Lecture 2: Subspaces, Sums, and Direct Sums (子空间，和 与 直和)

1. Motivation (动机)

在 Lecture 1 中，我们定义了向量空间（见 定义 1.20）。但在实际应用中，我们往往不需要研究整个庞大的空间，而是关注其中的“局部结构”。

• 子集 vs 子空间：子空间之于向量空间，类似于子集之于集合。但并非所有子集都是子空间，子集必须自身也满足向量空间的运算封闭性。
• 并集 vs 和：在集合论中，我们用“并集”来合并两个集合；但在向量空间中，两个子空间的并集通常不是子空间。因此，我们需要引入新的运算——“子空间的和”。

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2. Subspaces (子空间)

2.1 Definition

Definition 1.33 (Subspace) 向量空间 $V$ 的子集 $U$ 被称为 $V$ 的子空间，如果 $U$ 在 $V$ 的加法和标量乘法下，自身也构成一个向量空间。

2.2 Conditions for a Subspace

Note: 验证子空间时，不需要逐一验证 Lecture 1 中的 6 条公理（如加法交换律、结合律等），因为这些性质直接从父空间 $V$ “继承”。我们只需要验证 $U$ 是否对运算封闭以及包含必要的特殊元素（这正是 命题 1.34 的作用）。

Proposition 1.34 (子空间的三个判定条件) $V$ 的子集 $U$ 是子空间 $\iff$ $U$ 满足以下三个条件：

1. 加法单位元 (Additive Identity): $0 \in U$
2. 对加法封闭 (Closed under addition): $u, w \in U \implies u+w \in U$
3. 对标量乘法封闭 (Closed under scalar multiplication): $a \in F, u \in U \implies au \in U$

2.3 Examples & Typical Examples

常见反例与正例：

• $U = \{(x_1, x_2, 0) : x_1, x_2 \in F\}$ 是 $F^3$ 的子空间。（满足三条件）
• $U = \{(x_1, x_2, x_3, x_4) \in F^4 : x_3 = 5x_4 + b\}$ 是 $F^4$ 的子空间 $\iff b = 0$。（若 $b \neq 0$，则 $0 \notin U$）
• $U = \{f \text{ differentiable on } (0,3) : f'(2) = b\}$ 是子空间 $\iff b = 0$。

典型几何直观：

• $\{0\}$ 是最小的子空间，$V$ 是最大的子空间。
• $\mathbb{R}^2$ 的子空间：$\{0\}$、$\mathbb{R}^2$、过原点的直线。
• $\mathbb{R}^3$ 的子空间：$\{0\}$、$\mathbb{R}^3$、过原点的直线、过原点的平面。

2.4 Intersection of Subspaces

Theorem 1.40 任意多个子空间的交集仍然是子空间。 证明思路：利用子空间的三个判定条件（命题 1.34）。因为每个子空间都包含 $0$，所以交集包含 $0$；封闭性同理可从各个子空间的封闭性传递到交集。

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3. Sums of Subspaces (子空间的和)

3.1 Why not Union?

两个子空间的并集 $W_1 \cup W_2$ 很少是子空间（除非其中一个包含另一个）。因为并集无法保证加法封闭性（例如：一条直线上的向量加上另一条直线上的向量，结果可能落在两条直线之外）。

3.2 Definition of Sum

Definition 1.36 (Sum of Subspaces) 给定 $V$ 的子空间 $U_1, U_2, \dots, U_m$，它们的和定义为：

$$U_1 + U_2 + \dots + U_m = \{u_1 + u_2 + \dots + u_m : u_i \in U_i \text{ for each } i\}$$

3.3 Smallest Subspace Property

Proposition 1.40 (最小子空间) $U_1 + U_2 + \dots + U_m$ 是 $V$ 中包含 $U_1, U_2, \dots, U_m$ 的最小子空间。

概念类比：子空间的“和”在向量空间理论中，完全类比于集合论中的“并集”（包含它们的最小集合）。

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4. Direct Sums (直和)

4.1 Definition & Notation

Definition 1.41 (Direct Sum) 如果和 $U_1 + U_2 + \dots + U_m$ 中的每一个向量都能以唯一的方式表示为 $u_1 + u_2 + \dots + u_m$（其中 $u_i \in U_i$），则称该和为直和。

• 符号：使用 $\oplus$ 表示，如 $U_1 \oplus U_2 \oplus \dots \oplus U_m$。圆圈提醒我们：表示是唯一的（无冗余）。

4.2 Condition for a Direct Sum

Proposition 1.45 (零向量的唯一分解) $U_1 + \dots + U_m$ 是直和 $\iff$ 将 $0$ 写成 $u_1 + \dots + u_m$ 的唯一方式是 $u_1 = u_2 = \dots = u_m = 0$。 证明思路：

• ($\implies$) 由直和定义（定义 1.41）直接得出。
• ($\impliedby$) 假设有两种表示 $v = u_1 + \dots + u_m$ 和 $v = v_1 + \dots + v_m$，两式相减得 $0 = (u_1-v_1) + \dots + (u_m-v_m)$。由前提条件可知 $u_i - v_i = 0$，即 $u_i = v_i$，证明唯一性。

4.3 Direct Sum of Two Subspaces

Proposition 1.46 对于两个子空间 $U, W$：$U + W$ 是直和 $\iff U \cap W = \{0\}$（与 命题 1.45 一起构成直和的两个判据）。 证明思路：

• ($\implies$) 若 $v \in U \cap W$，则 $0 = v + (-v)$。由唯一性得 $v = 0$。
• ($\impliedby$) 若 $0 = u + w$，则 $u = -w$。这意味着 $u \in U \cap W = \{0\}$，故 $u=0, w=0$。

4.4 Important Caveat (重要易错点)

Note: 对于两个子空间，两两交集为 $\{0\}$ 是直和的充要条件。 但对于三个或更多子空间（$m > 2$），仅仅满足“两两交集为 $\{0\}$”是不够的！ 反例思考：在 $\mathbb{R}^2$ 中，取三条过原点但互不重合的直线 $U_1, U_2, U_3$。它们两两交集都是 $\{0\}$，但 $U_1 + U_2 + U_3$ 绝不是直和（因为 $\mathbb{R}^2$ 中任意向量可以用无数种方式由这三条直线上的向量相加得到）。

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5. 概念联系图 (Concept Map)

Vector Space V
 │
 ├── Subsets (子集) ──(不一定封闭)──> 不是子空间
 │
 ├── Subspaces (子空间) ──(满足3条件)──> 继承父空间公理
 │      │
 │      ├── Intersection (交集) ──(任意多个)──> 仍是子空间 (类比集合的交)
 │      │
 │      └── Sum (和) ──(包含它们的最小子空间)──> 仍是子空间 (类比集合的并)
 │             │
 │             └── Direct Sum (直和, ⊕) ──(要求元素表示唯一/0的分解唯一)
 │                    │
 │                    └── 两个子空间直和 ⟺ 交集仅为 {0} (类比集合的不相交并)

6. Examples

1.C.7. Prove or give a counterexample: If $U$ is a nonempty subset of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), then $U$ is a subspace of $\mathbb{R}^2$.

Solution:

This statement is false. We provide a counterexample.

Let $U = \{(x, y) \in \mathbb{R}^2 : x, y \in \mathbb{Z}\}$ (the set of points with integer coordinates).

• $U$ is nonempty since $(0, 0) \in U$.
• $U$ is closed under addition: if $(x_1, y_1), (x_2, y_2) \in U$, then $x_1, y_1, x_2, y_2 \in \mathbb{Z}$, so $(x_1 + x_2, y_1 + y_2) \in U$.
• $U$ is closed under taking additive inverses: if $(x, y) \in U$, then $x, y \in \mathbb{Z}$, so $-x, -y \in \mathbb{Z}$, which means $(-x, -y) \in U$.

However, $U$ is not a subspace of $\mathbb{R}^2$ because it is not closed under scalar multiplication. For example, $(1, 0) \in U$, but $\frac{1}{2}(1, 0) = (\frac{1}{2}, 0) \notin U$ since $\frac{1}{2} \notin \mathbb{Z}$.

Therefore, the conditions of being nonempty, closed under addition, and closed under taking additive inverses are not sufficient to guarantee that a subset is a subspace.

1.C.9. A function $f \colon \mathbb{R} \to \mathbb{R}$ is called periodic if there exists a positive number $p$ such that $f(x) = f(x + p)$ for all $x \in \mathbb{R}$. Is the set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ a subspace of $\mathbb{R}^\mathbb{R}$? Explain.

Solution:

No, the set of periodic functions is not a subspace of $\mathbb{R}^{\mathbb{R}}$.

To show this, we need to demonstrate that the set of periodic functions is not closed under addition.

Consider the following two periodic functions:

• $f(x) = \sin(x)$, which has period $2\pi$
• $g(x) = \sin(\sqrt{2}x)$, which has period $\frac{2\pi}{\sqrt{2}} = \sqrt{2}\pi$

Both $f$ and $g$ are periodic functions. However, their sum $h(x) = f(x) + g(x) = \sin(x) + \sin(\sqrt{2}x)$ is not periodic.

To see why, suppose $h$ has a period $p > 0$. Then for all $x \in \mathbb{R}$:

$$h(x + p) = h(x)$$ $$\sin(x + p) + \sin(\sqrt{2}(x + p)) = \sin(x) + \sin(\sqrt{2}x)$$

This would require both $\sin(x)$ and $\sin(\sqrt{2}x)$ to have $p$ as a common period. However, the ratio of their fundamental periods is:

$$\frac{2\pi}{\sqrt{2}\pi} = \sqrt{2}$$

Since $\sqrt{2}$ is irrational, there is no common period for these two functions. Therefore, $h(x) = \sin(x) + \sin(\sqrt{2}x)$ is not periodic.

Since the sum of two periodic functions is not necessarily periodic, the set of periodic functions is not closed under addition and thus is not a subspace of $\mathbb{R}^{\mathbb{R}}$.

1.C.14. Suppose $U = \{(x, -x, 2x) \in \mathbb{F}^3 : x \in \mathbb{F}\}$ and $W = \{(x, x, 2x) \in \mathbb{F}^3 : x \in \mathbb{F}\}$. Describe $U + W$ using symbols, and also give a description of $U + W$ that uses no symbols.

Solution: By the definition of the sum of subspaces, we have:

$$U + W = \{u + w : u \in U, w \in W\}$$

Substituting the definitions of $U$ and $W$:

$$U + W = \{(x, -x, 2x) + (y, y, 2y) : x, y \in \mathbb{F}\}$$ $$= \{(x+y, -x+y, 2x+2y) : x, y \in \mathbb{F}\}$$

Let $a = x+y$, $b = -x+y$, and $c = 2x+2y$. Notice that $c = 2(x+y) = 2a$. Since $x$ and $y$ can be any elements in $\mathbb{F}$, $a$ and $b$ can also independently take any values in $\mathbb{F}$ (we can solve for $x = \frac{a-b}{2}$ and $y = \frac{a+b}{2}$). Thus, using symbols, we can describe $U + W$ as:

$$U + W = \{(a, b, 2a) \in \mathbb{F}^3 : a, b \in \mathbb{F}\}$$

Description without symbols: $U + W$ is the plane in $\mathbb{F}^3$ consisting of all vectors whose third coordinate is exactly twice their first coordinate. (It is a 2-dimensional subspace of $\mathbb{F}^3$).

1.C.19. Prove or give a counterexample: If $V_1, V_2, U$ are subspaces of $V$ such that $V_1 + U = V_2 + U$, then $V_1 = V_2$.

Solution: This statement is FALSE. We provide a counterexample. Let $V = \mathbb{F}^2$. Let $U = \mathbb{F}^2$ (which is a subspace of $V$). Let $V_1 = \{(x, 0) \in \mathbb{F}^2 : x \in \mathbb{F}\}$ (the x-axis). Let $V_2 = \{(0, y) \in \mathbb{F}^2 : y \in \mathbb{F}\}$ (the y-axis). Clearly, $V_1$ and $V_2$ are subspaces of $V$, and $V_1 \neq V_2$.

By the definition of the sum of subspaces (Definition 1.36), since $U = \mathbb{F}^2$ already contains all possible vectors in $V$, adding any subspace to $U$ will just result in $U$ itself. Thus, $V_1 + U = \mathbb{F}^2$ and $V_2 + U = \mathbb{F}^2$. We have $V_1 + U = V_2 + U$, but $V_1 \neq V_2$. This contradicts the statement.

1.C.22. Suppose $U = \{(x, y, x + y, x - y, 2x) \in \mathbb{F}^5 : x, y \in \mathbb{F}\}$. Find three subspaces $W_1, W_2, W_3$ of $\mathbb{F}^5$, none of which equals $\{0\}$, such that $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$.

Solution: First, observe that $U$ is determined by two free variables $x$ and $y$. Thus, $\dim U = 2$. A basis for $U$ is given by the vectors $u_1 = (1, 0, 1, 1, 2)$ (setting $x=1, y=0$) and $u_2 = (0, 1, 1, -1, 0)$ (setting $x=0, y=1$). Since $\dim \mathbb{F}^5 = 5$, we need the sum of the dimensions of $W_1, W_2, W_3$ to be $5 - 2 = 3$. Since none of them can be $\{0\}$, each must have dimension exactly 1.

Let us choose the standard basis vectors $e_3 = (0, 0, 1, 0, 0)$, $e_4 = (0, 0, 0, 1, 0)$, and $e_5 = (0, 0, 0, 0, 1)$. Define:

• $W_1 = \text{span}(e_3) = \{(0, 0, z, 0, 0) : z \in \mathbb{F}\}$
• $W_2 = \text{span}(e_4) = \{(0, 0, 0, z, 0) : z \in \mathbb{F}\}$
• $W_3 = \text{span}(e_5) = \{(0, 0, 0, 0, z) : z \in \mathbb{F}\}$

None of these subspaces equals $\{0\}$.

To show that $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$, we use the condition for a direct sum (Proposition 1.45), which states that a sum of subspaces is a direct sum if and only if the only way to write $0$ as a sum of vectors from each subspace is by taking each vector to be $0$. Suppose $u + w_1 + w_2 + w_3 = 0$, where $u \in U$, $w_1 \in W_1$, $w_2 \in W_2$, $w_3 \in W_3$. Then there exist $x, y, z_1, z_2, z_3 \in \mathbb{F}$ such that:

$$(x, y, x+y, x-y, 2x) + (0, 0, z_1, 0, 0) + (0, 0, 0, z_2, 0) + (0, 0, 0, 0, z_3) = (0, 0, 0, 0, 0)$$

This gives the system of equations:

1. $x = 0$
2. $y = 0$
3. $x + y + z_1 = 0 \implies z_1 = 0$
4. $x - y + z_2 = 0 \implies z_2 = 0$
5. $2x + z_3 = 0 \implies z_3 = 0$

Thus, $u = 0, w_1 = 0, w_2 = 0, w_3 = 0$. By Proposition 1.45, the sum is a direct sum. Furthermore, the dimension of $U \oplus W_1 \oplus W_2 \oplus W_3$ is $2 + 1 + 1 + 1 = 5 = \dim \mathbb{F}^5$. Since a subspace of dimension equal to the whole space must be the whole space itself (Proposition 2.39), we conclude $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$.

1.C.24. A function $f \colon \mathbb{R} \to \mathbb{R}$ is called even if $f(-x) = f(x)$ for all $x \in \mathbb{R}$. A function $f \colon \mathbb{R} \to \mathbb{R}$ is called odd if $f(-x) = -f(x)$ for all $x \in \mathbb{R}$. Let $V_e$ denote the set of real-valued even functions on $\mathbb{R}$ and let $V_o$ denote the set of real-valued odd functions on $\mathbb{R}$. Show that $\mathbb{R}^\mathbb{R} = V_e \oplus V_o$.

Solution: First, we verify that $V_e$ and $V_o$ are subspaces of $\mathbb{R}^\mathbb{R}$ using the three conditions for a subspace (命题 1.34):

1. The zero function $0(x) = 0$ satisfies $0(-x) = 0(x)$ and $0(-x) = -0(x)$, so $0 \in V_e$ and $0 \in V_o$.
2. If $f, g \in V_e$, then $(f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x)$, so $V_e$ is closed under addition. Similarly for $V_o$.
3. If $f \in V_e$ and $c \in \mathbb{R}$, then $(cf)(-x) = c f(-x) = c f(x) = (cf)(x)$, so $V_e$ is closed under scalar multiplication. Similarly for $V_o$. Thus, $V_e$ and $V_o$ are subspaces of $\mathbb{R}^\mathbb{R}$.

Next, we show that $\mathbb{R}^\mathbb{R} = V_e + V_o$. For any arbitrary function $f \in \mathbb{R}^\mathbb{R}$, we can define two new functions:

$$f_e(x) = \frac{f(x) + f(-x)}{2}$$ $$f_o(x) = \frac{f(x) - f(-x)}{2}$$

We can verify that $f_e$ is even: $f_e(-x) = \frac{f(-x) + f(x)}{2} = f_e(x)$, so $f_e \in V_e$. We can verify that $f_o$ is odd: $f_o(-x) = \frac{f(-x) - f(x)}{2} = -\frac{f(x) - f(-x)}{2} = -f_o(x)$, so $f_o \in V_o$. Furthermore, $f_e(x) + f_o(x) = \frac{f(x) + f(-x) + f(x) - f(-x)}{2} = f(x)$. Thus, every function $f$ can be written as the sum of an even function and an odd function, meaning $\mathbb{R}^\mathbb{R} = V_e + V_o$.

Finally, we show that the sum is a direct sum by proving $V_e \cap V_o = \{0\}$ (Proposition 1.46). Suppose $f \in V_e \cap V_o$. Then for all $x \in \mathbb{R}$, we have both $f(-x) = f(x)$ and $f(-x) = -f(x)$. Equating the two gives $f(x) = -f(x)$, which implies $2f(x) = 0$, so $f(x) = 0$ for all $x \in \mathbb{R}$. Thus, $f$ is the zero function, meaning $V_e \cap V_o = \{0\}$. By Proposition 1.46, we conclude that $\mathbb{R}^\mathbb{R} = V_e \oplus V_o$.

7. Questions to Carry Forward (下节课前瞻)

1. 我们定义了子空间、和与直和，但如何衡量一个向量空间或子空间的“大小”？（引出维数 Dimension的概念）
2. 如何用一组向量来“生成”（span）一个子空间？
3. 什么是“线性无关”（Linear Independence）？它与直和的概念有什么内在联系？