2.2 Subspaces

Subspaces"/>

2.2 Subspaces

Subspace本身并不是一个很难的概念，就是一个自己也是space的子集合。其检验可以简化为Theorem 1， 即 α , β ∈ W , c ∈ F ⇔ c α + β ∈ W \alpha,\beta \in W,c\in F\Leftrightarrow c\alpha+\beta\in W α,β∈W,c∈F⇔cα+β∈W， 在很多其他的书中，这个定理用于subspace的定义。Example 6 中给出了一些例子，包括zero subspace，symmetric，Hermitan（self-adjoint）空间等。Example 7 说明homogeneous system的解的集合是一个subspace，并且还推广了一个矩阵的带数乘的分配律，即 A ( d B + C ) = d ( A B ) + A C A(dB+C)=d(AB)+AC A(dB+C)=d(AB)+AC，当然里面矩阵的乘法要有意义。
从Theorem 2开始，用一个很新的方式看待某一类特殊的subspace，即subspace spanned by S，Theorem 2首先说明，V的任意subspace的交集还是一个subspace，由此可知，包含S的subspace有一个最小的subspace（不断地取S的subspace的交集），并因此将subspace spanned by S定义出来：所有包含S的subspace的交。这一定义的好处是允许S有无限元素（实际上是无限维元素），当S只包含有限个vector时，subspace spanned by S是传统上大家比较熟悉的样子。Theorem 3即阐述了：subspace spanned by S实际上是S中vector所有的linear combination的集合，注意这里同样没有限制S是有限的，这就是这一种subspace定义方式的好处。
接下来有一个新的定义即集合的和或者说subspace的和，实质是每个集合中任取一个vector并相加而组成的集合。如果 W 1 , W 2 , … , W k W_1,W_2,\dots,W_k W1​,W2​,…,Wk​是 V V V的subspace，那么 W = W 1 + W 2 + ⋯ + W k W=W_1+W_2+\dots+W_k W=W1​+W2​+⋯+Wk​也是一个subspace（按定义可证），并且 W = subspace spanned by  W 1 ∪ ⋯ ∪ W k W=\text{subspace spanned by }W_1\cup\dots \cup W_k W=subspace spanned by W1​∪⋯∪Wk​，因为采用Theorem 3的类似证法，任意包含 W 1 ∪ ⋯ ∪ W k W_1\cup\dots \cup W_k W1​∪⋯∪Wk​的subspace都会包含 W W W。
最后是几个例子，其中Example 10和11比较精彩。Example 10提出了row-space的概念，其是矩阵的row vectors spanned的subspace，为未来引入矩阵的rank打下基础。Example 11是一个无限维的subspace的例子，并且为第四章介绍多项式打下基础。

Exercises

1. Which of the following sets of vectors α = ( a 1 , … , a n ) \alpha =(a_1,\dots,a_n) α=(a1​,…,an​) in R n R^n Rn are subspaces of R n ( n ≥ 3 ) R^n(n\geq3) Rn(n≥3)?

( a ) all α \alpha α such that a 1 ≥ 0 a_1\geq0 a1​≥0;
( b ) all α \alpha α such that a 1 + 3 a 2 = a 3 a_1+3a_2=a_3 a1​+3a2​=a3​;
( c ) all α \alpha α such that a 2 = a 1 2 a_2=a_1^2 a2​=a12​;
( d ) all α \alpha α such that a 1 a 2 = 0 a_1a_2=0 a1​a2​=0;
( e ) all α \alpha α such that a 2 a_2 a2​ is rational.
Solution:
(a) No, since ( 1 , 0 , … , 0 ) (1,0,\dots,0) (1,0,…,0) satisfies, but − ( 1 , 0 , … , 0 ) = ( − 1 , 0 , … , 0 ) -(1,0,\dots,0)=(-1,0,\dots,0) −(1,0,…,0)=(−1,0,…,0) doesn’t.
(b) Yes, for α = ( a 1 , … , a n ) , β = ( b 1 , … , b n ) \alpha=(a_1,\dots,a_n ),\beta=(b_1,\dots,b_n) α=(a1​,…,an​),β=(b1​,…,bn​), if a 1 + 3 a 2 = a 3 , b 1 + 3 b 2 = b 3 a_1+3a_2=a_3,b_1+3b_2=b_3 a1​+3a2​=a3​,b1​+3b2​=b3​, then any c ∈ R c\in R c∈R, the vector c α + β = ( c a 1 + b 1 , … , c a n + b n ) c\alpha+\beta=(ca_1+b_1,\dots,ca_n+b_n) cα+β=(ca1​+b1​,…,can​+bn​) has the property
( c a 1 + b 1 ) + 3 ( c a 2 + b 2 ) = c ( a 1 + 3 a 2 ) + b 1 + 3 b 2 = c a 3 + b 3 (ca_1+b_1 )+3(ca_2+b_2 )=c(a_1+3a_2 )+b_1+3b_2=ca_3+b_3 (ca1​+b1​)+3(ca2​+b2​)=c(a1​+3a2​)+b1​+3b2​=ca3​+b3​
( c ) No, since ( 1 , 1 , … , 0 ) (1,1,\dots,0) (1,1,…,0) satisfies, but 2 ( 1 , 1 , … , 0 ) = ( 2 , 2 , … , 0 ) 2(1,1,\dots,0)=(2,2,\dots,0) 2(1,1,…,0)=(2,2,…,0) doesn’t.
(d) No, since ( 1 , 0 , … , 0 ) (1,0,\dots,0) (1,0,…,0) and ( 0 , 1 , … , 0 ) (0,1,\dots,0) (0,1,…,0) satisfies, but ( 1 , 1 , … , 0 ) = ( 1 , 0 , … , 0 ) + ( 0 , 1 , … , 0 ) (1,1,\dots,0)=(1,0,\dots,0)+(0,1,\dots,0) (1,1,…,0)=(1,0,…,0)+(0,1,…,0) doesn’t.
(e) No, since ( 0 , 1 , … , 0 ) (0,1,\dots,0) (0,1,…,0) satisfies, but 2 ( 0 , 1 , … , 0 ) = ( 0 , 2 , … , 0 ) \sqrt{2} (0,1,\dots,0)=(0,\sqrt{2},\dots,0) 2 ​(0,1,…,0)=(0,2 ​,…,0) doesn’t.

2. Let V V V be the (real) vector space of all functions f f f from R R R into R R R. Which of the following sets of functions are subspaces of V V V?

( a ) all f f f such that f ( x 2 ) = f ( x ) 2 f(x^2)=f(x)^2 f(x2)=f(x)2;
( b ) all f f f such that f ( 0 ) = f ( 1 ) f(0)=f(1) f(0)=f(1);
( c ) all f f f such that f ( 3 ) = 1 + f ( − 5 ) f(3)=1+f(-5) f(3)=1+f(−5);
( d ) all f f f such that f ( − 1 ) = 0 f(-1)=0 f(−1)=0;
( e ) all f f f which are continuous.
Solution:
(a) No, consider f ( x ) = 1 f(x)=1 f(x)=1 if x ≥ 0 x\geq 0 x≥0 and f ( x ) = − 1 f(x)=-1 f(x)=