Chapter 1.C Subspaces 子空间
这一主要讨论的子空间相关的内容，涉及子空间的定义和判定、子空间的和是最小的包含子空间以及直和。
1.32 Definition subspace 子空间定义
A subset U of V is called a subspace of V if U is also a vector space(using the same addition and scalar multiplication as on V )
U作为V的子空间自然也满足在V中的加法和乘法 eg1.33
1.34 Conditions for a subspace 满足子空间的条件
A subset U of V is a subspace of V if and only if U satisfies the following three conditions:
- additive identity : 0 ∈ U
- closed under addition : u,w ∈ U implies u+v ∈ U
- closed under scalar multiplication:a ∈ F and u ∈ U implies au ∈ U
The other parts of the definition of a vector space, such as associativity and commutativity, are automatically satisfied for U because they hold on the larger space V .
此处证明三个条件非常简单，这里只用满足三个条件便可以证明子空间，至于U中的其他性质，因为U是V的子空间，而V中已经证明过相关性质，所以U自然也满足
作者在这里举了一些例子，其中包含一些函数子空间的例子，作者在这里强调了微积分部分的线性结构，本质上就是泛函分析，感兴趣可以去原书eg1.35看。
Sums of Subspaces 子空间的和
1.36 Definition sum of subsets 子空间的和的定义
Suppose U₁ ,..., Uₘ are subsets of V . The sum of U₁ ,..., Uₘ , denoteed U₁ +···+ Uₘ , is the set of all possible sums of elements of U₁ ,..., Uₘ . More precisely, U₁ +···+ Uₘ = { u₁ +···+ uₘ : u₁ ∈ U₁ ,..., uₘ ∈ Uₘ }.
eg1.37很好理解，eg1.38或许有些难理解，这里解释一下：
U = {x,x,y,y ∈ F⁴ :x,y ∈ F⁴ }, W = {x,x,x,y ∈ F⁴ :x,y ∈ F⁴ }
这里的 U+W 是如何等于(x,x,y,z)的呢，按正常计算，U+W 应该是（2x,2x,x+y,2y）,在这里令x'=2x,y'=x+y,z'=2y,就可以得出（x',x',y',z'）了，书上并没有这个过程，所有读者可能有点疑惑。
1.39 Sum of subspaces is the smallest containing subspace 子空间的和是最小的包含子空间
Suppose U₁ ,..., Uₘ are subspaces of V . Then U₁ +···+ Uₘ is the smallest subspace of V containing U₁ ,..., Uₘ .
这里要证明的有三点，一是子空间，二是有包含关系，三是最小。要证子空间很简单，只要满足上面的三个条件即可。令 uⱼ ∈ Uⱼ , wⱼ ∈ Uⱼ , 那么 u₁ +···+ uₘ ∈ U₁ +···+ Uₘ , w₁ +···+ wₘ ∈ U₁ +···+ Uₘ , 所以（ u₁ +···+ uₘ ） + （ w₁ +···+ wₘ ）=（ u₁ + w₁ ）+···+（ uₘ + wₘ ） ∈ U₁ +···+ Uₘ , 由此证加法封闭，乘法证法相同；0 ∈ uⱼ ，则0 +···+ 0 = 0 ∈ U₁ +···+ Uₘ ，由此加法元也证得，因此满足子空间条件。
要证包含，只需证（ U₁ ∪···∪ Uₘ ）⊂ U₁ +···+ Uₘ ，只需证( U₁ +···+ Uₘ ) ⊃ U₁ 且···且 ( U₁ +···+ Uₘ ) ⊃ Uₘ ，而这是显然的，由此包含关系证得。
要证最小，书上是这样证明的：
Conversely, every subspace of V containing U₁ ,..., Uₘ contains U₁ +···+ Uₘ (because subspaces must contain all finite sums of their elements)
这里可以用数轴的概念来辅助理解，在一条数轴上的[0，1]区间，0是最小元素只需证明0在∀ x ∈ [0,1]的条件下都满足0 ≦ x , 类比到这里就是∀ V ⊃ U₁ ∪···∪ Uₘ ，证 V ⊃ U₁ +···+ Uₘ ⊃ U₁ ∪···∪ Uₘ ，我们可以知道 U₁ ,..., Uₘ ∈ V ，再由 V 是一个子空间可以推出 U₁ +···+ Uₘ ∈ V 从而得出 U₁ +···+ Uₘ ⊂ V 。
Direct Sums 直和
1.40 Definition direct sum 直和的定义
Suppose U₁ ,..., Uₘ are subspaces of V .
- The sum U₁ +···+ Uₘ is called a direct sum if each element of U₁ +···+ Uₘ can be written in only one way as a sum u₁ +···+ uₘ , where each uⱼ is in Uⱼ .
- If U₁ +···+ Uₘ is a direct sum, then U₁ ⊕···⊕ Uₘ denotes U₁ +···+ Uₘ , with the ⊕ notation serving as an indication that this is a direct sum.
简单来说，直和的意思就是 U₁ +···+ Uₘ 中的元素只能由属于 Uⱼ 的元素 uⱼ 来以 u₁ +···+ uₘ 的形式表示，并且这种表示方式只有一种。eg1.41 eg1.42 eg1.43
如果每次都要用这种方式来判断直和的话就会变得非常麻烦，所以这里就有一种更加简单的判断直和的方式。
1.44 Condition for a direct sum 满足直和的条件
Suppose U₁ +···+ Uₘ are subspaces of V . Then U₁ +···+ Uₘ is a direct sum if and only if the only way to write 0 as a sum u₁ +···+ uₘ , where each uⱼ is in Uⱼ , is by taking each uⱼ equal to 0.
既然判断直和的办法是判断 U₁ +···+ Uₘ 中的元素只能由 u₁ +···+ uₘ 得出，并且只要是子空间就有加法元0存在，那么只用判断 U₁ +···+ Uₘ 中的加法元0只能由 u₁ +···+ uₘ 得出，并且每个 uⱼ 的值只能为0。
证明如下：
Suppose U₁ +···+ Uₘ is a direct sum.
Now suppose that the only way to write 0 as a sum u₁ +···+ uₘ , where each uⱼ is in Uⱼ , is by taking each uⱼ equal to 0. To show that U₁ +···+ Uₘ is a direct sum, let v ∈ U₁ +···+ Uₘ . We can write
v = u₁ +···+ uₘ for some u₁ ∈ U₁ ,..., uₘ ∈ Uₘ . To show that this representation is unique, suppose we also have
v = v₁ +···+ vₘ ,
where v₁ ∈ U₁ ,..., vₘ ∈ Uₘ . Subtracting these two equations, we have0 = ( u₁ - v₁ )+···+( uₘ - vₘ ). Because u₁ - v₁ ∈ U₁ ,..., uₘ - vₘ ∈ Uₘ , the equation above implies that each uⱼ - vⱼ equals 0. Thus u₁ = v₁ ,..., uₘ = vₘ , as desired.
1.45 Direct sum of two subspaces 两个子空间的直和
Suppose U and W are subspaces of V . Then U + W is a direct sum if and only if U ∩ W = {0}.
证明如下:
First suppose that U + W is a direct sum. If v ∈ U ∩ W , then 0 = v + ( -v ), where v ∈ U and -v ∈ W . By the unique representation of 0 as the sum of a vector in U and a vector in W , we have v = 0. Thus U ∩ W = {0}, completing the proof in one direction.
To prove the other direction, now suppose U ∩ W = {0}. To prove that U + W is a direct sum, suppose u ∈ U , w ∈ W and
0 = u + w To complete the proof, we need only show that u = w (by 1.44). The equation above implies that u = -w ∈ W . Thus u ∈ U ∩ W . Hence u = 0, which by the equation above implies that w = 0, completing the proof.