目录

Linear Algebra Done Right 4th 习题的参考解答及思考(第2章)

第2章

章节2A

练习2A.1

题目:

Find a list of four distinct vectors in \( \mathbf{F}^3 \) whose span equals \( \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \).

解答:

任意\( (x, y, z) \in \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \),有\( x + y + z = 0 \),可得\( z = -x - y \),进而可得\( (x, y, z) = (x, y, -x -y) = x(1, 0, -1) + y(0, 1, -1) \),这里\( (x, y, z \)可以看成\( (1, 0, -1), (0, 1, -1) \)的线性组合,对应的系数是\( x, y \),但是题目要求要有四个不同的向量,我们这里只有两个,可以想办法在\( (1, 0, -1), (0, 1, -1) \)的基础上再额外分化出来两个向量,令\( x = x_1 + 2x_2 \),这里永远可以做到的,不管\( x \)是什么值,我们永远可以找到两个值\( x_1 + 2x_2 \)满足\( x = x_1 + 2x_2 \),同理,我们可以令\( y = y_1 + 2y_2 \),则\( (x, y, z) = (x_1 + 2x_2)(1, 0, -1) + (y_1 + 2y_2)(0, 1, -1) = x_1(1, 0, -1) + x_2(2, 0, -2) + y_1(0, 1, -1) + y_2(0, 2, -2) \),我们断言\( (1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2) \)张成\( \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \),上面的过程其实已经证明了这点,只不过逻辑没那么顺,下面我们按正常的逻辑来证明。

先证明\( \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \subseteq \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) \):

\( \forall (x, y, z) \in \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \),有\( x + y + z = 0 \),可得\( z = -x - y \),我们取\( x_1 = 1, x_2 = \dfrac{x - x_1}{2} \),则\( x = x_1 + 2x_2 \),取\( y_1 = 1, y_2 = \dfrac{y - y_1}{2} \),则\( y = y_1 + 2y_2 \),此时和上面一样,易得\( (x, y, z) = x_1(1, 0, -1) + x_2(2, 0, -2) + y_1(0, 1, -1) + y_2(0, 2, -2) \),于是有\( (x, y, z) \in \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) \),综上,有\( \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \subseteq \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) \)。

接着证明\( \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) \subseteq \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \):

\( \forall v \in \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) \),有\( \exists a_1, a_2, a_3, a_4 \in \mathbf{F}, v = a_1(1, 0, -1) + a_2(2, 0, -2) + a_3(0, 1, -1) + a_4(0, 2, -2) = (a_1 + 2a_2, a_3 + 2a_4, -a_1 - 2a_2 - a_3 - 2a_4) \),这里3个坐标分量相加等于\( 0 \),于是有\( v \in \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \),综上,有\( \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) \subseteq \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \)。

至此可得,\( \text{span}((1, 0, -1), (2, 0, -2), (0, 1, -1), (0, 2, -2)) = \{ (x, y, z) \in \mathbf{F}^3 : x + y + z = 0 \} \)。

练习2A.2

题目:

Prove or give a counterexample: If \( v_1, v_2, v_3, v_4 \) spans \( V \). then the list \( v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 \) also spans \( V \).

解答:

命题成立,证明如下:

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

练习2A.3

题目:

Suppose \( v_1, \dots, v_m \) is a list of vectors in \( V \). For \( k \in \{ 1, \dots, m \} \), let \( w_k = v_1 + \dots + v_k \). Show that \( \text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m) \).

证明:

先证明\( \text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m) \):

\( \forall v \in \text{span}(v_1, \dots, v_m), \exists a_1, \dots, a_m, v = a_1v_1 + \dots + a_mv_m = a_1w_1 + (a_2 - a_1)w_2 + (a_3 - (a_1 + a_2))w_3 + \dots + (a_m - (a_1 + \dots + a_{m - 1}))w_m \),有\( v \in \text{span}(w_1, \dots, w_m) \),综上,有\( \text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m) \)。

接着证明\( \text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m) \):

\( \forall v \in \text{span}(w_1, \dots, w_m), \exists a_1, \dots, a_m, v = a_1w_1 + \dots + a_mw_m = (a_1 + \dots + a_m)v_1 + (a_2 + \dots + a_m)v_2 + \dots + amv_m \),有\( v \in \text{span}(v_1, \dots, v_m) \),综上,\( \text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m) \)。

至此可得,\( \text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m) \)。

证毕。

练习2A.4

题目:

  1. Show that a list of length one in a vector space is linearly independent if and only if the vector in the list is not 0.
  2. Show that a list of two vectors in a vector space is linearly independent if and only if neither of the two vectors in the list is a scalar multiple of the other.

证明:

证明1:

令单向量列表为\( v \)。

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1B.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。

证毕。

练习2A.5

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2A.6

题目:

Show that the list \( (2, 3, 1), (1, -1, 2), (7, 3, c) \) is linearly dependent in \( \mathbf{F}^3 \) if and only if \( c = 8 \).

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2A.7

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2A.8

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2A.9

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2A.10

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2A.11

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2A.12

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2A.13

题目:

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

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

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2A.14

题目:

Suppose \( v_1, \dots, v_m \) is a list of vectors in \( V \). For \( k \in \{ 1, \dots, m \} \), let \( w_k = v_1 + \dots + v_k \). Show that the list \( v_1, \dots, v_m \) is linearly independent if and only if the list \( w_1, \dots, w_m \) is linearly independent.

证明:

必要性:

\( \forall a_1, \dots, a_m \in \mathbf{F} \)满足 \( a_1w_1 + \dots + a_mw_m = 0 \),有\( (a_1 + \dots + a_m)v_1 + (a_2 + \dots + a_m)v_2 + \dots + amv_m = 0 \),此时由\( v_1, \dots, v_m \)线性无关,可得 \( (a_1 + \dots + a_m) = 0, (a_2 + \dots + a_m) = 0, \dots, a_m = 0 \),解得\( a_1 = \dots = a_m = 0 \),综上,有\( w_1, \dots, w_m \)线性无关。

充分性:

\( \forall a_1, \dots, a_m \in \mathbf{F} \)满足 \( a_1v_1 + \dots + a_mv_m = 0 \),有\( a_1w_1 + (a_2 - a_1)w_2 + (a_3 - (a_1 + a_2))w_3 + \dots + (a_m - (a_1 + \dots + a_{m - 1}))w_m = 0 \),由\( w_1, \dots, w_m \)线性无关,可得 \( a_1 = 0, a_2 - a_1 = 0, (a_3 - (a_1 + a_2)) = 0, \dots, (a_m - (a_1 + \dots + a_{m - 1})) = 0 \),解得\( a_1 = \dots = a_m = 0 \),综上,有\( v_1, \dots, v_m \)线性无关。

证毕。

练习2A.15

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2.22,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2A.16

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\( x_1, \dots, x_4 \)张成\( \mathcal{P}_4(\mathbf{F}) \),则根据2.22,可得\( 5 \leq 4 \),矛盾,因此假设不成立,不存在由\( 4 \)个多项式组成的列表\( x_1, \dots, x_4 \)张成\( \mathcal{P}_4(\mathbf{F}) \)。

练习2A.17

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\( v_1, v_2, \dots, v_{m - 1} \)不能张成\( V \)(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2A.13,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\( w_1, \dots, w_m \)张成\( V \),但\( v_1, \dots, v_{m + 1} \)线性无关,而\( m + 1 \leq m \),这和2.22矛盾,因此假设不成立,V是无限维的。

证毕。

练习2A.18

题目:

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

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2A.17,可得\( \mathbf{F}^{\infty} \)无限维。

证毕。

练习2A.19

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2A.17,可得目标线性空间是无限维的。

证毕。

练习2A.20

题目:

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

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2.22,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。

章节2B

练习2B.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\( \{ 0, v \} \)(\( v \neq 0 \))且对应的数域\( \mathbf{F} \)具有形式\( \{ 0, \lambda \} \)(\( \lambda \neq 0 \)),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2B.2

题目:

Verify all the assertions in Example 2.27.

Example 2.27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 \( \mathbf{F}^2 \). Note that this list has length two, which is the same as the length of the standard basis of \( \mathbf{F}^2 \). In the next section, we will see that this is not a conincidence.
  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 \( \mathcal{P}_m(\mathbf{F}) \), called the standard 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:

先证明\( 1, z, \dots, z^m \)线性无关,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。

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2B.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\( u_1, u_2, u_3 \)线性无关并且张成\( U \)。

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2.33的证明(注:看证明的内容,而不是仅仅看2.33定理本身的内容),可得\( \mathbf{R}^5 = U \bigoplus W \)。

练习2B.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\( u_1, u_2, u_3 \)线性无关并且张成\( U \)。

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2.33的证明,可得\( \mathbf{C}^5 = U \bigoplus W \)。

练习2B.5

题目:

Suppose \( V \) is finite-dimensional and \( U, W \) are subspaces of \( V \) such that \( V = U + W \). Prove that there exists a basis of \( V \) consisting of vectors in \( U \cup W \).

附注:

题目想要我们证明的是,如果两个子空间\( U, W \)能通过和操作生成\( V \)本身的话,则我们能在\( U, W \)中寻找到基,即一组完成的基向量,而不需要在\( U, W \)之外去找基向量。

证明:

这里直接简单点,提前用2.C中的2.43证明的内容,这里不会造成循环论证的问题(你甚至可以直接把证明的内容抄过来)。

根据2.43证明的内容,可得\( U + W \)存在基\( v_1, \dots, v_m, u_1, \dots, u_j, w_1 \dots, w_m \)满足里面的向量全部\( \in U \cup W \)(要么\( \in U \cap W \),要么\( \in U \),要么\( \in W \)),而\( V = U + W \),因此该基也是\( V \)的基。

证毕。

练习2B.6

题目:

Prove or give a counterexample: If \( p_0, p_1, p_2, p_3 \) is a list in \( \mathcal{P}_3(\mathbf{F}) \) such that none of the polynomials \( p_0, p_1, p_2, p_3 \) has degree 2, then \( p_0, p_1, p_2, p_3 \) is not a basis of \( \mathcal{P}_3(\mathbf{F}) \).

解答:

结论是错误的,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2B.7

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2B.8

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\( v_1, v_2 \)只有前两个坐标分量不为\( 0 \),无法张成\( U \),不是\( U \)的基。

练习2B.9

题目:

Suppose \( v_1, \dots, v_m \) is a list of vectors in \( V \). For \( k \in \{ 1, \dots, m \} \), let \( w_k = v_1 + \dots + v_k \). Show that \( v_1, \dots, v_m \) is a basis of \( V \) if and only if \( w_1, \dots, w_m \) is a basis of \( V \).

证明:

根据练习2A.3,可得\( \text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m) \),根据练习2A.14,可得\( v_1, \dots, v_m \)线性无关当且仅当\( w_1, \dots, w_m \)线性无关。

如果\( v_1, \dots, v_m \)为\( V \)的基,即它线性无关且张成\( V \),则\( w_1, \dots, w_m \)也线性无关且张成\( V \),也就是说\( w_1, \dots, w_m \)也是\( V \)的基,反之亦然。

证毕。

练习2B.10

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 \( V \).

证明:

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1.46,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。

证毕。

练习2B.11

题目:

Suppose \( V \) is a real vector space. Show that if \( v_1, \dots, v_n \) is a basis of \( V \) (as a real vector space), then \( v_1, \dots, v_n \) is also a basis of the complexification \( V_{\mathbf{C}} \) (as a complex vector space). (See Exercise 8 in Section 1B for the definition of the complexification \( V_{\mathbf{C}} \).)

证明:

注:\( v_1, \dots, v_n \)在\( V_{\mathbf{C}} \)下对应的向量为 \( v_1 + i0, \dots, v_n + i0 \),这里还是继续简记为\( v_1, \dots, v_n \)。

我们先证明\( v_1, \dots, v_n \)在\( V_{\mathbf{C}} \)下线性无关:

\( \forall a_1 + b_1i, \dots, a_n + b_ni \in \mathbf{C} \)满足 \( (a_1 + b_1i)v_1 + \dots (a_n + b_ni)v_n = 0 \),可得\( ((a_1v_1 - b_10) + i(a_10 + b_1v_1)) + \dots + ((a_nv_n - b_n0) + i(a_n0 + b_nv_n)) = (a_1v_1 + i(b_1v_1)) + \dots + (a_nv_n + i(b_nv_n)) = (a_1v_1 + \dots + a_nv_n) + i(b_1v_1 + \dots + b_nv_n) = 0 \),于是可得\( a_1v_1 + \dots + a_nv_n = 0, b_1v_1 + \dots + b_nv_n = 0 \),由\( v_1, \dots, v_n \)在\( V \)下线性无关,可得\( a_1 = \dots = a_n = 0, b_1 = \dots = b_n = 0 \),进而可得\( a_1 + b_1i = \dots = a_n + b_ni = 0 \),综上,有\( v_1, \dots, v_n \)在\( V_{\mathbf{C}} \)下线性无关。

接着证明\( v_1, \dots, v_n \)在\( V_{\mathbf{C}} \)下张成\( V_{\mathbf{C}} \):

显然有\( \text{span}(v_1, \dots, v_n) \subseteq V_{\mathbf{C}} \),我们还得证明\( V_{\mathbf{C}} \subseteq \text{span}(v_1, \dots, v_n) \), \( \forall u + iv \in V_{\mathbf{C}} \),由\( u \in V, v \in V \)以及\( v_1, \dots, v_n \)为\( V \)的基,可得\( \exists a_1, \dots, a_n, b_1, \dots, b_n \)使得 \( u = a_1v_1 + \dots a_nv_n, v = b_1v_1 + \dots b_nv_n \),于是有\( u + iv = (a_1v_1 + \dots a_nv_n) + i(b_1v_1 + \dots b_nv_n) = ((a_1v_1) + i(b_1v_1)) + \dots + ((a_nv_n) + i(b_nv_n)) = (a_1 + b_1i)v_1 + \dots + (a_n + b_ni)v_n \),这意味着\( u + iv \in \text{span}(v_1, \dots, v_n) \),综上,有\( \text{span}(v_1, \dots, v_n) \subseteq V_{\mathbf{C}} \),于是有\( \text{span}(v_1, \dots, v_n) = V_{\mathbf{C}} \)。

至此可得,\( v_1, \dots, v_n \)是\( V_{\mathbf{C}} \)的基。

证毕。

章节2C

练习2C.1

题目:

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

证明:

根据2.37,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\( \dim U = 1 \),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,最后如果\( \dim U = 2 \),则根据2.39,可得\( U = V \),

证毕。

练习2C.2

题目:

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

证明:

根据2.37,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\( \dim U = 1 \),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。

如果\( \dim U = 3 \),则根据2.39,可得\( U = V \)。

证毕。

练习2C.3

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2.38,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2.33的证明,可得\( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \)。

练习2C.4

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:

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

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2.38,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2.33的证明,可得\( \mathcal{P}_4(\mathbf{R}) = U \bigoplus W \)。

练习2C.5

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2.38,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2.33的证明,可得\( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \)。

练习2C.6

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2.33的证明,可得\( \mathcal{P}_4(\mathbf{F}) = U \bigoplus W \)。

练习2C.7

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2.33的证明,可得\( \mathcal{P}_4(\mathbf{R}) = U \bigoplus W \)。

练习2C.8

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2A.12,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:\( \dim \text{span}(v_1 + w, \dots, v_m + w) = m \)。

证毕。

练习2C.9

题目:

Suppose \( m \) is a positive integer and \( p_0, p_1, \dots, p_m \in \mathcal{P}(\mathbf{F}) \) are such that each \( p_k \) has degree \( k \). Prove that \( p_0, p_1, \dots, p_m \) is a basis of \( \mathcal{P}_m(\mathbf{F}) \).

证明:

先证明\( p_0, \dots, p_m \)线性无关, \( \forall a_0, \dots, a_m \in \mathbf{F} \)满足\( a_0p_0 + \dots + a_mp_m = 0 \),则有\( a_0 = \dots = a_m = 0 \),这是因为低次项多项式的绝对值增长速度比高次多项式慢至少一个数量级,因此不同次数的多项式之间不可能相互抵消称为\( 0 \)多项式,唯一的可能就是系数全为\( 0 \)。

接着由\( p_0, \dots, p_m \)列表的长度为\( m = \dim \mathcal{P}_m(\mathbf{F}) \) 以及2.38,可得\( p_0, \dots, p_m \)为\( \mathcal{P}_m(\mathbf{F}) \)的基。

证毕。

练习2C.10

题目:

Suppose \( m \) is a positive integer. For \( 0 \leq k \leq m \), let \( p_k(x) = x^k(1 - x)^{m - k} \). Show that \( p_0, \dots, p_m \) is a basis of \( \mathcal{P}_m(\mathbf{F}) \). (The basis in this exercise leads to what are called Bernstein polynomials. you can do a web search to learn how Bernstein polynomials are used to approximate continuous functions on [0, 1].)

解答:

易得\( \forall 0 \leq k \leq m, p_k(x) = x^k(1 - x)^{m - k} \)中的\( (1 - x)^{m - k} \)包含\( 0 \leq i \leq {m - k} \)次的项,进而可得\( p_k \)包含\( k \leq i \leq m \)次的项,特别的,\( p_0 \)包含\( 0 \leq i \leq m \)次的项, \( p_1 \)包含\( 1 \leq i \leq m \)次的项,相对于\( p_0 \)少了个\( 0 \)次项,而\( 0 \)次以上的项相抵消,最多只能得到\( -\infty \)次的\( 0 \)多项式,是得不到\( 0 \)次多项式的,否则一个多项式就不能通过系数唯一决定了,而我们知道多项式是可以通过系数唯一决定的,矛盾,综上,可得\( p_1 \notin \text{span}(p_0) \),同理可得\( \forall 2 \leq k \leq m, p_k \notin \text{span}(p_0, \dots, p_{k - 1}) \),再根据2.19,可得\( p_0, \dots, p_m \)线性无关,最后根据2.38,可得\( p_0, \dots, p_m \)为\( \mathcal{P}_m(\mathbf{F}) \)的基。

练习2C.11

题目:

Suppose \( U \) and \( W \) are both four-dimensional subspaces of \( \mathbf{C}^6 \). Prove that there exist two vectors in \( U \cap W \) such that neither of these vectors is a scalar multiple of the other.

证明:

根据2.43,可得\( \dim(U \cap W) = \dim U + \dim W - \dim(U + W) = 8 - \dim \mathbf{C}^6 = 2 \),因此\( U \cap W \)所有基的长度均为\( 2 \),而一个长度为\( 2 \)的列表为基当且仅当列表中的任意一个向量都不是另外一个向量的标量倍,于是任取\( U \cap W \)的任意一个基就是题目要的两个向量了。

证毕。

练习2C.12

题目:

Suppose that \( U \) and \( W \) are subspaces of \( \mathbf{R}^8 \) such that \( \dim U = 3, \dim W = 5 \), and \( U + W = \mathbf{R}^8 \). Prove that \( \mathbf{R}^8 = U \oplus W \).

证明:

根据2.34,可得\( \dim(U \cap W) = \dim U + \dim W - \dim(U + W) = 8 - \dim \mathbf{R}^8 = 0 \),进而可得\( U \cap W = \{ 0 \} \),此时根据1.46,可得\( \mathbf{R}^8 = U + W = U \oplus W \)。

证毕。

练习2C.13

题目:

Suppose \( U \) and \( W \) are both five-dimensional subspaces of \( \mathbf{R}^9 \). Prove that \( U \cap W \neq \{ 0 \} \).

证明:

根据2.34,可得\( \dim(U \cap W) = \dim U + \dim W - \dim(U + W) = 10 - \dim(U + W) \),而\( U + W \subseteq \mathbf{R}^9 \),根据2.37,可得\( \dim(U + W) \leq \dim \mathbf{R}^9 = 9 \),于是可得\( \dim(U \cap W) = 10 - \dim(U + W) \geq 10 - 9 = 1 \),这意味着\( U \cap W \neq \{ 0 \} \)。

证毕。

练习2C.14

题目:

Suppose \( V \) is a ten-dimensional vector space and \( V_1, V_2, V_3 \) are subspaces of \( V \) with \( \dim V_1 = \dim V_2 = \dim V_3 = 7 \). Prove that \( V_1 \cap V_2 \cap V_3 \neq \{ 0 \} \).

证明:

根据2.34以及2.37,可得\( \dim(V_1 \cap V_2 \cap V_3) = \dim(V_1 \cap V_2) + \dim V_3 - \dim((V_1 \cap V_2) + V_3) = 7 + \dim(V_1 \cap V_2) - \dim((V_1 \cap V_2) + V_3) = 7 + (\dim V_1 + \dim V_2 - \dim(V_1 + V_2)) - \dim((V_1 \cap V_2) + V_3) = 7 + 7 + 7 - \dim(V_1 + V_2) - \dim((V_1 \cap V_2) + V_3) \geq 7 + 7 + 7 - \dim V - \dim V = 21 - 10 - 10 = 1 \),这意味着\( V_1 \cap V_2 \cap V_3 \neq \{ 0 \} \)。

证毕。

练习2C.15

题目:

Suppose \( V \) is finite-dimensional and \( V_1, V_2, V_3 \) are subspaces of \( V \) with \( \dim V_1 + \dim V_2 + \dim V_3 > 2 \dim V \). Prove that \( V_1 \cap V_2 \cap V_3 \neq \{ 0 \} \).

证明:

根据2.34以及2.37,可得\( \dim(V_1 \cap V_2 \cap V_3) = \dim(V_1 \cap V_2) + \dim V_3 - \dim((V_1 \cap V_2) + V_3) = \dim V_3 + \dim(V_1 \cap V_2) - \dim((V_1 \cap V_2) + V_3) = \dim V_3 + (\dim V_1 + \dim V_2 - \dim(V_1 + V_2)) - \dim((V_1 \cap V_2) + V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 + V_2) - \dim((V_1 \cap V_2) + V_3) > 2 \dim V - \dim(V_1 + V_2) - \dim((V_1 \cap V_2) + V_3) > 2 \dim V - \dim V - \dim V = 0 \),这意味着\( V_1 \cap V_2 \cap V_3 \neq \{ 0 \} \)。

证毕。

练习2C.16

题目:

Suppose \( V \) is finite-dimensional and \( U \) is a subspace of \( V \) with \( U \neq V \). Let \( n = \dim V \) and \( m = \dim U \). Prove that there exist \( n - m \) subspaces of \( V \), each of dimension \( n - 1 \), whose intersection equals \( U \).

证明:

任取\( U \)的基\( u_1, \dots, u_m \),根据2.32,我们可以拓展该基以成为\( V \)的基,记新增的向量为\( w_1, \dots, w_{n - m} \),由于\( U \neq V \),我们有\( \dim U < \dim V \),进而可得新增的向量数目一定不为\( 0 \),即\( n - m \geq 1 \),此时我们从\( V \)的基\( u_1, \dots, u_m, w_1, \dots, w_{n - m} \)中依次移除\( w_1, \dots, w_{n - m} \)中的某个向量,具体的,\( \forall 1 \leq i \leq n - m \),令\( B_i \)为\( u_1, \dots, u_m, w_1, \dots, w_{n - m} \)移除向量\( w_i \)后得到的列表,可得\( B_i \)线性无关。

接下来,我们证明这\( n - m \)个\( B_i \)张成的向量空间的交集为\( U \),具体的,\( \forall 1 \leq i \leq n - m \),令\( V_i = \text{span} (B_i) \),有\( V_i \)为\( V \)的子空间且\( \dim V_i = n - 1 \),我们证明\( \bigcap_{i = 1}^{i = n - m} V_i = U \):

先证明\( \bigcap_{i = 1}^{i = n - m} V_i \subseteq U \), \( \forall v \in \bigcap_{i = 1}^{i = n - m} V_i \),有\( \forall 1 \leq i \leq n - m, v \in V_i \),可得\( \exists a_{1i}, \dots, a_{(n - 1)i} \in \mathbf{F} \),使得\( v = a_{1i}u_1 + \dots + a_{mi}u_m + a_{(m + 1)i}w_1 + \dots + a_{(m + (i - 1))i}w_{i - 1} + a_{(m + i)i}w_{i + 1} + \dots a_{(n - 1)i}w_{n - m} \),于是有\( \forall 1 \leq i \leq n - m - 1, a_{1i}u_1 + \dots + a_{mi}u_m + a_{(m + 1)i}w_1 + \dots + a_{(m + (i - 1))i}w_{i - 1} + a_{(m + i)i}w_{i + 1} + \dots + a_{(n - 1)i}w_{n - m} = a_{1(i + 1)}u_1 + \dots + a_{m(i + 1)}u_m + a_{(m + 1)(i + 1)}w_1 + \dots + a_{(m + ((i + 1) - 1))(i + 1)}w_{(i + 1) - 1} + a_{(m + (i + 1))(i + 1)}w_{(i + 1) + 1} + \dots a_{(n - 1)(i + 1)}w_{n - m} \),取\( a_{(m + i)i} = 0, a_{(m + (i + 1))(i + 1)} = 0 \),由\( u_1, \dots, u_m, w_1, \dots, w_{n - m} \)线性无关,移项后可得所有对应系数都相等,特别的,有\( a_{(m + i)(i + 1)} = a_{(m + i)i} = 0, a_{(m + (i + 1))i} = a{(m + (i + 1))(i + 1)} = 0 \),由于这两个等式对于所有\( 1 \leq i \leq n - m - 1 \)都成立,可得\( \forall 1 \leq i \leq n - m, a_{(m + 1)i} = \dots = a_{(n - 1)i} = 0 \),加上\( \forall 1 \leq i \leq n - m, a_{1i} = a_{1(i + 1)}, \dots, a_{mi} = a_{m(i + 1)} \) (前面说了,对应系数都相等),于是我们取\( a_{11}, \dots, a_{m1} \)作为\( v \)的系数就行了,可得\( v = a_{11}u_1 + \dots + a_{m1}u_m \in U \)。

接着证明\( U \subseteq \bigcap_{i = 1}^{i = n - m} V_i \), \( \foral v \in U \),可得\( \exists a_1, \dots, a_m \in \mathbf{F} \),使得\( v = a_1u_1 + \dots + a_mu_m \),而各个\( B_i \)的基均包含向量\( u_1, \dots, u_m \),于是有\( \forall 1 \leq i \leq n - m, v \in B_i \),可得,有\( v \in \bigcap_{i = 1}^{i = n - m} V_i \)。

至此可得,\( \bigcap_{i = 1}^{i = n - m} V_i = U \)。

证毕。

练习2C.17

题目:

Suppose that \( V_1, \dots, V_m \) are finite-dimensional subspaces of \( V \). Prove that \( V_1 + \dots + V_m \) is finite-dimensional and \( \dim(V_1 + \dots + V_m) \leq \dim(V_1) + \dots + \dim(V_m) \). (The inequality above is an equality if and only if \( V_1 + \dots + V_m \) is a direct sum, as will be shown in 3.94.)

证明:

使用数学归纳法来证明,当\( m = 1 \)时,我们有\( V_1 + \dots + V_m = V_m \),于是有\( \dim(V_1 + \dots + V_m) = \dim V_m \leq \dim(V_m) = \dim(V_1) + \dots + \dim(V_m) \),成立。

归纳假设当\( m = k \)时成立,当\( m = k + 1 \)时, \( V_1 + \dots + V_{k + 1} = (V_1 + \dots + V_k) + V_{k + 1} \),根据2.43,可得\( \dim(V_1 + \dots + V_{k + 1}) = \dim((V_1 + \dots + V_k) + V_{k + 1}) \leq \dim(V_1 + \dots + V_k) + \dim(V_{k + 1}) \),根据归纳假设,可得\( \dim(V_1 + \dots + V_k) \leq \dim V_1 + \dots + \dim V_k \),于是结合前面可得,\( \dim((V_1 + \dots + V_k) + V_{k + 1}) \leq \dim(V_1 + \dots + V_k) + \dim(V_{k + 1}) \leq \dim V_1 + \dots + \dim V_k + \dim V_{k + 1} \),成立。

综上,归纳完毕,命题成立。

证毕。

练习2C.18

题目:

Suppose \( V \) is finite-dimensional with \( \dim V = n \geq 1 \). Prove that there exist one-dimensional subspaces \( V_1, \dots, V_n \) of \( V \) such that \( V = V_1 \oplus \dots \oplus V_n \).

证明:

任取\( v_1, \dots, v_n \)为\( V \)的基, \( \forall 1 \leq i \leq n \),令\( V_i = \text{span}(v_i) \),可得\( \dim V_i = 1 \)。

我们要证明\( V = V_1 + \dots + V_n \),显然有\( V_1 + \dots + V_n \subseteq V \),为了证明\( V = V_1 + \dots + V_n \),我们还得证明\( V \subseteq V_1 + \dots + V_n \), \( \forall v \in V \),由\( v_1, \dots, v_n \)为\( V \)的基,可得存在 唯一的 \( a_1, \dots, a_n \in \mathbf{F} \)满足 \( v = a_1v_1 + \dots + a_nv_n \),这里\( \forall 1 \leq i \leq n, a_iv_1 \in V_i \),于是有\( v \in V_1 + \dots + V_n \),至此可得,\( V = V_1 + \dots + V_n \),同时由于用于表示\( v \)的系数是唯一的,可得\( V = V_1 \oplus \dots \oplus V_n \)。

证毕。

练习2C.19

题目:

Explain why you might guess, motivated by analogy with the formula for the number of elements in the union of three finite sets, that if \( V_1, V_2, V_3 \) are subspaces of a finite-dimensional vector space, then \( \dim(V_1 + V_2 + V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_2) - \dim(V_1 \cap V_3) - \dim(V_2 \cap V_3) + \dim(V_1 \cap V_2 \cap V_3) \). Then either prove the formula above or give a counterexample.

解答:

你提前看一眼下一题就知道不成立了。

首先解释下,为什么我们会猜上述等式成立,这是由于根据章节2C结尾的表格,我们发现集合基数和并集的关系和向量空间维度和和操作的关系之间具有很强的相似性,进而由于3个集合的基数和并集会满足和上述类似的等式,于是我们推测3个向量空间的维度和和操作会满足上述等式。

然而实际上该等式不成立,反例:

令\( V = \mathbf{R}^2, U_1 = \{ (x, 0) : x \in \mathbf{R} \}, U_2 = \{ (0, x) : x \in \mathbf{R} \}, U_3 = \{ (x, x) : x \in \mathbf{R} \} \),这里\( U_1 + U_2 + U_3 = V \),可得\( \dim(U_1 + U_2 + U_3) = 2 \),但\( \dim U_1 + \dim U_2 + \dim U_3 = 1 + 1 + 1 = 3, \dim(U_1 \cap U_2) = \dim(U_2 \cap U_3) = \dim(U_1 \cap U_3) = \dim(U_1 \cap U_2 \cap U_3) = 0 \),可得\( \dim(U_1 + U_2 + U_3) = 2 \neq \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_2) - \dim(V_1 \cap V_3) - \dim(V_2 \cap V_3) + \dim(V_1 \cap V_2 \cap V_3) = 1 + 1 + 1 - 0 - 0 - 0 - 0 = 3 \)。

练习2C.20

Prove that if \( V_1, V_2 \) and \( V_3 \) are subspaces of a finite-dimensional vector space, then

\[ \begin{aligned} \dim(V_1 + V_2 + V_3) &= \dim V_1 + \dim V_2 + \dim V_3 \\ &\qquad - \dfrac{\dim(V_1 \cap V_2) + \dim(V_1 \cap V_3) + \dim(V_2 \cap V_3)}{3} \\ &\qquad - \dfrac{\dim((V_1 + V_2) \cap V_3) + \dim((V_1 + V_3) \cap V_2) + \dim((V_2 + V_3) \cap V_1)}{3} \end{aligned}. \]

(The formula may seem strange because the right side does not look like an integer.)

证明:

根据2.43,可得\( \dim(V_1 + V_2 + V_3) = \dim((V_1 + V_2) + V_3) = \dim(V_1 + V_2) + \dim V_3 - \dim((V_1 + V_2) \cap V_3) = (\dim V_1 + \dim V_2 - \dim(V_1 \cap V_2)) + \dim V_3 - \dim((V_1 + V_2) \cap V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_2) - \dim((V_1 + V_2) \cap V_3) \)。

类似的,可得\( \dim(V_1 + V_2 + V_3) = \dim(V_1 + (V_2 + V_3)) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_2 \cap V_3) - \dim((V_2 + V_3) \cap V_1) \) 以及\( \dim(V_1 + V_2 + V_3) = \dim((V_1 + V_3) + V_2) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_3) - \dim((V_1 + V_3) \cap V_2) \) (注:\( \dim(V_1 + V_2 + V_3) = \dim((V_1 + V_3) + V_2) \)是因为和操作满足结合律和交换律),进而可得\( \dim(V_1 + V_2 + V_3) = \dfrac{\dim((V_1 + V_2) + V_3) + \dim(V_1 + (V_2 + V_3)) + \dim((V_1 + V_3) + V_2)}{3} = \dim V_1 + \dim V_2 + \dim V_3 - \dfrac{\dim(V_1 \cap V_2) + \dim(V_1 \cap V_3) + \dim(V_2 \cap V_3)}{3} - \dfrac{\dim((V_1 + V_2) \cap V_3) + \dim((V_1 + V_3) \cap V_2) + \dim((V_2 + V_3) \cap V_1)}{3} \),等式成立。

证毕。