\bx = 0\bu_1 + \cdots +0\bu_n = \bzero\)。

Exercise 4(b)

證明當 \(\beta\) 是單位長垂直基底時﹐任意向量都有 \(\langle {\bf x}, {\bf y} \rangle_\beta = \langle {\bf x}, {\bf y} \rangle\)。

答:
已知 \(\langle {\bf x}, {\bf y} \rangle_\beta = \langle [{\bf x}]_\beta, [{\bf y}]_\beta \rangle\)。
令 \({\bf x}=c_1{\bf u}_1+ \cdots + c_n{\bf u}_n\)、 \({\bf y}=d_1{\bf u}_1+ \cdots + d_n{\bf u}_n\)，
其中 \(\beta={\{\bf u}_1,\cdots,{\bf u}_n\}\) 為在 \(\mathbb{R}^n\) 中的一組基底，\([{\bf x}]_\beta = ({c}_1,\cdots,{c}_n)\)、\([{\bf y}]_\beta = ({d}_1,\cdots,{d}_n)\)。
因為 \(\beta\) 是單位長垂直基底，所以 \[\langle {\bf u}_m, {\bf u}_n \rangle=\begin{cases} 1 & , m = n,\\ 0 & , m \neq n. \end{cases} \] 因此可得 \[\begin{aligned} \langle {\bf x}, {\bf y}\rangle &=(c_1{\bf u}_1 + \cdots + c_n{\bf u}_n) \cdot (d_1{\bf u}_1 + \cdots + d_n{\bf u}_n) \\ &= c_1d_1||{\bf u}_1^2|| + \cdots + c_nd_n||{\bf u}_n^2|| \\ &= c_1d_1 + \cdots + c_nd_n \\ &=({c}_1,\cdots,{c}_n) \cdot ({d}_1,\cdots,{d}_n)\\ &=\langle [{\bf x}]_\beta, [{\bf y}]_\beta \rangle \\ &=\langle {\bf x}, {\bf y} \rangle_\beta \end{aligned} \]
故得證 \(\langle {\bf x}, {\bf y} \rangle_\beta = \langle {\bf x}, {\bf y} \rangle\)。