## 证明区间[0, 1]上的所有实值连续函数构成的实向量空间是无限维的。

Question:

Show that the real vector space consisting of all real-valued continuous function on the interval $$[0, 1]$$ is infinite-dimensional.

Solution:

Since we know the subspace of finite-dimensional vector space is a finite-dimensional vector space, if a space’s subspace is infinite-dimensional, this space will be infinite-dimensional.

Now think about $$P(\mathbb{F})$$ on the interval [0, 1]. Take a list of polynomials of the $$P(\mathbb{F})$$. Let m be the maximum power of this list of polynomials. Then the highest power of polynomials in the span of this polynomial list is $$m$$. Therefor the polynomials whose highest power is $$m + 1$$ are not belong to this span. Thus there not exist any list can span $$P(\mathbb{F})$$. Hence $$P(\mathbb{F})$$ is infinite-dimensional.

Since $$p(\mathbb{F})$$ is a subspace of the vector space V (which is consists of all real-valued continuous functions on the interval $$[0, 1]$$), V is infinite-dimensional.

## 证明F^∞是无限维的

Question:

Prove $$\mathbb{F}^\infty$$ is infinite-dimensional.

Solution:

## 设v1, v2, v3, v4张成V，证明v1 – v2, v2 – v3, v3 – v4, v4也张成V.

Question:

Suppose $$v_1, v_2, v_3, v_4$$ spans $$V$$.

Prove that $$v_1 – v_2$$, $$v_2 – v_3$$, $$v_3 – v_4$$, $$v_4$$ also spans $$V$$.

Solution:

Let $$v \in V$$. Since $$(v_1, v_2, v_3, v_4)$$ spans $$V$$, we know that there exist scalars $$a_1, a_2, a_3, a_4 \in \mathbb{F}$$ such that:

$$a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4 = v$$.

We seek coefficient $$b_1, b_2, b_3, b_4 \in \mathbb{F}$$ such that:

$$b_1 (v_1 – v_2) + b_2 (v_2 – v_3) + b_3 (v_3 – v_4) + b_4 v_4 = v = a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4$$.

Using distributive property, we get:

$$b_1 v_1 + (b_2 – b_1)v_2 + (b_3 – b_2)v_3 + (b_4 – b_3)v_4 = a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4$$.

So we get:

$$b_1 = a_1, b_i – b_{i – 1}= a_i$$.    $$(i \in [2, 4])$$

Now we claim that for each $$b_i$$, there exist $$b_i = \sum_{k = 1}^i a_k$$. Clearly this is correct for $$b_1$$, since $$b_1 = a_1$$. Now we assume $$j \in [2, 3]$$ and $$b_j = \sum_{k = 1}^j a_k$$. Since $$b_i – b_{i – 1}= a_i$$, we can deduce:

$$b_{j + 1} = b_j + a_{j+1} = \sum_{k = 1}^j a_k+ a_{j + 1} = \sum_{k = 1}^{j + 1} a_k$$.

So the formula is proved by induction.

This proved that:

$$a_1(v_1 – v_2) + (a_1 + a_2)(v_2 – v_3) + (a_1 + a_2 + a_3)(v_4 – v_3) + (a_1+a_2+a_3+a_4)v_4 = v$$.

Since for each $$v \in span(v_1, v_2, v_3, v_4)$$ there exist this formula, the list $$v_1 – v_2, v_2 – v_3, v_3 – v_4, v_4$$ spans $$V$$.

## 设Ue是R上的偶函数集合，Uo是R的奇函数集合，证明R^R = Ue与Uo的直和

Foreword:

We use the operator “$$\oplus$$” to express “direct sum”.

Question:

Solution:

Let

$$U_e = \{f: \mathbb{R} \rightarrow \mathbb{R}$$ : such that f is even$$\}$$,

$$U_o = \{f: \mathbb{R} \rightarrow \mathbb{R}$$ : such that f is odd$$\}$$.

For each function $$f(x) \in \mathbb{R}^\mathbb{R}$$, we can divide it into:

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

$$f_o(x) = \frac{f(x) – f(-x)}{2}$$.

Clearly $$f(x) = f_o(x) + f_e(x)$$.

Therefor $$\mathbb{R}^\mathbb{R} = U_e + U_o$$.

Since $$U_e \cap U_o = f_0(x) = 0$$, $$\mathbb{R}^\mathbb{R} = U_e \oplus U_o$$.

## 证明或给出反例：如果U1, U2, W是V的子空间，使得U1 + W = U2 + W，则U1 = U2.

Question:

Prove or give a counterexample if $$U_1, U_2, W$$ are esubspaces of $$V$$ such that $$U_1 + W = U_2 + W$$, then $$U_1 = U_2$$.

Solution:

Let $$V = \mathbb{Z}^2$$, $$U_1 = 4\mathbb{Z}^2$$ and $$U_2 = W = 2\mathbb{Z}^2$$.

Clearly $$U_1 + W = U_2 + W = 2\mathbb{Z}^2$$, but $$U_1 \neq U_2$$.

For another example, we let $$U_1, U_2, W$$ be the spans of $$(0, 1), (1, 1), (1, 0)$$ respectively. Geometrically, $$U_1, U_2, W$$ are three mutually distinct lines. In this situation, we also have $$U_1 + W = U_2 + W = \mathbb{R}^2$$ but $$U_1 \neq U_2$$.

## V的子空间加法运算有单位元吗？哪些子空间有加法逆元？

Question:

Does the operation of addition on the subspaces of $$V$$ have an additive identity? Which subspaces have additive inverses?

Solution:

Obviously the subspace $$\{0\}$$ is an additive identity for the operation of addition on the subspace of $$V$$. Every subspace of $$V$$ plus $$\{0\}$$ is the subspace itself.

Assume $$U$$, $$V$$ are two subspaces of $$V$$. If $$V$$ is the additive inverse of $$U$$, $$U + V = \{0\}$$. Since both of them should be contained in the subspace $$U + V$$, they have to satisfy that $$U = V = \{0\}$$.

## 设U是V的子空间，求U + U

Question:

Suppose that $$U$$ is a subspace of $$V$$. What is $$U + U$$?

Solution:

$$U + U = U$$.

Proof:

Take $$u, v \in U$$. Then every element in $$U + U$$ can be written as $$u + v$$. Since $$U$$ is a subspace of $$V$$, U is closed under addition. Therefor $$u + v \in U$$. Hence $$U + U \subset U$$.

If we take $$w = u + 0 \in U + U$$, we can get every element in $$U$$ ($$w \in U$$). Therefor $$U \subset U + U$$.

Thus $$U = U + U$$.

## 证明V的三个子空间的并是V的子空间，当且仅当其中一个子空间包含另外两个子空间

Foreword++:

Foreword:

I strongly recommend you to read the previous blog which discussed a similar problem. You can go to that blog by clicking link: 证明V的两个子空间的并是V的子空间当且仅当其中一个子空间包含另一个子空间. I will skip the same part of this proof which is discussed in that proof.

Question:

Prove that the union of three subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces of $$V$$ contains the others.

Solution:

Let $$A$$, $$B$$, $$C$$ be the three subspaces of V.

Part 1:

Assume $$A \cup B \cup C = A$$ or $$B$$ or $$C$$.

Clearly $$A \cup B \cup C$$ is one of these three vector spaces of $$A$$, $$B$$, $$C$$. Therefor $$A \cup B \cup C$$ is the subspaces of $$V$$.

Part 2:

Assume $$A \cup B \cup C \neq A$$ or $$B$$ or $$C$$.

Firstly, consider the condition:

$$A \cup B \cup C = A \cup B$$ or $$B \cup C$$ or $$A \cup C$$.

We have discussed this situation in the previous blog, you can read it by clicking this link: 证明V的两个子空间的并是V的子空间当且仅当其中一个子空间包含另一个子空间.

Secondly, consider the situation that neither of these three subspaces contained in the others.

## 证明V的两个子空间的并是V的子空间当且仅当其中一个子空间包含另一个子空间

Question:

Prove that the union of two subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces of $$V$$ is contained in the other.

Solution:

Assume two set A, B are subspaces of $$V$$.

Part 1:

Assume $$A \cup B = A$$ or $$B$$.

Clearly $$A \cup B = A$$ or $$B \in V$$.

Therefor if one subspace of $$V$$ is contained in the other, the union of two subspaces is a subspace of $$V$$.

Part 2:

Assume $$A \cup B \neq A$$ or $$B$$.

Now we take $$a \in A$$ but not in $$B$$, and $$b \in B$$ but not in $$A$$.

Since $$a \in A$$, we know $$-a \in A$$.

Assume $$a + b \in A$$.

As we know, $$A$$ is a subspace of $$V$$. So $$A$$ is closed under addition.

Therefor, $$(a + b) + (-a)$$ should be in A.

But in fact, $$(a + b) + (-a) = b$$. We have assumed that $$b \notin A$$. We have reached a contradiction on the assumption that $$a + b \in A$$. Thus $$a + b \notin A$$.

If we assume $$a + b \in B$$, we will also reach a contradiction like this.

Therefor, $$a + b$$ is not in A and B.

i.e. $$a + b \notin A \cup B$$.

Then the set $$A \cup B$$ isn’t closed under addition.

Therefor if $$A \cup B \neq A$$ or $$B$$, $$A \cup B$$ is not a subspace of V.

Now we have proved that the union of two subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces of $$V$$ is contained in the other.

## R到R的周期函数构成的集合是R^R的子空间吗？

Question:

A function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ 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 $$\mathbb{R} \rightarrow \mathbb{R}$$ a subspace of $$\mathbb{R}^{\mathbb{R}}$$? Explain.

Solution:

The answer to this question is YES. Now we will prove it.

Let $$V$$ be the set of all periodic functions from $$\mathbb{R} \rightarrow \mathbb{R}$$.

Part 1:

Let $$f_0(x) = 0$$. Clearly $$f_0 \in V$$.

Part 2:

Take $$f, g \in V$$, $$f(x + m) = f(x)$$, $$g(x + n) = g(x)$$.

Clearly, \begin{aligned}(f + g)(x + lcm(m, n)) = & f(x + lcm(m, n)) + g(x + lcm(m, n)) \\ = & f(x) + g(x) \\ = & (f + g)(x)\end{aligned}.

Thus $$f + g \in V$$.

Part 3:

Take $$f \in V$$, and $$a \in R$$.

We have $$(af)(x + p) = a \cdot f(x + p) = a \cdot f(x) = (af)(x)$$.

Thus $$af \in V$$.

Therefor the set of all periodic functions from $$\mathbb{R} \rightarrow \mathbb{R}$$ is a subspace of $$\mathbb{R}^{\mathbb{R}}$$.