Linear Algebra Done Right 4th 习题的参考解答及思考(第3章)
第3章
章节3A
练习3A.1
题目:
Suppose \( b, c \in \mathbf{R} \). Define \( T: \mathbf{R}^3 \to \mathbf{R}^2 \) by \( T(x, y, z) = (2x - 4y + 3z + b, 6x + cxyz) \). Show that \( T \) is linear if and only if \( b = c = 0 \).
证明:
必要性:
如果\( T \)为线性映射,则\( \forall u = (u_1, u_2, u_3), v = (v_1, v_2, v_3) \in \mathbf{R}^3 \),有\( T(u + v) = T(u) + T(v) \),可得\( 2(u_1 + v_1) - 4(u_2 + v_2) + 3(u_3 + v_3) + b, 6(u_1 + v_1) + c(u_1 + v_1)(u_2 + v_2)(u_3 + v_3) = (2u_1 - 4u_2 + 3u_3 + b, 6u_1 + cu_1u_2u_3) + (2v_1 - 4v_2 + 3v_3 + b, 6v_1 + cv_1v_2v_3) \),可得\( 2b = b, c(u_1u_2u_3 + v_1v_2v_3) = c(u_1 + v_1)(u_2 + v_2)(u_3 + v_3) \),由\( 2b = b \),可得\( b = 0 \),假设\( c \neq 0 \),则由\( c(u_1u_2u_3 + v_1v_2v_3) = c(u_1 + v_1)(u_2 + v_2)(u_3 + v_3) \),可得\( (u_1u_2u_3 + v_1v_2v_3) = (u_1 + v_1)(u_2 + v_2)(u_3 + v_3) \)对所有 \( u = (u_1, u_2, u_3), v = (v_1, v_2, v_3) \in \mathbf{R}^3 \)都成立,特别的,我们令\( u = (1, 0, 0), v = (0, 1, 0) \),则\( (u_1u_2u_3 + v_1v_2v_3) = 2 \neq 0 = (u_1 + v_1)(u_2 + v_2)(u_3 + v_3) \)矛盾,因此假设不成立,可得\( c = 0 \)。
充分性:
如果\( b = c = 0 \),则\( T(x, y, z) = (2x - 4y + 3z, 6x) \),此时\( \forall u = (u_1, u_2, u_3), v = (v_1, v_2, v_3) \in \mathbf{R}^3 \),可得\( T(u + v) = 2(u_1 + v_1) - 4(u_2 + v_2) + 3(u_3 + v_3), 6(u_1 + v_1) \),而\( T(u) + T(v) = (2u_1 - 4u_2 + 3u_3, 6u_1) + (2v_1 - 4v_2 + 3v_3, 6v_1) = 2(u_1 + v_1) - 4(u_2 + v_2) + 3(u_3 + v_3), 6(u_1 + v_1) = T(u + v) \),可得\( T(u + v) = T(u) + T(v) \),类似的,易得\( \forall \lambda \in \mathbf{R}, v \in \mathbf{R}^3 \),有\( T(\lambda v) = \lambda T(v) \),综上,有\( T \)为线性映射。
证毕。
练习3A.2
题目:
Suppose \( b, c \in \mathbf{R} \). Define \( T: \mathcal{P}(\mathbf{R}) \to \mathbf{R}^2 \) by \[ Tp = (3p(4) + 5p'(6) + bp(1)p(2), \int_{-1}^{2} x^3p(x) \ dx + c \sin p(0)). \] Show that \( T \) is linear if and only if \( b = c = 0 \).
证明:
必要性:
如果\( T \)为线性映射,则\( \forall p, q \in \mathcal{P}(\mathbf{R}) \), \( T(p + q) = T(p) + T(q) \),可得\( 3(p + q)(4) + 5(p' + q')(6) + b((p + q)(1))((p + q)(2)) = (3p(4) + 5p'(6) + bp(1)p(2)) + (3q(4) + 5q'(6) + bq(1)q(2)) \) 以及\( \int_{-1}^{2} x^3(p + q)(x) \ dx + c \sin (p + q)(0) = (\int_{-1}^{2} x^3p(x) \ dx + c \sin p(0)) + (\int_{-1}^{2} x^3q(x) \ dx + c \sin q(0)) \)。
由前者解得\( b((p + q)(1))((p + q)(2)) = bp(1)p(2) + bq(1)q(2) = b(p(1)p(2) + q(1)q(2)) \),假设\( b \neq 0 \),则\( ((p + q)(1))((p + q)(2)) = p(1)p(2) + q(1)q(2) \),举两个特别的\( p, q \),易得矛盾,因此假设不成立,可得\( b = 0 \)。
由后者解得\( c \sin (p + q)(0) = c \sin p(0) + c \sin q(0) = c(\sin p(0) + \sin q(0)) \),同理,在假设\( c \neq 0 \)的情况下举反例易得矛盾,因此假设不成立,可得\( c = 0 \)。
充分性:
如果\( b = c = 0 \),则\( Tp = (3p(4) + 5p'(6), \int_{-1}^{2} x^3p(x) \ dx) \),易验证\( \forall p, q \in \mathcal{P}(\mathbf{R}), \forall \lambda \in \mathbf{R} \),有\( T(p + q) = Tp + Tq, T(\lambda p) = \lambda Tp \),可得\( T \)为线性映射。
证毕。
练习3A.3
Suppose that \( T \in \mathcal{L}(\mathbf{F}^n, \mathbf{F}^m) \). Show that there exist scalars \( A_{j, k} \in \mathbf{F} \) for \( j = 1, \dots, m \) and \( k = 1, \dots, n \) such that \[ T(x_1, \dots, x_n) = (A_{1, 1}x_1 + \dots + A_{1, n}x_n, \dots, A_{m, 1}x_1 + \dots + A_{m, n}x_n) \] for every \( (x_1, \dots, x_n) \in \mathbf{F}^n \). (This exercise shows that the linear map has the form promised in the second to last item of Example 3.3.)
证明:
一个线性映射由它对所有基向量的映射唯一决定,因此我们只需要关注对基向量的映射即可。
令\( u_1, \dots, u_n \)为\( \mathbf{F}^n \)的标准基,令\( v_1, \dots, v_m \)为\( \mathbf{F}^m \)的标准基, \( \forall 1 \leq i \leq n \),令\( T(u_i) = w_i \),则\( w_i \in \mathbf{F}^m \),可得\( \exists A_{1, i}, \dots, A_{m, i} \in \mathbf{F} \),使得\( T(u_i) = w_i = A_{1, i}v_1 + \dots + A_{m, i}v_m \),进而\( \forall (x_1, \dots, x_n) \in \mathbf{F}^n \),由\( T \)为线性映射,可得\( T((x_1, \dots, x_n)) = x_1 T(u_1) + \dots + x_n T(u_n) = x_1(A_{1, 1}v_1 + \dots + A_{m, 1}v_m) + \dots + x_n(A_{1, n}v_1 + \dots + A_{m, n}v_m) = (A_{1, 1}x_1 + \dots + A_{1, n}x_n, \dots, A_{m, 1}x_1 + \dots + A_{m, n}x_n) \)。
证毕。
练习3A.4
Suppose \( T \in \mathcal(V, W) \) and \( v_1, \dots, v_m \) is a list of vectors in \( V \) such that \( Tv_1, \dots, Tv_m \) is a linearly independent list in \( W \). Prove that \( v_1, \dots, v_m \) is linearly independent.
证明:
假设\( v_1, \dots, v_m \)线性相关,则存在不全为\( 0 \)的系数\( a_1, \dots, a_m \)满足 \( a_1v_1 + \dots + a_mv_m = 0 \),任取其中一个不为\( 0 \)的系数\( a_k \),则\( a_k v_k = -a_1v_1 - \dots - a_{k - 1}v_{k - 1} - a_{k + 1}v_{k + 1} - \dots - a_mv_{m} \),由\( T \)为线性映射,可得\( a_k Tv_k = -a_1Tv_1 - \dots - a_{k - 1}Tv_{k - 1} - a_{k + 1}Tv_{k + 1} - \dots - a_mTv_{m} \),移项可得\( a_1Tv_1 + \dots + a_mTv_m = 0 \),但这里\( a_k \neq 0 \),这意味着\( Tv_1, \dots, Tv_m \)线性相关,矛盾,因此假设不成立,有\( v_1, \dots, v_m \)线性相关。
证毕。
练习3A.5
题目:
Prove that \( \mathcal{L}(V, W) \) is a vector space, as was asserted in 3.6.
证明:
看着定义1.20一个个验证过去。
证明满足交换律:
\( \forall S, T \in \mathcal{L}(V, W) \),有\( \forall v \in V, (S + V)(v) = Sv + Tv = Tv + Sv = (T + S)v \),可得\( S + T = T + S \)。
证明满足结合律:
\( \forall T_1, T_2, T_3 \in \mathcal{L}(V, W) \),有\( \forall v \in V, ((T_1 + T_2) + T_3)(v) = (T_1v + T_2v) + T_3v = T_1v + (T_2v + T_3v) = (T_1 + (T_2 + T_3))(v) \),可得\( (T_1 + T_2) + T_3 = T_1 + (T_2 + T_3) \)。
证明加法单位元的存在性:
构造\( 0 \in \mathcal{L}(V, W) \),\( \forall v \in V \),令\( 0(v) = 0 \),则易得\( \forall S \in \mathcal{L}(V, W) \), \( S + 0 = S \)。
证明加法逆元的存在性:
\( \forall S \in \mathcal{L}(V, W) \),构造\( T \in \mathcal{L}(V, W) \), \( \forall v \in V \),令\( T(v) = -S(v) \),易得\( S + T = 0 \)。
证明乘法单位元的存在性:
\( \forall S \in \mathcal{L}(V, W) \),有\( \forall v \in V \),\( (1S)(v) = 1(Sv) = Sv \),可得\( 1S = S \)。
证明满足分配律:
\( \forall S, T \in \mathcal{L}(V, W), \forall a, b \in \mathbf{F} \)。
\( \forall v \in V \),有\( (a(S + T))(v) = a((S + T)(v)) = a(Sv + Tv) = aSv + aTv \),可得\( a(S + T) = aS + aT \)。
\( \forall v \in V \),有\( ((a + b)S)(v) = (a + b)(Sv) = aSv + bSv \),可得\( (a + b)S = aS + bS \)。
综上,根据定义1.20,可得\( \mathcal{L}(V, W) \)为向量空间。
证毕。
练习3A.6
题目:
Prove that multiplication of linear maps has the associative, identity, and distributive properties asserted in 3.8.
3.8的内容:
associativity
\( (T_1T_2)T_3 = T_1(T_2T_3) \) whenever \( T_1, T_2 \), and \( T_3 \) are linear maps such that the products make sense (meaning \( T_3 \) maps into the domain of \( T_2 \) and \( T_2 \) maps into the domain of \( T_1 \)).
identity
\( TI = IT = T \) whenever \( T \in \mathcal{L}({V, W}) \); here the first \( I \) is the identity operator on \( V \), and the second \( I \) is the identity operator on \( W \).
distributive properties
\( (S_1 + S_2)T = S_1T + S_2T \) and \( S(T_1 + T_2) = ST_1 + ST_2 \) whenever \( T, T_1, T_2 \in \mathcal{L}(U, V) \) and \( S, S_1, S_2 \in \mathcal{L}(V, W) \).
证明:
证明满足结合律:
线性映射的乘法就是函数的复合,而我们知道函数的复合是复合结合律的,简单写的话,到这里就可以说结合律成立了,详细点的话,如下:
\( \forall T_1 \in \mathcal{L}(V_1, V_2), T_2 \in \mathcal{L}(V_2, V_3), T_3 \in \mathcal{L}(V_3, V_4) \),我们有\( \forall v \in V_1, ((T_1T_2)T_3)(x) = (T_1T_2)(T_3x) = T_1(T_2(T_3x)) = T_1((T_2T_3)(x)) = (T_1(T_2T_3))(x) \),综上,有\( (T_1T_2)T_3 = T_1(T_2T_3) \)。
证明乘法单位元的存在性:
构造映射\( I_1 \in \mathcal{L}(V, V), \forall v \in V, I_1(v) = v \),构造映射\( I_2 \in \mathcal{L}(W, W), \forall w \in W, I_2(w) = w \),易得\( \forall v \in \mathcal{L}(V, W), TI = IT = T \)。
证明满足分配律:
\( \forall T, T_1, T_2 \in \mathcal{L}(U, V), \forall S, S_1, S_2 \in \mathcal{L}(V, W) \):
\( \forall u \in U, ((S_1 + S_2)T)(u) = (S_1 + S_2)(Tu) = S_1(Tu) + S_2(Tu) = (S_1T)(u) + (S_2T)(u) \),综上,有\( (S_1 + S_2)T = S_1T + S_2T \)。
\( \forall u \in U, S(T_1 + T_2)(u) = S((T_1 + T_2)(u)) = S(T_1u + T_2u) = S(T_1u) + S(T_2u) = (ST_1)(u) + (ST_2)(u) \),综上,有\( S(T_1 + T_2) = ST_1 + ST_2 \)。
证毕。
练习3A.7
题目:
Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if \( \dim V = 1 \) and \( T \in \mathcal{L}(V) \), then there exists \( \lambda \in \mathbf{F} \) such that \( Tv = \lambda v \) for all \( v \in V \).
证明:
给定任意满足题述条件的线性映射\( T \),因为\( \dim V = 1 \),可得\( V \)的基中的向量数量为\( 1 \),任取\( V \)的基\( w \),则\( \forall v \in V \),存在\( \alpha \in \mathbf{F} \),使得\( v = \alpha w \),因为\( T \in \mathcal{L}(V) \),有\( Tw \in V \),可得存在\( \lambda \in \mathbf{F} \),使得\( Tw = \lambda w \),进而可得\( Tv = T(\alpha w) = \alpha (Tw) = \alpha (\lambda w) = \lambda (\alpha w) = \lambda v \)。
证毕。
练习3A.8
题目:
Give an example of a function \( \varphi: \mathbf{R}^2 \to \mathbf{R} \) such that \( \varphi(av) = a \varphi(v) \) for all \( a \in \mathbf{R} \) and all \( v \in \mathbf{R}^2 \) but \( \varphi \) is not linear. (This exercise and the next exercise show that neither homogeneity nor additivity alone is enough to imply that a function is a linear map.)
解答:
构造\( \varphi: \mathbf{R}^2 \to \mathbf{R} \), \( \forall (x, y) \in \mathbf{R}^2 \),如果\( y \neq 0 \),则令\( \varphi((x, y)) = \dfrac{x^2}{y} \),如果\( y = 0 \),则令\( \varphi((x, y)) = 0 \),于是\( \forall a \in \mathbf{R}, (x, y) \in \mathbf{R}^2, \varphi(a(x, y)) = \varphi((ax, ay)) = \dfrac{a^2x^2}{ay} = a\dfrac{x^2}{y} = a \varphi((x, y)) \),但是\( \varphi((1, 2) + (3, 4)) = \varphi((4, 6)) = \dfrac{16}{6}, \varphi((1, 2)) = \dfrac{1}{2}, \varphi((3, 4)) = \dfrac{9}{4} \),可得\( \varphi((1, 2)) + \varphi((3, 4)) = \dfrac{11}{4} = \dfrac{33}{12} \neq \dfrac{32}{12} = \dfrac{16}{6} = \varphi((1, 2) + (3, 4)) \),这意味着\( \varphi \)非线性。
练习3A.9
题目:
Give an example of a function \( \varphi: \mathbf{C} \to \mathbf{C} \) such that \( \varphi(w + z) = \varphi(w) + \varphi(z) \) for all \( w, z \in \mathbf{C} \) but \( \varphi \) is not linear. (Here C is thought of as a complex vector space. There also exists a function \( \varphi: \mathbf{R} \to \mathbf{R} \) such that \( \varphi \) satisfies the additivity condition above but \( \varphi \) is not linear. However, showing the existence of such a function involves considerably more advanced tools.)
解答:
构造\( \varphi: \mathbf{C} \to \mathbf{C} \), \( \forall a + bi \in \mathbf{C} \),令\( \varphi(a + bi) = a \),则\( \forall w = (a + bi), z = (c + di) \in \mathbf{C} \),有\( \varphi(w + z) = \varphi((a + c) + (b + d)i) = a + c = \varphi(w) + \varphi(z) \),但是\( i \varphi(1) = i \neq \varphi(i \times 1) = \varphi(i) = 0 \),这意味着\( \varphi \)非线性。