目录

Linear Algebra Done Right习题的参考解答及思考(第2章)

第2章

练习2.A

练习2.A.1

题目:

Support \( v_1, v_2, v_3, v_4 \) spans \( V \). Prove that the list \( v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 \) also spans \( V \).

证明:

由于\( v_1, v_2, v_3, v_4 \)生成\( V \),因此\( \forall v \in V \), \( \exists a_1, a_2, a_3, a_4 \in \mathbf{F}, v = a_1v_1 + a_2v_2 + a_3v_4 + a_4v_4 = a_1(v_1 - v_2) + (a_1 + a_2)(v_2 - v_3) + (a_1 + a_2 + a_3)(v_3 - v_4) + (a_1 + a_2 + a_3 + a_4)v_4 \),也就说任意\( v \in V \)均可以表达成\( v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 \)的线性组合,故\( V \subseteq \text{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) \)。

反之,\( \forall v \in \text{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) \),有\( \exists a_1, a_2, a_3, a_4, v = a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4v_4 = a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + (a_4 - a_3)v_4 \in \text{span}(v_1, v_2, v_3, v_4) = V \),因此\( \text{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) \subseteq V \)

综上,\( \text{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) = V \)。

证毕。

练习2.A.2

题目:

Verify the assertions in Example 2.18.

Example 2.18的内容:

  1. A list \( v \) of one vector \( v \in V \) is linearly independent if and only if \( v \neq 0 \).
  2. A list of two vectors in \( V \) is linearly independent if and only if neither vector is a scalar multiple of the other.
  3. \( (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) \) is linearly independent in \( \mathbf{F}^4 \).
  4. The list \( 1, z, \dots, z^m \) is linearly independent in \( \mathcal{P}(\mathbf{F}) \) for each nonnegative interger \( m \).

证明:

证明1:

必要性:

如果单向量列表\( v \)线性无关,此时假设\( v = 0 \),则任取一个\( a \neq 0 \in \mathbf{F} \),有\( av = 0 \),这和列表\( v \)线性无关矛盾,因此假设不成立,有\( v \neq 0 \)。

充分性:

如果\( v \neq 0 \),此时假设单向量列表\( v \)线性相关,则\( \exists a_1 \neq 0 \in \mathbf{F}, a_1v = 0 \),然而根据练习1.B.2, \( a_1v = 0 \)意味着\( a_1 = 0 \)或\( v = 0 \),这和两者都\( \neq 0 \)矛盾,因此假设不成立,有列表\( v \)线性无关。

证明2:

令双向量列表为\( v_1, v_2 \)。

必要性:

如果列表\( v_1, v_2 \)线性无关,则\( v_1 \neq 0, v_2 \neq 0 \),假设其中一个向量是另外一个向量标量倍,不妨设\( \exists a \in \mathbf{F}, v_1 = av_2 \),则有\( a \neq 0 \)以及\( \dfrac{1}{a}v_1 - v_2 = \dfrac{1}{a}(av_2) - v_2 = v_2 - v_2 = 0 \),这和列表\( v_1, v_2 \)线性无关矛盾,因此假设不成立,有列表中任意一个向量都不是另外一个向量的标量倍。

充分性:

如果列表\( v_1, v_2 \)中任意一个向量都不是另外一个向量的标量倍,则假设列表\( v_1, v_2 \)线性相关,此时\( \exists a_1, a_2 \in \mathbf{F} \)满足 \( a_1v_1 + a_2v_2 = 0 \)且\( a_1, a_2 \)不全为\( 0 \),不妨设\( a_1 \neq 0 \)(\( a_2 \neq 0 \)情况的证明类似),可得\( v_1 = \dfrac{-a_2}{a_1}v_2 \),也就是\( v_1 \)是\( v_2 \)的标量倍,矛盾,因此假设不成立,有列表\( v_1, v_2 \)线性无关。

证明3:

假设\( (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) \)线性相关,则\( \exists a_1, a_2, a_3 \)满足 \( a_1(1, 0, 0, 0) + a_2(0, 1, 0, 0) + a_3(0, 0, 1, 0) = 0 \) 且\( a_1, a_2, a_3 \)不全为\( 0 \),不妨设\( a_1 \neq 0 \),则\( a_1(1, 0, 0, 0) + a_2(0, 1, 0, 0) + a_3(0, 0, 1, 0) = (a_1, 0, 0, 0) + (0, a_2, 0, 0) + (0, 0, a_3, 0) = (a_1, a_2, a_3, 0) \neq 0 \),矛盾,因此假设不成立,有\( (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) \)线性无关。

证明4:

令\( m \)为任意非负整数,假设列表\( 1, z, \dots, z^m \)线性相关(注意,这里\( z \)是多项式函数的变量),则\( \exists a_0, a_1, \dots, a_m \in \mathbf{F} \)满足 \( a_0 + a_1z + \dots + a_mz^m = 0 \)(注意,这里右边的\( 0 \)是函数值一律为\( 0 \)的多项式函数)且\( a_0, a_1, \dots, a_m \)不全为\( 0 \),令\( k \)为这些不为\( 0 \)的系数中下标最大的那个,于是有\( \forall k + 1 \leq i \leq m, a_i = 0 \),进而有\( a_0 + a_1z + \dots + a_mz^m = a_0 + a_1z + \dots + a_kz^k = 0 \),改写下\( a_0 + a_1z + \dots + a_kz^k = 0 \),得到\( z^k = \dfrac{-a_0}{a_k} + \dfrac{-a_1}{a_k}z + \dots + \dfrac{-a_{k - 1}}{a_k}z^{k - 1} \),这意味着我们能将\( k \)次的多项式\( z^k \)表达成等式右边\( k - 1 \)次的多项式,矛盾,因此假设不成立,列表\( 1, z, \dots, z^m \)线性无关。

证毕。

练习2.A.3

题目:

Find a number \( t \) such that \( (3, 1, 4), (2, -3, 5), (5, 9, t) \) is not linearly independent in \( \mathbf{R}^3 \).

解答:

设\( a_1, a_2, a_3 \in \mathbf{R} \),则\( a_1(3, 1, 4) + a_2(2, -3, 5) + a_3(5, 9, t) = (3a_1 + 2a_2 + 5a_3, a_1 - 3a_2 + 9a_3, 4a_1 + 5a_2 + ta_3) = 0 \),可得\( a_1 = -3a_3, a_2 = 2a_3, (t - a)a_3 = 0 \),因此令\( t = 2, a_3 = 1, a_1 = -3, a_2 = 2 \),有\( a_1(3, 1, 4) + a_2(2, -3, 5) + a_3(5, 9, t) = 0 \),又\( a_1, a_2, a_3 \)不全为\( 0 \),因此\( (3, 1, 4), (2, -3, 5), (5, 9, t) \)在\( \mathbf{R}^3 \)中线性相关。

练习2.A.4

题目:

Verify the assertion in the second bullet point in Example 2.20.

Example 2.20第二点的内容:

The list \( (2, 3, 1), (1, -1, 2), (7, 3, c) \) is linearly dependent in \( \mathbf{F}^3 \) if and only if \( c = 8 \), as you should verify.

证明:

必要性:

如果\( (2, 3, 1), (1, -1, 2), (7, 3, c) \)在\( \mathbf{F}^3 \)中线性相关,则存在不全为\( 0 \)的\( a_1, a_2, a_3 \in \mathbf{F} \)使得 \( a_1(2, 3, 1) + a_2(1, -1, 2) + a_3(7, 3, c) = 0 \),可得\( a_1 = -2a_3, a_2 = -3a_3, (c - 8)a_3 = 0 \),由\( (c - 8)a_3 = 0 \),可得\( a_3 = 0 \)或\( c - 8 = 0 \):但如果\( a_3 = 0 \),则\( a_1 = a_2 = 0 \),这和它们不全为\( 0 \)矛盾,因此只剩下\( c - 8 = 0 \)这种可能性了,进而可得\( c = 8 \)。

充分性:

如果\( c = 8 \),则令\( a_3 = 1, a_1 = -2 , a_2 = -3 \),可得\( a_1(2, 3, 1) + a_2(1, -1, 2) + a_3(7, 3, c) = 0 \),又\( a_1, a_2, a_3 \)不全为\( 0 \),因此\( (2, 3, 1), (1, -1, 2), (7, 3, c) \)在\( \mathbf{F}^3 \)中线性相关。

证毕。

练习2.A.5

题目:

  1. Show that if we think of \( \mathbf{C} \) as a vector space over \( \mathbf{R} \), then the list \( (1 + i, 1 - i) \) is linearly independent.
  2. Show that if we think of \( \mathbf{C} \) as a vector space over \( \mathbf{C} \), then the list \( (1 + i, 1 - i) \) is linearly dependent.

证明:

证明1:

假设\( (1 + i, 1 - i) \)线性相关,则\( \exists a_1, a_2 \in \mathbf{R} \),\( a_1, a_2 \)不全为\( 0 \),使得\( a_1(1 + i) + a_2(1 - i) = 0 \),可得\( (a_1 + a_2) + (a_1 - a_2)i = 0 \),进而可得\( a_1 + a_2 = -(a_1 - a_2)i \),这里左边两个实数相加的结果是实数,这意味着右边也得是实数,于是可得\( -(a_1 - a_2) = 0, a_1 + a_2 = -(a_1 - a_2)i = -(0)i = 0 \),进而有\( a_1 = a_2, a_1 = -a_2 \),这意味着\( -a_2 = a_2 \),假设\( a_2 \neq 0 \),则两边同乘\( \dfrac{1}{a_2} \)可得\( -1 = 1 \),矛盾,因此\( a_2 \neq 0 \)的假设不成立,有\( a_2 = 0 \),但这样子的话,会有\( a_1 = a_2 = 0 \),这和\( a_1, a_2 \)不全为\( 0 \)矛盾,因此一开始的假设不成立,有\( (1 + i, 1 - i) \)线性无关。

证毕。

证明2:

令\( a_1 = 1 - i, a_2 = -(1 + i) \),可得\( a_1(1 + i) + a_2(1 - i) = 0 \),从而\( (1 + i, 1 - i) \)线性相关。

证毕。

练习2.A.6

题目:

Suppose \( v_1, v_2, v_3, v_4 \) is linearly independent in \( V \). Prove that the list

\[ v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 \]

is also linearly independent.

证明:

\( \forall a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足\( a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4v_4 = 0 \),可得\( a_1v_1 + (-a_1 + a_2)v_2 + (-a_2 + a_3)v_3 + (-a_3 + a_4)v_4 = 0 \),因为\( v_1, v_2, v_3, v_4 \)线性无关,因此\( a_1 = -a_1 + a_2 = -a_2 + a_3 = -a_3 + a_4 = 0 \):

  1. 由\( a_1 = -a_1 + a_2 = 0 \),可得\( a_2 = 2a_1 = 2 \times 0 = 0 \)。
  2. 由\( -a_1 + a_2 = -a_2 + a_3 \)以及\( a_2 = 2a_1 \),可得\( a_3 = 3a_1 = 3 \times 0 = 0 \)。
  3. 由\( -a_2 + a_3 = -a_3 + a_4 \)以及\( a_2 = 2a_1, a_3 = 3a_1 \),可得\( a_4 = 4a_1 = 4 \times 0 = 0 \)。

综上,有\( a_1 = a_2 = a_3 = a_4 = 0 \),可得\( v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 \)线性无关。

证毕。

练习2.A.7

题目:

Prove or give a counterexample: If \( v_1, v_2, \dots, v_m \) is a linearly independent list of vectors in \( V \), then

\[ 5v_1 - 4v_2, v_2, v_3, \dots, v_m \]

is linearly independent.

解答:

令\( a_1 = a_3 = a_4 = \dots = a_m = 0, a_2 = 4 \),则可得\( a_1(5v_1 - 4v_2) + a_2v_2 + a_3v_3 + \dots + a_mv_m = 0 \),这里\( a_2 \neq 0 \),这意味着\( 5v_1 - 4v_2, v_2, v_3, \dots, v_m \)线性相关。

练习2.A.8

题目:

Prove or give a counterexample: If \( v_1, v_2, \dots, v_m \) is a linearly independent list of vectors in \( V \) and \( \lambda \in \mathbf{F} \) with \( \lambda \neq 0 \), then \( \lambda v_1, \lambda v_2, \dots, \lambda v_m \) is linearly independent.

证明:

\( \forall a_1, a_2, \dots, a_m \in \mathbf{F} \)满足\( a_1(\lambda v_1) + a_2(\lambda v_2) + \dots + a_m(\lambda v_m) = 0 \),可得\( (a_1 \lambda)v_1 + (a_2 \lambda)v_2 + \dots + (a_m \lambda)v_m = 0 \),因为\( v_1, v_2, \dots, v_m \)线性无关,因此\( a_1 \lambda = a_2 \lambda = \dots = a_m \lambda = 0 \),再由\( \lambda \neq 0 \),可得\( a_1 = a_2 = \dots = a_m = 0 \),综上可得\( \lambda v_1, \lambda v_2, \dots, \lambda v_m \)线性无关。

证毕。

练习2.A.9

题目:

Prove or give a counterexample: If \( v_1, \dots, v_m \) and \( w_1, \dots, w_m \) are linearly independent lists of vectors in \( V \), then \( v_1 + w_1, \dots, v_m + w_m \) is linearly independent.

解答:

不成立,举个反例,令\( V = \mathbf{R}^2 \),令\( v_1 = (0, 1) , v_2 = (1, 0), w_1 = (0, -1), w_2 = (-1, 0) \),则\( v_1, v_2 \)和\( w_1, w_2 \)均线性无关,但\( v_1 + w_1, v_2 + w_2 = (0, 0), (0, 0) \)线性相关。

练习2.A.10

题目:

Suppose \( v_1, \dots, v_m \) is linearly independent in \( V \) and \( w \in V \). Prove that if \( v_1 + w, \dots, v_m + w \) is linearly dependent, then \( w \in \text{span}(v_1, \dots, v_m) \).

证明:

由\( v_1 + w, \dots, v_m + w \)线性相关,可得\( \exists \)不全为\( 0 \)的数\( a_1, \dots, a_m \in \mathbf{F} \),使得\( a_1(v_1 + w) + \dots + a_m(v_m + w) = 0 \),不妨设\( a_1 \neq 0 \),则 \( a_1(v_1 + w) = -a_2(v_2 + w) - \dots - a_m(v_m + w) \),可得\( a_1v_1 + a_1w = (-a_2v_2 - \dots - a_mv_m) + (-a_2 - \dots - a_m)w \),进而可得\( w = \dfrac{-a_1v_1 - \dots - a_mv_m}{a_1 + \dots + a_m} = \dfrac{-a_1}{a_1 + \dots + a_m}v_1 - \dots - \dfrac{a_m}{a_1 + \dots + a_m}v_m \),这意味着\( w \in \text{span}(v_1, \dots, v_m) \)。

证毕。

练习2.A.11

题目:

Suppose \( v_1, \dots, v_m \) is linearly independent in \( V \) and \( w \in V \). Show that \( v_1, \dots, v_m, w \) is linearly independent if and only if

\[ w \notin \text{span}(v_1, \dots, v_m) \].

证明:

必要性:

如果\( v_1, \dots, v_m, w \)线性无关,此时假设\( w \in \text{span}(v_1, \dots, v_m) \),则\( \exists a_1, \dots, a_m \in \mathbf{F} \)使得 \( w = a_1v_1 + \dots + a_mv_m \),进而可得\( a_1v_1 + \dots + a_mv_m - w = 0 \),这里\( w \)的系数\( -1 \)不为\( 0 \),这和\( v_1, \dots, v_m, w \)线性无关矛盾,因此假设不成立,有\( w \notin \text{span}(v_1, \dots, v_m) \)。

充分性:

如果\( w \notin \text{span}(v_1, \dots, v_m) \),则\( \forall a_1, \dots, a_m, a_{m + 1} \in \mathbf{F} \)满足 \( a_1v_1 + \dots + a_mv_m + a_{m + 1}w = 0 \) (1) ,可得\( a_{m + 1}w = -a_1v_1 - \dots - a_mv_m \),这里由于\( w \notin \text{span}(v_1, \dots, v_m) \),因此必须有\( a_{m + 1} = 0 \) (否则两边乘以\( \dfrac{1}{a_{m + 1}} \)会得到\( w \in \text{span}(v_1, \dots, v_m) \),矛盾),将\( a_{m + 1} = 0 \)代回到(1)可得,\( a_1v_1 + \dots + a_mv_m = 0 \),进而由\( v_1, \dots, v_m \)线性无关,可得\( a_1 = \dots = a_m = 0 \),综上,有\( a_1 = \dots = a_m = a_{m + 1} = 0 \),这意味着\( v_1, \dots, v_m, w \)线性无关。

证毕。

练习2.A.12

题目:

Explain why there does not exist a list of six polynomials that is linearly independent in \( \mathcal{P}_4(\mathbf{F}) \).

解答:

我们知道\( \text{span}(1, z, \dots, z^4) = \mathcal{P}_4(\mathbf{F}) \),假设存在由\( 6 \)个多项式组成的线性无关列表\( x_1, \dots, x_6 \),则根据2.23,可得\( 6 \leq 5 \),矛盾,因此假设不成立,不存在由\( 6 \)个多项式组成的线性无关列表。

练习2.A.13

题目:

Explain why no list of four polynomials spans \( \mathcal{P}_4(\mathbf{F}) \).

解答:

我们知道\( \text{span}(1, z, \dots, z^4) = \mathcal{P}_4(\mathbf{F}) \) 且\( 1, z, \dots, z^4 \)线性无关,假设存在由\( 4 \)个多项式组成的列表\( x_1, \dots, x_4 \)生成\( \mathcal{P}_4(\mathbf{F}) \),则根据2.23,可得\( 4 \leq 5 \),矛盾,因此假设不成立,不存在由\( 4 \)个多项式组成的列表\( x_1, \dots, x_4 \)生成\( \mathcal{P}_4(\mathbf{F}) \)。

练习2.A.14

题目:

Prove that \( V \) is infinite-dimensional if and only if there is a sequence \( v_1, v_2, \dots \) of vectors in \( V \) such that \( v_1, \dots, v_m \) is linearly independent for every positive integer \( m \).

证明:

必要性:

如果\( V \)无限维,则我们构造序列\( v_1, v_2, \dots \),\( \forall i \geq 1, v_i \in V \),归纳假设\( \forall 1 \leq m' < m, v_{m'} \)已定义且\( v_1, \dots, v_{m'} \)线性无关,我们定义\( v_m \),因为\( V \)无限维,因此\( v_1, v_2, \dots, v_{m - 1} \)不能生成\( V \)(否则\( V \)就是有限维,矛盾),进而存在\( v \in V \setminus \text{span}(v_1, v_2, \dots, v_{m - 1}) \),令\( v_m = v \),由归纳假设,得\( v_1, v_2, \dots, v_{m - 1} \)线性无关,此时根据练习2.A.11,可得\( v_1, v_2, \dots, v_m \)也线性无关,归纳完毕,我们得到了一个序列\( v_1, v_2, \dots \)满足\( \forall m \geq 1, v_1, \dots, v_m \)线性无关。

充分性:

如果存在序列\( v_1, v_2, \dots \)满足\( \forall i \geq 1, v_i \in V \)且 \( \forall m \geq 1, v_1, \dots, v_m \)线性无关,此时假设\( V \)有限维,则存在列表\( w_1, \dots, w_m \)生成\( V \),但\( v_1, \dots, v_{m + 1} \)线性无关,而\( m + 1 > m \),这和2.23矛盾,因此假设不成立,V是无限维的。

证毕。

练习2.A.15

题目:

Prove that \( \mathbf{F}^{\infty} \) is infinite-dimensional.

注:

\( \mathbf{F}^{\infty} \)的定义见1.22。

证明:

构造序列\( x_1, x_2, \dots \),令\( x_1 = (1, 0, \dots) \),令\( x_2 = (0, 1, \dots) \),以此类推,可得\( \forall m \geq 1, x_m \in \mathbf{F}^{\infty} \)且\( x_1, \dots, x_m \)线性无关,根据练习2.A.14,可得\( \mathbf{F}^{\infty} \)无限维。

证毕。

练习2.A.16

题目:

Prove that the real vector space of all continuous real-valued functions on the interval \( [0, 1] \) is infinite-dimensional.

证明:

构造序列\( f_1, f_2, \dots \), \( \forall i \geq 1 \),令\( f_i = z^{i - 1} \),可得\( \forall m \geq 1, f_m \)为\( [0, 1] \)上的实值连续函数,且\( f_1, \dots, f_m \)线性无关,根据练习2.A.14,可得目标线性空间是无限维的。

证毕。

练习2.A.17

题目:

Suppose \( p_0, p_1, \dots, p_m \) are polynomials in \( \mathcal{P}_m(\mathbf{F}) \) such that \( p_j(2) = 0 \) for each \( j \). Prove that \( p_0, p_1, \dots, p_m \) is not linearly independent in \( \mathcal{P}_m(\mathbf{F}) \).

注:

如果用2.39的话,则证明会简化很多,但是2.39是2.C的内容,这里选择不用,虽然用也没问题,不会有循环依赖的问题。

证明:

令\( V = \{ p \in \mathcal{P}_m({\mathbf{F}}) : p(2) = 0 \} \),易得\( V \)是向量空间,我们证明\( \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) = V \):

先证明\( \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) \subseteq V \), \( \forall p \in \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) \),有\( \exists a_1, a_2, \dots, a_m \in \mathbf{F} \)使得 \( p = a_1(z - 2) + a_2(z^2 - 2^2) + \dots + a_m(z^m - 2^m) \),明显有\( p \in \mathcal{P}_m({\mathbf{F}}) \),除此之外,还有\( p(2) = a_1(2 - 2) + a_2(2^2 - 2^2) + \dots + a_m(2^m - 2^m) = 0 \),于是有\( p \in V \),综上,\( \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) \subseteq V \)。

接着证明\( V \subseteq \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) \), \( \forall p \in V \),我们有\( p \in \mathcal{P}_m({\mathbf{F}}), p(2) = 0 \),由\( \text{span}(1, z, \dots, z^m) = \mathcal{P}_m(\mathbf{F}) \),可得\( \exists a_0, a_1, \dots, a_m \in \mathbf{F} \)使得 \( p = a_0 + a_1z + \dots + a_mz^m = a_0 + (a_1(z - 2) + a_2(z^2 - 2^2) + \dots + a_m(z^m - 2^m)) + (2a_1 + 2^2a_2 + \dots + 2^ma_m) = (a_0 + 2a_1 + 2^2a_2 + \dots + 2^ma_m) + (a_1(z - 2) + a_2(z^2 - 2^2) + \dots + a_m(z^m - 2^m)) \) (1) ,又\( p(2) = 0 \),可得\( p(2) = a_0 + 2a_1 + 2^2a_2 + \dots + 2^ma_m = 0 \),回代\( a_0 + 2a_1 + 2^2a_2 + \dots + 2^ma_m = 0 \)到(1),可得\( p = a_1(z - 2) + a_2(z^2 - 2^2) + \dots + a_m(z^m - 2^m) \),这意味着\( p \in \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) \),,综上,\( V \subseteq \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) \)。

至此,我们证明了\( \text{span}(z - 2, z^2 - 2^2, \dots, z^m - 2^m) = V \),易得\( \text{span}(p_0, p_1, \dots, p_m) \subseteq V \),此时假设\( p_0, p_1, \dots, p_m \)线性无关,则根据2.23,可得\( m + 1 \leq m \),矛盾,因此假设不成立,有\( p_0, p_1, \dots, p_m \)线性相关。

证毕。

练习2.B

练习2.B.1

题目:

Find all vector spaces that have exactly one basis.

注:

这里的have exactly one basis指的是向量空间只有一个基,而不是组成基的向量数为1。

解答:

假设\( v_1, v_2, \dots, v_m \)为\( V \)的基,则易证\( v_1 + v_2, v_2, \dots, v_m \)也为\( V \)的基,因此如果\( V \)的基的长度\( \geq 2 \),则\( V \)的基的数量一定\( \geq 2 \)(注:组成基的向量一定不是\( 0 \)向量,因此\( v_1 + v_2 \neq v_1 \)),故如果\( V \)的基的数量为\( 1 \),则它基的长度一定\( \leq 1 \)(注:一个向量空间的基的长度是唯一的),因此它基的长度只有\( 0 \)和\( 1 \)这两种可能性:

  1. 如果基的长度为\( 0 \),则\( V = \{ 0 \} \)。
  2. 如果基的长度为\( 1 \),则由于任意向量空间均包含\( 0 \),因此\( 0 \in V \),除此之外,我们断言\( V \)至多还有额外的一个元素\( v \),即\( V = \{ 0, v \} \),假设\( V \)还有两个不同的元素\( v \neq 0, v' \neq 0 \),此时由于\( V \)的基的长度为\( 1 \),易得\( v, v' \)均是\( V \)长度为\( 1 \)的基,这和\( V \)只有一个基矛盾,因此\( V \)只能有一个额外的元素\( v \),当然,满足这点还不够,这里如果数域\( \mathbf{F} \)有两个以上的元素,则取\( \lambda \)满足\( \lambda \neq 0, \lambda \neq 1 \),可得\( \lambda v \)又是\( V \)的另外一个基,也矛盾,因此这种情况下还得要求\( \mathbf{F} \)至多有两个元素,其中一个元素一定是\( 0 \),同时考虑到数域要求有乘法逆元,而\( 0 \)没有乘法逆元,因此该数域还得有额外一个非\( 0 \)元素\( \lambda \),也就是\( \mathbf{F} = \{ 0, \lambda \} \),这里只要定义合适的加法,乘法运算,\( \{ 0, \lambda \} \)就可以是数域,具体的,令\( 0 + \lambda = \lambda + 0 = \lambda, \lambda + \lambda = 0, 0 + 0 = 0, 0 \times \lambda = \lambda \times 0 = 0, \lambda \times \lambda = \lambda, 0 \times 0 = 0 \),令\( 0 \)为加法单位元,\( \lambda \)为乘法单位元,容易验证\( \{ 0, \lambda \} \)为数域,注意这里\( \lambda \)的加法逆元\( -\lambda = \lambda \),再令\( \lambda \)为向量数乘的单位元,则会有\( \lambda v = v \),这意味着数乘无法产生新的向量,从而无法通过数乘\( v \)得到新的基,而\( 0v = 0 \)也无法产生新的基,可得\( V = \{ 0, v \} \)在数域\( \{ 0, \lambda \} \)下只有一个基\( v \)。

综上,只有两种情况向量空间\( V \)会只有一个基:

  1. \( V = \{ 0 \} \)。
  2. \( V \)具有形式\( \{ 0, v \} \)且对应的数域\( \mathbf{F} \)具有形式\( \{ 0, \lambda \} \),其中,\( \mathbf{F} \)内运算规则以及\( \mathbf{F} \)和\( V \)之间的运算规则见上面的讨论。

练习2.B.2

题目:

Verify all the assertions in Example 2.28.

Example 2.28的内容:

  1. The list \( (1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1) \) is a basis of \( \mathbf{F}^n \), called the standard basis of \( \mathbf{F}^n \).
  2. The list \( (1, 2), (3, 5) \) is a basis of \( \mathbf{F}^2 \).
  3. The list \( (1, 2, -4), (7, -5, 6) \) is linearly independent in \( \mathbf{F}^3 \) but is not a basis of \( \mathbf{F}^3 \) because it does not span \( \mathbf{F}^3 \).
  4. The list \( (1, 2), (3, 5), (4, 13) \) spans \( \mathbf{F}^2 \) but is not a basis of \( \mathbf{F}^2 \) because it is not linearly independent.
  5. The list \( (1, 1, 0), (0, 0, 1) \) is a basis of \( \{ (x, x, y) \in \mathbf{F}^3 : x, y \in \mathbf{F} \} \).
  6. The list \( (1, -1, 0), (1, 0, -1) \) is a basis of \( \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \).
  7. The list \( 1, z, \dots, z^m \) is a basis of \( \mathcal{P}_m(\mathbf{F}) \).

验证1:

首先证明\( (1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1) \)线性无关, \( \forall a_1, a_2, \dots, a_n \in \mathbf{F} \)满足 \( a_1(1, 0, \dots, 0) + a_2(0, 1, 0, \dots, 0) + \dots + a_n(0, \dots, 0, 1) = 0 \),可得\( a_1 = a_2 = \dots = a_jn = 0 \),从而\( (1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1) \) 线性无关。

接着证明\( \text{span}((1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1)) = \mathbf{F}^n \),显然有\( \text{span}((1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1)) \subseteq \mathbf{F}^n \),我们还得证明\( \mathbf{F}^n \subseteq \text{span}((1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1)) \), \( \forall v = (v_1, v_2, \dots, v_n) \in \mathbf{F}^n \),有\( v = v_1(1, 0, \dots, 0) + v_2(0, 1, 0, \dots, 0) + \dots + v_n(0, \dots, 0, 1) \),可得\( v \in \text{span}((1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1)) \),进而可得\( \mathbf{F}^n \subseteq \text{span}((1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1)) \),综上可得,\( \text{span}((1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1)) = \mathbf{F}^n \)。

综上,有\( (1, 0, \dots, 0), (0, 1, 0, \dots, 0), \dots, (0, \dots, 0, 1) \)为\( \mathbf{F}^n \)的基

验证2:

首先证明\( (1, 2), (3, 5) \)线性无关, \( \forall a_1, a_2 \in \mathbf{F} \)满足\( a_1(1, 2) + a_2(3, 5) = 0 \),可得\( a_1 + 3a_2 = 0, 2a_1 + 5a_2 = 0 \),解得\( a_1 = a_2 = 0 \),于是有\( (1, 2), (3, 5) \)线性无关。

接着证明\( \text{span}((1, 2), (3, 5)) = \mathbf{F}^2 \),显然有\( \text{span}((1, 2), (3, 5)) \subseteq \mathbf{F}^2 \),我们还得证明\( \mathbf{F}^2 \subseteq \text{span}((1, 2), (3, 5)) \), \( \forall v = (v_1, v_2) \in \mathbf{F}^2 \),假设\( \exists a_1, a_2 \in \mathbf{F} \)满足\( a_1(1, 2) + a_2(3, 5) = v \),可得\( a_1 + 3a_2 = v_1, 2a_1 + 5a_2 = v_2 \),解得\( a_2 = 2v_1 - v_2, a_1 = -5v_1 + 3v_2 \),这意味着\( v \in \text{span}((1, 2), (3, 5)) \),至此可得,\( \mathbf{F}^2 \subseteq \text{span}((1, 2), (3, 5)) \),综上可得,\( \text{span}((1, 2), (3, 5)) = \mathbf{F}^2 \)。

综上,有\( (1, 2), (3, 5) \)为\( \mathbf{F}^2 \)的基。

验证3:

首先证明\( (1, 2, -4), (7, -5, 6) \)线性无关, \( \forall a_1, a_2 \in \mathbf{F} \)满足\( a_1(1, 2, -4) + a_2(7, -5, 6) = 0 \),可得\( a_1 + 7a_2 = 0, 2a_1 - 5a_2 = 0, -4a_1 + 6a_2 = 0 \),解得\( a_1 = a_2 = 0 \),于是有\( (1, 2, -4), (7, -5, 6) \)线性无关。

接着证明\( \text{span}((1, 2, -4), (7, -5, 6)) \neq \mathbf{F}^3 \),令\( v = (8, 0, 0) \in \mathbf{F}^3 \),假设\( v \in \text{span}((1, 2, -4), (7, -5, 6)) \),则\( \exists a_1, a_2 \in \mathbf{F} \)满足\( a_1(1, 2, -4) + a_2(7, -5, 6) = v \),可得\( a_1 + 7a_2 = 8, 2a_1 - 5a_2 = 0, -4a_1 + 6a_2 = 0 \),由\( 2a_1 - 5a_2 = 0, -4a_1 + 6a_2 = 0 \)解得\( a_1 = a_2 = 0 \),但是这样的话,会有\( a_1 + 7a_2 = 0 \neq 8 \),矛盾,因此假设不成立,有\( v \notin \text{span}((1, 2, -4), (7, -5, 6)) \),这意味着\( \text{span}((1, 2, -4), (7, -5, 6)) \neq \mathbf{F}^3 \),进而\( (1, 2, -4), (7, -5, 6) \)不是\( \mathbf{F}^3 \)的基。

验证4:

首先证明\( \text{span}((1, 2), (3, 5), (4, 13)) = \mathbf{F}^2 \),明显有\( \text{span}((1, 2), (3, 5), (4, 13)) \subseteq \mathbf{F}^2 \),我们还得证明\( \mathbf{F}^2 \subseteq \text{span}((1, 2), (3, 5), (4, 13)) \) \( \forall v = (v_1, v_2) \in \mathbf{F}^2 \),假设\( \exists a_1, a_2, a_3 \in \mathbf{F} \)满足 \( a_1(1, 2) + a_2(3, 5) + a_3(4, 13) = v \),可得\( a_1 + 3a_2 + 4a_3 = v_1, 2a_1 + 5a_2 + 13a_3 = v_2 \),解得\( a_1 = -5v_1 + 3v_2 - 19a_3, a_2 = 2v_1 - v_2 + 5a_3 \),其中\( a_3 \)为自由变量,取\( a_3 = 1 \),则\( a_1 = -5v_1 + 3v_2 - 19, a_2 = 2v_1 - v_2 + 5 \),可得\( v \in \text{span}((1, 2), (3, 5), (4, 13)) \),综上,有\( \mathbf{F}^2 \subseteq \text{span}((1, 2), (3, 5), (4, 13)) \),至此可得,\( \text{span}((1, 2), (3, 5), (4, 13)) = \mathbf{F}^2 \)。

接着证明\( (1, 2), (3, 5), (4, 13) \)线性相关,根据上面的讨论,令\( a_3 = 1, a_1 = -5 \times 0 + 3 \times 0 - 19 \times 1 = -19, a_2 = 2 \times 0 - 0 + 5 \times 1 = 5 \),则可得\( a_1(1, 2) + a_2(3, 5) + a_3(4, 13) = 0 \),这意味着\( (1, 2), (3, 5), (4, 13) \)线性相关。

验证5:

令\( V = \{ (x, x, y) \in \mathbf{F}^3 : x, y \in \mathbf{F} \} \)。

首先证明\( (1, 1, 0), (0, 0, 1) \)线性无关, \( \forall a_1, a_2 \in \mathbf{F} \)满足\( a_1(1, 1, 0) + a_2(0, 0, 1) = 0 \),可得\( a_1 = a_2 = 0 \),于是有\( (1, 1, 0), (0, 0, 1) \)线性无关。

接着证明\( \text{span}((1, 1, 0), (0, 0, 1)) = V \):

先证明\( \text{span}((1, 1, 0), (0, 0, 1)) \subseteq V \), \( \forall a_1, a_2 \in \mathbf{F} \),有\( a_1(1, 1, 0) + a_2(0, 0, 1) = (a_1, a_1, a_2) \in V \),可得\( \text{span}((1, 1, 0), (0, 0, 1)) \subseteq V \)。

再证明\( V \subseteq \text{span}((1, 1, 0), (0, 0, 1)) \), \( \forall v = (x, x, y) \in V \),令\( a_1 = x, a_2 = y \) ,可得\( a_1(1, 1, 0) + a_2(0, 0, 1) = v \),进而\( v \in \text{span}((1, 1, 0), (0, 0, 1)) \),于是有\( V \subseteq \text{span}((1, 1, 0), (0, 0, 1)) \)。

综上,可得\( \text{span}((1, 1, 0), (0, 0, 1)) = V \)。

至此,我们有\( (1, 1, 0), (0, 0, 1) \)为\( V \)的基。

验证6:

令\( V = \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \)。

首先证明\( (1, -1, 0), (1, 0, -1) \)线性无关, \( \forall a_1, a_2 \in \mathbf{F} \)满足\( a_1(1, -1, 0) + a_2(1, 0, -1) = 0 \),可得\( a_1 + a_2 = 0, -a_1 = 0, -a_2 = 0 \),解得\( a_1 = a_2 = 0 \),于是有\( (1, -1, 0), (1, 0, -1) \)线性无关。

接着证明\( \text{span}((1, -1, 0), (1, 0, -1)) = V \):

先证明\( \text{span}((1, -1, 0), (1, 0, -1)) \subseteq V \), \( \forall a_1, a_2 \in \mathbf{R} \),有\( a_1(1, -1, 0) + a_2(1, 0, -1) = (a_1 + a_2, -a_1, -a_2) \),这里三个坐标\( (a_1 + a_2) + (-a_1) + (-a_2) = 0 \),可得\( a_1(1, -1, 0) + a_2(1, 0, -1) = (a_1 + a_2, -a_1, -a_2) \in V \),于是有\( \text{span}((1, -1, 0), (1, 0, -1)) \subseteq V \)。

再证明\( V \subseteq \text{span}((1, -1, 0), (1, 0, -1)) \), \( \forall v = (x, y, z) \in V \),有\( x + y + z = 0 \),进而有\( x = -y - z \),令\( a_1 = -y, a_2 = -z \),可得\( a_1(1, -1, 0) + a_2(1, 0, -1) = (a_1 + a_2, -a_1, -a_2) = (-y - z, y, z) = (x, y, z) \),于是有\( v \in \text{span}((1, -1, 0), (1, 0, -1)) \),至此可得\( V \subseteq \text{span}((1, -1, 0), (1, 0, -1)) \)。

综上,可得\( \text{span}((1, -1, 0), (1, 0, -1)) = V \)。

至此,我们有\( \text{span}((1, -1, 0), (1, 0, -1)) \)为\( V \)的基。

验证7:

我们在练习2.A.2的4中已经证明了\( 1, z, \dots, z^m \)线性无关了,这里直接用。

接着证明\( \text{span}(1, z, \dots, z^m) = \mathcal{P}_m(\mathbf{F}) \),显然有\( \text{span}(1, z, \dots, z^m) \subseteq \mathcal{P}_m(\mathbf{F}) \),还得证明\( \mathcal{P}_m(\mathbf{F}) \subseteq \text{span}(1, z, \dots, z^m) \), \( \forall p \in \mathcal{P}_m(\mathbf{F}) \),根据\( \mathcal{P}_m(\mathbf{F}) \)的定义,有\( p = a_0 + a_1z + \dots + a_mz^m \),可得\( p \in \text{span}(1, z, \dots, z^m) \),于是有\( \mathcal{P}_m(\mathbf{F}) \subseteq \text{span}(1, z, \dots, z^m) \),至此可得\( \text{span}(1, z, \dots, z^m) = \mathcal{P}_m(\mathbf{F}) \)。

综上,我们有\( 1, z, \dots, z^m \)为\( \mathcal{P}_m(\mathbf{F}) \)的基。

练习2.B.3

题目:

  1. Let \( U \) be the subspace of \( \mathbf{R}^5 \) defined by \( U = \{ (x_1, x_2, x_3, x_4, x_5) \in \mathbf{R}^5 : x_1 = 3x_2 \text{ and } x_3 = 7x_4 \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathbf{R}^5 \).
  3. Find a subspace \( W \) of \( \mathbf{R}^5 \) such that \( \mathbf{R}^5 = U \bigoplus W \).

解答:

解答1:

令\( u_1 = (3, 1, 0, 0, 0), u_2 = (0, 0, 7, 1, 0), u_3 = (0, 0, 0, 0, 1) \)。

我们要证明\( u_1, u_2, u_3 \)为\( U \)的基,这得证明\( u_1, u_2, u_3 \)线性无关并且生成\( U \)。

先证明\( u_1, u_2, u_3 \)线性无关, \( \forall a_1, a_2, a_3 \in \mathbf{R}^5 \)满足 \( a_1u_1 + a_2u_2 + a_3u_3 = 0 \),可得\( a_1 = a_2 = a_3 = 0 \),这意味着\( u_1, u_2, u_3 \)线性无关。

接着证明\( \text{span}(u_1, u_2, u_3) = U \):

先证明\( \text{span}(u_1, u_2, u_3) \subseteq U \), \( \forall a_1, a_2, a_3 \in \mathbf{R}^5 \),有\( a_1u_1 + a_2u_2 + a_3u_3 = (3a_1, a_1, 7a_2, a_2, a_3) \in U \),可得\( \text{span}(u_1, u_2, u_3) \subseteq U \)。

再证明\( U \subseteq \text{span}(u_1, u_2, u_3) \), \( \forall v = (x_1, x_2, x_3, x_4, x_5) \in U \),有\( x_1 = 3x_2, x_3 = 7x_4 \),令\( a_2 = x_2, a_1 = 3x_2, a_4 = x_4, a_3 = 7x_4 \),可得\( a_1u_1 + a_2u_2 + a_3u_3 = v \),进而可得\( v \in \text{span}(u_1, u_2, u_3) \),至此有\( U \subseteq \text{span}(u_1, u_2, u_3) \)。

综上可得,\( \text{span}(u_1, u_2, u_3) = U \)。

至此,我们确定了\( u_1, u_2, u_3 \)为\( U \)的基。

解答2:

额外的,令\( u_4 = (1, 0, 0, 0, 0), u_5 = (0, 0, 1, 0, 0) \)。

我们证明\( u_1, u_2, u_3, u_4, u_5 \)为\( \mathbf{R}^5 \)的基:

先证明\( u_1, u_2, u_3, u_4, u_5 \)线性无关, \( \forall a_1, a_2, a_3, a_4, a_5 \in \mathbf{R}^5 \)满足 \( a_1u_1 + a_2u_2 + a_3u_3 + a_4u_4 + a_5u_5 = 0 \),可得\( 3a_1 + a_4 = 0, a_1 = 0, 7a_2 + a_5 = 0, a_2 = 0, a_3 = 0 \),解得\( a_1 = a_2 = a_3 = a_4 = a_5 = 0 \),这意味着\( u_1, u_2, u_3, u_4, u_5 \)线性无关。

接着证明\( \text{span}(u_1, u_2, u_3, u_4, u_5) = \mathbf{R}^5 \),显然有\( \text{span}(u_1, u_2, u_3, u_4, u_5) \subseteq \mathbf{R}^5 \),还得证明\( \mathbf{R}^5 \subseteq \text{span}(u_1, u_2, u_3, u_4, u_5) \), \( \forall v = (x_1, x_2, x_3, x_4, x_5) \in \mathbf{R}^5 \),令\( a_1 = x_2, a_4 = -3x_2 + x_1, a_2 = x_4, a_5 = -7x_4 + x_3, a_3 = x_5 \),可得\( a_1u_1 + a_2u_2 + a_3u_3 + a_4u_4 + a_5u_5 = v \),于是有\( \mathbf{R}^5 \subseteq \text{span}(u_1, u_2, u_3, u_4, u_5) \),至此可得,\( \text{span}(u_1, u_2, u_3, u_4, u_5) = \mathbf{R}^5 \)。

综上,拓展后的\( u_1, u_2, u_3, u_4, u_5 \)为\( \mathbf{R}^5 \)的基。

解答3:

令\( W = \text{span}(u_4, u_5) \),易得\( W \)为\( \mathbf{R}^5 \)的子空间,再根据2.34的证明(注:看证明的内容,而不是2.34定理本身的内容),可得\( \mathbf{R}^5 = U \bigoplus W \)。

练习2.B.4

题目:

  1. Let \( U \) be the subspace of \( \mathbf{C}^5 \) defined by \( U = \{ (z_1, z_2, z_3, z_4, z_5) \in \mathbf{C}^5 : 6z_1 = z_2 \text{ and } z_3 + 2z_4 + 3z_5 = 0 \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathbf{C}^5 \).
  3. Find a subspace \( W \) of \( \mathbf{C}^5 \) such that \( \mathbf{C}^5 = U \bigoplus W \).

解答:

解答1:

令\( u_1 = (1, 6, 0, 0, 0), u_2 = (0, 0, 2, -1, 0), u_3 = (0, 0, 3, 0, -1) \),

我们要证明\( u_1, u_2, u_3 \)为\( U \)的基,这得证明\( u_1, u_2, u_3 \)线性无关并且生成\( U \)。

先证明\( u_1, u_2, u_3 \)线性无关, \( \forall a_1, a_2, a_3 \in \mathbf{R}^5 \)满足 \( a_1u_1 + a_2u_2 + a_3u_3 = 0 \),可得\( a_1 = 0, 2a_2 + 3a_3 = 0, a_3 = 0 \),解得\( a_1 = a_2 = a_3 = 0 \),这意味着\( u_1, u_2, u_3 \)线性无关。

接着证明\( \text{span}(u_1, u_2, u_3) = U \):

先证明\( \text{span}(u_1, u_2, u_3) \subseteq U \), \( \forall a_1, a_2, a_3 \in \mathbf{R}^5 \),有\( a_1u_1 + a_2u_2 + a_3u_3 = (a_1, 6a_1, 2a_2 + 3_a3, -a_2, -a_3) \in U \),可得\( \text{span}(u_1, u_2, u_3) \subseteq U \)。

再证明\( U \subseteq \text{span}(u_1, u_2, u_3) \), \( \forall v = (z_1, z_2, z_3, z_4, z_5) \in U \),有\( 6z_1 = z_2, z_3 + 2z_4 + 3z_5 = 0 \),可得\( v = (z_1, 6z_1, -2z_4, - 3z_5, z_4, z_5) \),令\( a_1 = z_1, a_2 = -z_4, a_3 = -z_5 \),可得\( a_1u_1 + a_2u_2 + a_3u_3 = (z_1, 6z_1, -2z_4 - 3z_5, z_4, z_5) = v \),进而可得\( v \in \text{span}(u_1, u_2, u_3) \),至此有\( U \subseteq \text{span}(u_1, u_2, u_3) \)。

综上可得,\( \text{span}(u_1, u_2, u_3) = U \)。

至此,我们确定了\( u_1, u_2, u_3 \)为\( U \)的基。

解答2:

额外的,令\( u_4 = (1, 0, 0, 0, 0), u_5 = (0, 0, 1, 0, 0) \)。

我们证明\( u_1, u_2, u_3, u_4, u_5 \)为\( \mathbf{C}^5 \)的基:

先证明\( u_1, u_2, u_3, u_4, u_5 \)线性无关, \( \forall a_1, a_2, a_3, a_4, a_5 \in \mathbf{C}^5 \)满足 \( a_1u_1 + a_2u_2 + a_3u_3 + a_4u_4 + a_5u_5 = 0 \),可得\( a_1 + a_4 = 0, 6a_1 = 0, 2a_2 + 3a_3 + a_5 = 0, -a_2 = 0, -a_3 = 0 \),解得\( a_1 = a_2 = a_3 = a_4 = a_5 = 0 \),这意味着\( u_1, u_2, u_3, u_4, u_5 \)线性无关。

接着证明\( \text{span}(u_1, u_2, u_3, u_4, u_5) = \mathbf{C}^5 \),显然有\( \text{span}(u_1, u_2, u_3, u_4, u_5) \subseteq \mathbf{C}^5 \),还得证明\( \mathbf{C}^5 \subseteq \text{span}(u_1, u_2, u_3, u_4, u_5) \), \( \forall v = (z_1, z_2, z_3, z_4, z_5) \in \mathbf{C}^5 \),令\( a_1 = \dfrac{z_2}{6}, a_4 = -\dfrac{z_2}{6} + z_1, a_2 = -z_4, a_3 = -z_5, a_5 = 2z_4 + 3z_5 + z_3 \),可得\( a_1u_1 + a_2u_2 + a_3u_3 + a_4uz_4 + a_5u_5 = v \),于是有\( \mathbf{C}^5 \subseteq \text{span}(u_1, u_2, u_3, u_4, u_5) \),至此可得,\( \text{span}(u_1, u_2, u_3, u_4, u_5) = \mathbf{C}^5 \)。

综上,拓展后的\( u_1, u_2, u_3, u_4, u_5 \)为\( \mathbf{C}^5 \)的基。

解答3:

令\( W = \text{span}(u_4, u_5) \),易得\( W \)为\( \mathbf{C}^5 \)的子空间,再根据2.34的证明,可得\( \mathbf{C}^5 = U \bigoplus W \)。

练习2.B.5

题目:

Prove or disprove: there exists a basis \( p_0, p_1, p_2, p_3 \) of \( \mathcal{P}_3(\mathbf{F}) \) such that none of the polynomials \( p_0, p_1, p_2, p_3 \) has degree 2.

解答:

结论是正确的,比如\( 1 + x^3, x + x^3, x^2 + x^3, x^3 \)就是\( \mathcal{P}_3(\mathbf{F}) \)的基,但是基中没有一个多项式的次数是2。

我们证明下\( 1 + x^3, x + x^3, x^2 + x^3, x^3 \)是\( \mathcal{P}_3(\mathbf{F}) \)的基:

首先证明\( 1 + x^3, x + x^3, x^2 + x^3, x^3 \)线性无关, \( \forall a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_1(1 + x^3) + a_2(x + x^3) + a_3(x^2 + x^3) + a_4x^3 = 0 \),可得\( a_1 + a_2x + a_3x^2 + (a_1 + a_2 + a_3 + a_4)x^3 = 0 \),因为\( 1, x, x^2, x_3 \)为\( \mathcal{P}_3(\mathbf{F}) \)的基,可得\( a_1 = 0, a_2 = 0, a_3 = 0, a_1 + a_2 + a_3 + a_4 = 0 \),解得\( a_1 = a_2 = a_3 = a_4 = 0 \),有\( 1 + x^3, x + x^3, x^2 + x^3, x^3 \)线性无关。

接着证明\( \text{span}(1 + x^3, x + x^3, x^2 + x^3, x^3) = \mathcal{P}_3(\mathbf{F}) \),显然有\( \text{span}(1 + x^3, x + x^3, x^2 + x^3, x^3) \subseteq \mathcal{P}_3(\mathbf{F}) \),还得证明\( \mathcal{P}_3(\mathbf{F}) \subseteq \text{span}(1 + x^3, x + x^3, x^2 + x^3, x^3) \), \( \forall p = a_0 + a_1x + a_2x^2 + a_3x^3 \in \mathcal{P}_3(\mathbf{F}) \),可得\( p = a_0(1 + x^3) + a_1(x + x^3) + a_2(x^2 + x^3) + (-a_0 - a_1 - a_2 + a_3)x^3 \),这意味着\( p \in \text{span}(1 + x^3, x + x^3, x^2 + x^3, x^3) \),至此有\( \mathcal{P}_3(\mathbf{F}) \subseteq \text{span}(1 + x^3, x + x^3, x^2 + x^3, x^3) \),综上可得,\( \text{span}(1 + x^3, x + x^3, x^2 + x^3, x^3) = \mathcal{P}_3(\mathbf{F}) \)。

综上,\( 1 + x^3, x + x^3, x^2 + x^3, x^3 \)是\( \mathcal{P}_3(\mathbf{F}) \)的基。

练习2.B.6

题目:

Suppose \( v_1, v_2, v_3, v_4 \) is a basis of \( V \). Prove that

\[ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \]

is also a basis of V.

证明:

首先证明\( v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \)线性无关, \( \forall a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_1(v_1 + v_2) + a_2(v_2 + v_3) + a_3(v_3 + v_4) + a_4v_4 = 0 \),可得\( a_1v_1, (a_1 + a_2)v_2, (a_2 + a_3)_v3 + (a_3 + a_4)v_4 = 0 \),由\( v_1, v_2, v_3, v_4 \)为\( V \)的基,可得\( a_1 = a_1 + a_2 = a_2 + a_3 = a_3 + a_4 = 0 \),解得\( a_1 = a_2 = a_3 = a_4 = 0 \),于是有\( v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \)线性无关。

接着证明\( \text{span}(v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4) = V \),显然有\( \text{span}(v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4) \subseteq V \),还得证明\( V \subseteq \text{span}(v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4) \), \( \forall v \in V \),由\( v_1, v_2, v_3, v_4 \)为\( V \)的基,可得\( \exists a_1, a_2, a_3, a_4 \in \mathbf{F} \)使得 \( v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 \),进而可得\( v = a_1(v_1 + v_2) + (-a_1 + a_2)(v_2 + v_3) + (a_1 - a_2 + a_3)(v_3 + v_4) + (-a_1 + a_2 - a_3 + a_4)v_4 \),这意味着\( v \in \text{span}(v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4) \),于是有\( V \subseteq \text{span}(v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4) \)。至此可得,\( \text{span}(v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4) = V \)。

综上,\( v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \)为\( V \)的基。

证毕。

练习2.B.7

题目:

Prove or give a counterexample: If \( v_1, v_2, v_3, v_4 \) is a basis of \( V \) and \( U \) is a subspace of \( V \) such that \( v_1, v_2 \in U \) and \( v_3 \notin U \) and \( v_4 \notin U \), then \( v_1, v_2 \) is a basis of \( U \).

解答:

结论不成立,反例:

令\( V = \mathbf{R}^4, v_1 = (1, 0, 0, 0), v_2 = (0, 1, 0, 0), v_3 = (0, 0, 1, 0), v_4 = (0, 0, 1, 1) \),易得\( v_1, v_2, v_3, v_4 \)为\( V \)的基。

令\( U = \{ (x_1, x_2, x_3, 0) : x_1, x_2, x_3 \in \mathbf{R} \} \),可得\( v_3 \notin U, v_4 \notin U \),但\( v_1, v_2 \)只有前两个坐标不为\( 0 \),无法生成\( U \),不是\( U \)的基。

练习2.B.8

题目:

Suppose \( U \) and \( W \) are subspaces of \( V \) such that \( V = U \bigoplus W \). Suppose also that \( u_1, \dots, u_m \) is a basis of \( U \) and \( w_1, \dots, w_n \) is a basis of \( W \). Prove that

\[ u_1, \dots, u_m, w_1, \dots, w_n \]

is a basis of \( V \).

解答:

首先证明\( u_1, \dots, u_m, w_1, \dots, w_n \)线性无关, \( \forall a_1, \dots, a_m, b_1, \dots b_n \in \mathbf{F} \)满足 \( a_1u_1 + \dots + a_mu_m + b_1w_1 + \dots + b_nw_n = 0 \),可得\( a_1u_1 + \dots + a_mu_m = -b_1w_1 - \dots - b_nw_n \),令\( v = a_1u_1 + \dots + a_mu_m = -b_1w_1 - \dots - b_nw_n \),由于\( u_1, \dots, u_m \)为\( U \)的基,可得\( v \in U \),再由\( w_1, \dots, w_n \)为\( W \)的基,可得\( v \in W \),于是有\( v \in U \cap W \),根据1.45,有\( U \cap W = \{ 0 \} \),于是有\( v = 0 \),可得\( a_1u_1 + \dots + a_mu_m = -b_1w_1 - \dots - b_nw_n = 0 \),分别由\( u_1, \dots, u_m \)以及\( w_1, \dots, w_n \)线性无关,解得\( a_1 = \dots = a_m = 0 \)以及\( b_1 = \dots = b_n = 0 \),至此可得\( u_1, \dots, u_m, w_1, \dots, w_n \)线性无关。

接着证明\( \text{span}(u_1, \dots, u_m, w_1, \dots, w_n) = V \),显然有\( \text{span}(u_1, \dots, u_m, w_1, \dots, w_n) \subseteq V \),还得证明\( V \subseteq \text{span}(u_1, \dots, u_m, w_1, \dots, w_n) \), \( \forall v \in V \),由\( V = U \bigoplus W \),可得\( v \in U \bigoplus W \),进而有\( \exists u \in U, w \in W \)使得\( v = u + w \),由\( u \in U \)以及\( u_1, \dots, u_m \)为\( U \)的基,可得\( \exists a_1, \dots, a_m \in \mathbf{F} \)使得\( u = a_1u_1 + \dots + a_mu_m \),同理可得\( \exists b_1, \dots, b_n \in \mathbf{F} \)使得\( u = b_1u_1 + \dots + b_nu_n \),进而可得\( v = u + w = a_1u_1 + \dots + a_mu_m + b_1u_1 + \dots + b_nu_n \),这意味着\( v \in \text{span}(u_1, \dots, u_m, w_1, \dots, w_n) \),至此有\( V \subseteq \text{span}(u_1, \dots, u_m, w_1, \dots, w_n) \),综上可得,\( \text{span}(u_1, \dots, u_m, w_1, \dots, w_n) = V \)。

综上,\( u_1, \dots, u_m, w_1, \dots, w_n \)为\( V \)的基。

练习2.C

练习2.C.1

题目:

Suppose \( V \) is finite-dimensional and \( U \) is a subspace of ( V ) such that \( \dim U = \dim V \). Prove that \( U = V \).

附注:

用3.59可以直接推出结论成立,但是这是后面的内容,现在不能用。

证明:

明显的,有\( U \subseteq V \),我们还得证明\( V \subseteq U \), 令\( n = \dim U = \dim V \),任取\( U \)的基\( u_1, u_2, \dots, u_n \),我们证明该列表也是\( V \)的基,首先,我们有\( u_1 \in V, u_2 \in V, \dots, u_n \in V \),同时,由于该列表是\( U \)的基,因此组成该列表的向量线性无关,加上列表的长度为\( n = \dim V \),根据2.39,可得 \( u_1, u_2, \dots, u_n \)为\( V \)的基,进而\( V \)中的所有向量都可以表达为\( u_1, u_2, \dots, u_n \)的线性组合,可得\( V \subseteq U \),综上,有\( U = V \)。

证毕。

练习2.C.2

题目:

Show that the subspaces of \( \mathbf{R}^2 \) are precisely \( \{ 0 \}, \mathbf{R}^2 \) and all lines in \( \mathbf{R}^2 \) through the origin.

证明:

根据2.38,我们知道\( \mathbf{R}^2 \)的子空间\( U \)的维度只能是\( 0, 1, 2 \),这里如果\( \dim U = 0 \),则\( U = \{ 0 \} \),如果\( \dim U = 2 \),则根据练习2.C.1,可得\( U = V \),最后如果\( \dim U = 1 \),则\( U \)的基只会由一个向量组成,任取\( U \)的基\( u \),我们有\( U = \text{span}(u) = \{ au : a \in \mathbf{R} \} \),可得\( U \)为过原点的直线。

证毕。

练习2.C.3

题目:

Show that the subspaces of \( \mathbf{R}^3 \) are precisely \( \{ 0 \}, \mathbf{R}^3 \), all lines in \( \mathbf{R}^3 \) through the origin, and all planes in \( \mathbf{R}^3 \) through the origin.

证明:

根据2.38,我们知道\( \mathbf{R}^3 \)的子空间\( U \)的维度只能是\( 0, 1, 2, 3 \)。

如果\( \dim U = 0 \),则\( U = \{ 0 \} \)。

如果\( \dim U = 3 \),则根据练习2.C.1,可得\( U = V \)。

如果\( \dim U = 1 \),则\( U \)的基只会由一个向量组成,任取\( U \)的基\( u \),我们有\( U = \text{span}(u) = \{ au : a \in \mathbf{R} \} \),可得\( U \)为过原点的直线。

如果\( \dim U = 2 \),则\( U \)的基会由两个向量组成,任取\( U \)的基\( u_1, u_2 \),我们有\( U = \text{span}(u_1, u_2) = \{ a_1u_1 + a_2u_2 : a \in \mathbf{R} \} \),这里由于\( u_1, u_2 \)线性无关,因此\( u_1, u_2 \)不共线,进而\( \{ a_1u_1 + a_2u_2 : a \in \mathbf{R} \} \)为平面而非直线,加上\( 0u_1 + 0u_2 = 0 \),可得该平面过原点,综上,有\( U \)为过原点的平面。

证毕。

练习2.C.4

题目:

  1. Let \( U = \{ p \in \mathcal{P}_4(\mathbf{F}) : p(6) = 0 \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathcal{P}_4(\mathbf{F}) \).
  3. Find a subspaceia \( W \) of \( \mathcal{P}_4(\mathbf{F}) \) such that \( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \).

解答:

解答1:

\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3 \)为\( U \)的基,我们证明这点:

首先证明\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3 \)线性无关, \( \forall a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_1(z - 6) + a_2(z^2 - 6z) + a_3(z^3 - 6z^2) + a_4(z^4 - 6z^3) = 0 \),可得\( -6a_1 + (a_1 - 6a_2)z + (a_2 - 6a_3)z^2 + (a_3 - 6a_4)z^3 + a_4z^4 = 0 \),由\( 1, z, z^2, z^3, z^4 \)线性无关,可得\( -6a_1 = 0, a_1 - 6a_2 = 0, a_2 - 6a_3 = 0, a_3 - 6a_4 = 0, a_4 = 0 \),解得\( a_1 = a_2 = a_3 = 0 = a_4 = 0 \),至此有\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3 \)线性无关。

接着证明\( \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) = U \):

先证明\( \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) \subseteq U \), \( \forall p \in \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) \),有\( \exists a_1, a_2, a_3, a_4 \in \mathbf{F} \)使得 \( p = a_1(z - 6) + a_2(z^2 - 6z) + a_3(z^3 - 6z^2) + a_4(z^4 - 6z^3) \),可得\( p(6) = 0 \),又显然有\( p \in \mathcal{P}_4(\mathbf{F}) \),可得\( p \in U \),于是可得\( \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) \subseteq U \)。

再证明\( U \subseteq \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) \), \( \forall p \in U \),有\( p \in \mathcal{P}_4(\mathbf{F}), p(6) = 0 \),由\( p \in \mathcal{P}_4(\mathbf{F}) \)以及\( 1, z, z^2, z^3, z^4 \)为\( \mathcal{P}_4(\mathbf{F}) \)的基,可得\( \exists a_0, a_1, a_2, a_3, a_4 \in \mathbf{F} \)使得 \( p = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 \),可得

\[ \begin{aligned} p &= a_4(z^4 - 6z^3) + 6a_4z^3 + a_3(z^3 - 6z^2) + 6a_3z^2 + a_2(z^2 - 6z) + 6a_2z + a_1(z - 6) + 6a_1 \\ & = a_1(z - 6) + a_2(z^2 - 6z) + a_3(z^3 - 6z^2) + a_4(z^4 - 6z^3) + (6a_1 + 6a_2z + 6a_3z^2 + 6a_4z^3) \\ &= a_1(z - 6) + a_2(z^2 - 6z) + a_3(z^3 - 6z^2) + a_4(z^4 - 6z^3) \\ &\quad + (6a_1 + 6a_2(z - 6) + 6^2a_2 + 6a_3(z^2 - 6z) + 6^2a_3z + 6a_4(z^3 - 6z^2) + 6^2a_4z^2) \\ &= (a_1 + 6a_2)(z - 6) + (a_2 + 6a_3)(z^2 - 6z) + (a_3 + 6a_4)(z^3 - 6z^2) + a_4(z^4 - 6z^3) \\ &\quad + (6a_1 + 6^2a_2 + 6^2a_3z + 6^2a_4z^2) \\ &= (a_1 + 6a_2)(z - 6) + (a_2 + 6a_3)(z^2 - 6z) + (a_3 + 6a_4)(z^3 - 6z^2) + a_4(z^4 - 6z^3) \\ &\quad + (6a_1 + 6^2a_2 + 6^2a_3(z - 6) + 6^3a_3 + 6^2a_4(z^2 - 6z) + 6^3a_4z) \\ &= (a_1 + 6a_2 + 6^2a_3)(z - 6) + (a_2 + 6a_3 + 6^2a_4)(z^2 - 6z) + (a_3 + 6a_4)(z^3 - 6z^2) \\ &\quad + a_4(z^4 - 6z^3) + (6a_1 + 6^2a_2 + 6^3a_3 + 6^3a_4z) \\ &= (a_1 + 6a_2 + 6^2a_3)(z - 6) + (a_2 + 6a_3 + 6^2a_4)(z^2 - 6z) + (a_3 + 6a_4)(z^3 - 6z^2) \\ &\quad + a_4(z^4 - 6z^3) + (6a_1 + 6^2a_2 + 6^3a_3 + 6^3a_4(z - 6) + 6^4a_4) \\ &= (a_1 + 6a_2 + 6^2a_3 + 6^3a_4)(z - 6) + (a_2 + 6a_3 + 6^2a_4)(z^2 - 6z) \\ &\quad + (a_3 + 6a_4)(z^3 - 6z^2) + a_4(z^4 - 6z^3) + (6a_1 + 6^2a_2 + 6^3a_3 + 6^4a_4) \end{aligned} \],

由\( p(6) = 0 \),可得\( p(6) = 6a_1 + 6^2a_2 + 6^3a_3 + 6^4a_4 = 0 \),进而可得\( p = (a_1 + 6a_2 + 6^2a_3 + 6^3a_4)(z - 6) + (a_2 + 6a_3 + 6^2a_4)(z^2 - 6z) + (a_3 + 6a_4)(z^3 - 6z^2) + a_4(z^4 - 6z^3) \in \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) \),至此可得,\( U \subseteq \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) \)。

综上可得,\( \text{span}(z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3) = U \)。

综上,有\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3 \)为\( U \)的基。

解答2:

再加个多项式\( 1 \)到之前的列表里面去,得到 \( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3, 1 \),我们证明该列表是\( \mathcal{P}_4(\mathbf{F}) \)的基:首先证明\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3, 1 \)线性无关, \( \forall a_0, a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_0 + a_1(z - 6) + a_2(z^2 - 6z) + a_3(z^3 - 6z^2) + a_4(z^4 - 6z^3) = 0 \),可得\( a_0 - 6a_1 + (a_1 - 6a_2)z + (a_2 - 6a_3)z^2 + (a_3 - 6a_4)z^3 + a_4z^4 = 0 \),由\( 1, z, z^2, z^3, z^4 \)线性无关,可得\( a_0 - 6a_1 = 0, a_1 - 6a_2 = 0, a_2 - 6a_3 = 0, a_3 - 6a_4 = 0, a_4 = 0 \),解得\( a_0 = a_1 = a_2 = a_3 = 0 = a_4 = 0 \),至此有\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3, 1 \)线性无关。又列表的长度\( = \dim \mathcal{P}_4(\mathbf{F}) = 5 \),由2.39,可得\( z - 6, z^2 - 6z, z^3 - 6z^2, z^4 - 6z^3, 1 \)为\( \mathcal{P}_4(\mathbf{F}) \)的基。

解答3:

令\( W = \text{span}(1) \),易得\( W \)为\( \mathcal{P}_4(\mathbf{F}) \)的子空间,再根据2.34的证明,可得\( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \)。

练习2.C.5

题目:

  1. Let \( U = \{ p \in \mathcal{P}_4(\mathbf{R}) : p''(6) = 0 \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathcal{P}_4(\mathbf{R}) \).
  3. Find a subspace \( W \) of \( \mathcal{P}_4(\mathbf{R}) \) such that \( \mathcal{P}_4(\mathbf{R}) = U \bigoplus W \).

解答:

解答1:

\( 1, x, x^3 - 18x^2, x^4 - 12x^3 \)为\( U \)的基(注:这里没有次数为\( 2 \)的项是因为它求二次导后就变成\( 0 \)了,但是这里有次数为\( 1 \)的项\( x \)是因为\( p''(6) = 0 \)并不能给出\( 1 \)次项的任何信息,读者看后面的推导过程就知道什么意思了),我们证明这点:

首先证明\( 1, x, x^3 - 18x^2, x^4 - 12x^3 \)线性无关, \( \forall a_0, a_1, a_3, a_4 \in \mathbf{R} \)满足 \( a_0 + a_1x + a_3(x^3 - 18x^2) + a_4(x^4 - 12x^3) = 0 \),可得\( a_0 + a_1x - 18a_3x^2 + (a_3 - 12a_4)x^3 + a_4x^4 = 0 \),再由\( 1, x, x^2, x^3, x^4 \)线性无关,可得\( a_0 = a_1 = -18a_3 = a_3 - 12a_4 = a_4 = 0 \),解得\( a_0 = a_1 = a_3 = a_4 = 0 \),这意味着\( 1, x, x^3 - 18x^2, x^4 - 12x^3 \)线性无关。

接着证明\( \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) = U \):

先证明\( \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) \subseteq U \), \( \forall p = a_0 + a_1x + a_3(x^3 - 18x^2) + a_4(x^4 - 12x^3) \in \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) \),可得\( p = a_0 + a_1x - 18a_3x^2 + (a_3 - 12a_4)x^3 + a_4x^4 \in \mathcal{P}_4(\mathbf{R}) \),又易验证\( p''(6) = 0 \),因此有\( p \in U \),至此可得\( \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) \subseteq U \)。

再证明\( U \subseteq \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) \), \( \forall p \in U \),有\( p = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 \in \mathcal{P}_4(\mathbf{R}), p''(6) = 0 \),可得:

\[ \begin{aligned} p &= a_0 + a_3(x^3 - 18x^2) + a_4(x^4 - 12x^3) + (a_1x + a_2x^2) + (18a_3x^2 + 12a_4x^3) \\ &= a_0 + a_3(x^3 - 18x^2) + a_4(x^4 - 12x^3) + 12a_4(x^3 - 18x^2) + (a_1x + a_2x^2) + (18a_3x^2 + (12 \times 18)a_4x^2) \\ &= a_0 + (a_3 + 12a_4)(x^3 - 18x^2) + a_4(x^4 - 12x^3) + (a_1x + (a_2 + 18a_3 + (12 \times 18)a_4)x^2) \end{aligned} \]

又\( p''(6) = 0 \),可得\( p''(6) = 2(a_2 + 18a_3 + (12 \times 18)a_4) = 0 \),将\( 2(a_2 + 18a_3 + (12 \times 18)a_4) = 0 \)回代,可得 \( p = a_0 + a_1x + (a_3 + 12a_4)(x^3 - 18x^2) + a_4(x^4 - 12x^3) \in \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) \),至此可得\( U \subseteq \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) \)。

综上,可得\( \text{span}(1, x, x^3 - 18x^2, x^4 - 12x^3) = U \),加上之前证明的\( 1, x, x^3 - 18x^2, x^4 - 12x^3 \)线性无关,可得\( 1, x, x^3 - 18x^2, x^4 - 12x^3 \)为\( U \)的基。

解答2:

再加个多项式\( x^2 \)到之前的列表里面去,得到 \( 1, x, x^3 - 18x^2, x^4 - 12x^3, x^2 \),我们证明该列表是\( \mathcal{P}_4(\mathbf{R}) \)的基:首先证明\( 1, x, x^3 - 18x^2, x^4 - 12x^3, x^2 \)线性无关, \( \forall a_0, a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_0 + a_1x + a_2x^2 + a_3(x^3 - 18x^2) + a_4(x^4 - 12x^3) = 0 \),可得\( a_0 + a_1x + a_2x^2 - 18a_3x^2 + (a_3 - 12a_4)x^3 + a_4x^4 = 0 \),再由\( 1, x, x^2, x^3, x^4 \)线性无关,可得\( a_0 = a_1 = a_2 = -18a_3 = a_3 - 12a_4 = a_4 = 0 \),解得\( a_0 = a_1 = a_2 = a_3 = a_4 = 0 \),这意味着\( 1, x, x^3 - 18x^2, x^4 - 12x^3, x^2 \)线性无关。又列表的长度\( = \dim \mathcal{P}_4(\mathbf{R}) = 5 \),由2.39,可得\( 1, x, x^3 - 18x^2, x^4 - 12x^3, x^2 \)为\( \mathcal{P}_4(\mathbf{R}) \)的基。

解答3:

令\( W = \text{span}(x^2) \),易得\( W \)为\( \mathcal{P}_4(\mathbf{R}) \)的子空间,再根据2.34的证明,可得\( \mathcal{P}_4(\mathbf{R}) = U \bigoplus W \)。

练习2.C.6

题目:

  1. Let \( U = \{ p \in \mathcal{P}_4(\mathbf{F}) : p(2) = p(5) \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathcal{P}_4(\mathbf{F}) \).
  3. Find a subspace \( W \) of \( \mathcal{P}_4(\mathbf{F}) \) such that \( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \).

解答:

解答1:

\( 1, (z - 2)(z - 5) = z^2 - 7z + 10, z(z - 2)(z - 5) = z^3 - 7z^2 + 10z, z^2(z - 2)(z - 5) = z^4 - 7z^3 + 10z^2 \)为\( U \)的基,我们证明这点:

首先证明\( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2 \)线性无关, \( \forall a_0, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_0 + a_2(z^2 - 7z) + a_3(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) = 0 \),可得\( a_0 + (-7a_2 + 10a_3)z + (a_2 - 7a_3 + 10a_4)z^2 + (a_3 - 7a_4)z^3 + a_4z^4 = 0 \),由\( 1, z^2, z^3, z^4 \)线性无关,可得\( a_0 = -7a_2 + 10a_3 = a_2 - 7a_3 + 10a_4 = a_3 - 7a_4 = a_4 = 0 \),解得\( a_0 = a_2 = a_3 = a_4 = 0 \),于是可得\( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2 \)线性无关。

接着证明\( \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) = U \):

先证明\( \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) \subseteq U \), \( \forall p = a_0 + a_2(z^2 - 7z + 10) + a_3(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \in \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) \),我们有\( p = a_0 + (-7a_2 + 10a_3)z + (a_2 - 7a_3 + 10a_4)z^2 + (a_3 - 7a_4)z^3 + a_4z^4 \in \mathcal{P}_4(\mathbf{F}) \),又\( p(2) = a_0 = p(5) \),可得\( p \in U \)。

再证明\( U \subseteq \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) \), \( \forall p = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 \in U \),有\( p(2) = p(5) \),可得:

\[ \begin{aligned} p &= a_0 + a_2(z^2 - 7z + 10) + a_3(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \\ &\quad + (7a_2z - 10a_2) + (7a_3z^2 - 10a_3z) + (7a_4z^3 - 10a_4z^2) \\ &= (a_0 - 10a_2) + a_2(z^2 - 7z + 10) + a_3(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \\ &\quad + (7a_3 - 10a_4)z^2 + 7a_4z^3 + (7a_2 - 10a_3)z \\ &= (a_0 - 10a_2) + a_2(z^2 - 7z + 10) + a_3(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \\ &\quad + (7a_3 - 10a_4)(z^2 - 7z + 10) + 7a_4(z^3 - 7z^2 + 10z) + (7a_2 - 10a_3)z \\ &\quad + (7(7a_3 - 10a_4)z - 10(7a_3 - 10a_4)) + (49a_4z^2 - 70a_4z) \\ &= (a_0 - 10a_2 - 10(7a_3 - 10a_4)) + (a_2 + 7a_3 - 10a_4)(z^2 - 7z + 10) \\ &\quad + (a_3 + 7a_4)(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \\ &\quad + (7a_2 - 10a_3 + 7(7a_3 - 10a_4) - 70a_4)z + 49a_4z^2 \\ &= (a_0 - 10a_2 - 10(7a_3 - 10a_4) - 490a_4) + (a_2 + 7a_3 - 10a_4 + 49a_4)(z^2 - 7z + 10) \\ &\quad + (a_3 + 7a_4)(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \\ &\quad + (7a_2 - 10a_3 + 7(7a_3 - 10a_4) - 70a_4 + (7 \times 49)a_4)z \end{aligned} \]

又\( p(2) = p(5) \),可得\( 2(7a_2 - 10a_3 + 7(7a_3 - 10a_4) - 70a_4 + (7 \times 49)a_4) = 5(7a_2 - 10a_3 + 7(7a_3 - 10a_4) - 70a_4 + (7 \times 49)a_4) \),于是可得\( 3(7a_2 - 10a_3 + 7(7a_3 - 10a_4) - 70a_4 + (7 \times 49)a_4) = 0 \),进一步可得\( (7a_2 - 10a_3 + 7(7a_3 - 10a_4) - 70a_4 + (7 \times 49)a_4) = 0 \),回代到\( p \),可得\( p = (a_0 - 10a_2 - 10(7a_3 - 10a_4) - 490a_4) + (a_2 + 7a_3 - 10a_4 + 49a_4)(z^2 - 7z + 10) + (a_3 + 7a_4)(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) \),进而有\( p \in \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) \),至此可得\( U \subseteq \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) \)。

综上,有\( \text{span}(1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2) = U \)。

至此,我们有\( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2 \)为\( U \)的基。

解答2:

再加个多项式\( z \)到之前的列表里面去,得到 \( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2, z \),我们证明该列表是\( \mathcal{P}_4(\mathbf{F}) \)的基:首先证明\( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2, z \)线性无关, \( \forall a_0, a_1, a_2, a_3, a_4 \in \mathbf{F} \)满足 \( a_0 + a_1z + a_2(z^2 - 7z) + a_3(z^3 - 7z^2 + 10z) + a_4(z^4 - 7z^3 + 10z^2) = 0 \),可得\( a_0 + a_1z + (-7a_2 + 10a_3)z + (a_2 - 7a_3 + 10a_4)z^2 + (a_3 - 7a_4)z^3 + a_4z^4 = 0 \),由\( 1, z, z^2, z^3, z^4 \)线性无关,可得\( a_0 = a_1 = -7a_2 + 10a_3 = a_2 - 7a_3 + 10a_4 = a_3 - 7a_4 = a_4 = 0 \),解得\( a_0 = a_1 = a_2 = a_3 = a_4 = 0 \),于是可得\( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2, z \)线性无关。又列表的长度\( = \dim \mathcal{P}_4(\mathbf{F}) = 5 \),由2.39,可得\( 1, z^2 - 7z + 10, z^3 - 7z^2 + 10z, z^4 - 7z^3 + 10z^2, z \)为\( \mathcal{P}_4(\mathbf{F}) \)的基。

解答3:

令\( W = \text{span}(z) \),易得\( W \)为\( \mathcal{P}_4(\mathbf{F}) \)的子空间,再根据2.34的证明,可得\( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \)。

练习2.C.7

题目:

  1. Let \( U = \{ p \in \mathcal{P}_4(\mathbf{F}) : p(2) = p(5) = p(6) \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathcal{P}_4(\mathbf{F}) \).
  3. Find a subspace \( W \) of \( \mathcal{P}_4(\mathbf{F}) \) such that \( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \).

解答:

解答1:

\( 1, (z - 2)(z - 5)(z - 6), z(z - 2)(z - 5)(z - 6) \)为\( U \)的基,证明类似上面几题,不证了。

解答2:

还缺少\( 1 \)次、\( 2 \)次的多项式,故添加两个多项式\( z, z^2 \)到之前的列表里面去,得到列表\( 1, (z - 2)(z - 5)(z - 6), z(z - 2)(z - 5)(z - 6), z, z^2 \) 为\( \mathcal{P}_4(\mathbf{F}) \)的基,证明类似上面几题,不证了。

解答3:

令\( W = \text{span}(z, z^2) \),易得\( W \)为\( \mathcal{P}_4(\mathbf{F}) \)的子空间,再根据2.34的证明,可得\( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \)。

练习2.C.8

题目:

  1. Let \( U = \{ p \in \mathcal{P}_4(\mathbf{R}) : \int_{-1}^{1} p = 0 \} \). Find a basis of \( U \).
  2. Extend the basis in part 1 to a basis of \( \mathcal{P}_4(\mathbf{R}) \).
  3. Find a subspace \( W \) of \( \mathcal{P}_4(\mathbf{R}) \) such that \( \mathcal{P}_4(\mathbf{R}) = U \bigoplus W \).

解答:

解答1:

\( x, 3x^2 - 1, x^3, 5x^4 - 1 \)为\( U \)的基,证明类似上面几题,不证了。

解答2:

还缺少\( 0 \)次的多项式,故添加两个多项式\( 1 \)到之前的列表里面去,得到列表\( x, 3x^2 - 1, x^3, 5x^4 - 1, 1 \) 为\( \mathcal{P}_4(\mathbf{R}) \)的基,证明类似上面几题,不证了。

解答3:

令\( W = \text{span}(1) \),易得\( W \)为\( \mathcal{P}_4(\mathbf{R}) \)的子空间,再根据2.34的证明,可得\( \mathcal{P}_4(\mathbf{R}) = U \bigoplus W \)。

练习2.C.9

题目:

Suppose \( v_1, \dots, v_m \) is linearly independent in \( V \) and \( w \in V \). Prove that

\[ \dim \text{span}(v_1 + w, \dots, v_m + w) \geq m - 1 \].

证明:

分情况讨论:

如果\( v_1 + w, \dots, v_m + w \)线性无关,则\( \dim \text{span}(v_1 + w, \dots, v_m + w) = m \geq m - 1 \)。

如果\( v_1 + w, \dots, v_m + w \)线性相关,则根据练习2.A.10,可得\( w \in \text{span}(v_1, \dots, v_m) \),进而易得\( \text{span}(v_1 + w, \dots, v_m + w) = \text{span}(v_1, \dots, v_m) \),于是有\( \dim \text{span}(v_1 + w, \dots, v_m + w) = \dim \text{span}(v_1, \dots, v_m) = m \geq m - 1 \)。

实际上,从上面来看,我们可以得到更强的结论:

\[ \dim \text{span}(v_1 + w, \dots, v_m + w) \geq m \]

证毕。

未完待续,慢慢更新