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 $$\mathbb{R}^{\mathbb{R}}$$.

给出R^2的一个非空子集U的例子，使得U对于加法是封闭并且具有加法逆元的，但U不是R^2的子空间

Question:

Give a example of a nonempty subset $$U$$ of $$\mathbb{R}^2$$ such that $$U$$ is closed under addition and under taking additive inverses but U is not a subspace of $$\mathbb{R}^2$$.

Solution:

Consider the subset $$\mathbb{Z}^2$$. This set is closed under addition, as we will get an integer if we add one integer to another integer. But if we assume $$a = \sqrt{2}$$ and $$u = (1, 1) \in \mathbb{Z}^2$$, we have

$$a \cdot u = \sqrt{2} \cdot (1, 1) = (\sqrt{2}, \sqrt{2}) \notin \mathbb{Z}^2$$.

给出R^2的一个非空子集U的例子，使得U在标量乘法下是封闭的，但U不是R^2的子空间。

Question:

Give an example of a non-empty subset $$U$$ in $$R^2$$ such that $$U$$ is closed under scalar multiplication, but $$U$$ is not a linear subspace of $$R^2$$.

Solution:

Let $$U$$ = {$$(x, y) | x = 0 or y = 0$$}.

Now we take $$u \in U, a \in \mathbb{R}$$.

If $$u = (x, 0)$$, then $$a \cdot u = (ax, 0) \in U$$.

If $$u = (0, y)$$, then $$a \cdot u = (0, ay) \in U$$.

Therefor, U is closed under scalar multiplication.

If a subset U is a subspace, it need to satisfy:

1. $$0 \in U$$.        (This is what we called additive identity)
2. If $$u, v \in U$$, $$u + v \in U$$.        (i.e. $$U$$ is closed under addition)
3. U is closed under scalar multiplication.

Now we have proved that $$U$$ is satisfied the 3. To prove $$U$$ isn’t a subspace of $$R^2$$, we need to prove that U is failed to satisfy 1 or 2.

Note that if we take $$u \in U$$ and $$a = 0$$, we have $$a \cdot u = 0$$. Since we have proved that U is closed under scalar multiplication, now we prove $$0 \in U$$.

Because 2 implies 1 if U is a non-empty space, we have to find a U which isn’t closed under addition.

For $$U$$ = {$$(x, y)$$ | $$x = 0$$ or $$y = 0$$}, we find that if $$u = (x, 0)$$, $$v = (0, y)$$, $$u + v = (x, y) \notin U$$.

Now we find a example {$$(x, y)$$ | $$x = 0$$ or $$y = 0$$} that isn’t a vector space of $$R^2$$.

Is {(a, b, c) in F^3 : a^3 = b^3} a subspace of F^3? (the F is R and C)

There are two question to be solved:

1. Is {$$(a, b, c) \in \mathbb{R}^3 : a^3 = b^3$$} a subspace of $$\mathbb{R}^3$$?

2. Is {$$(a, b, c) \in \mathbb{C}^3 : a^3 = b^3$$} a subspace of $$\mathbb{C}^3$$?

Solution to question 1:

Let $$V$$ = {$$(a, b, c) \in \mathbb{R}^3 : a^3 = b^3$$}.

Obviously $$0 = (0, 0, 0) \in V$$.

Take $$u, w \in V, u = (u_1, u_2, u_3)$$ and $$w = (w_1, w_2, w_3)$$.

We have $$u_1^3 = u_2^3, w_1^3 = w_2^3$$.

Thus $$u_1 = u_2, w_1 = w_2$$        (1-2-1).

Now we have $$u + w = (u_1 + w_1, u_2 + w_2, u_3 + w_3)$$.

According to formulas (1-2-1), we get $$u_1 + w_1 = u_2 + w_2$$

(i.e. $$(u_1 + w_1)^3 = (u_2 + w_2)^3$$).

Therefor $$u + w \in V$$.

Part 3, closed under scalar multiplication:

Take $$u \in V, u = (u_1, u_2, u_3),$$ and $$a \in \mathbb{R}$$.

We have $$au = a(u_1, u_2, u_3) = (au_1, au_2, au_3)$$.

Clearly, $$(au_1)^3 = (au_2)^3 \leftrightarrow u_1^3 = u_2^3$$.

Therefor $$au \in V$$.

Therefor {$$(a, b, c) \in \mathbb{R}^3 : a^3 = b^3$$} is a subspace of $$\mathbb{R}^3$$.

Solution to question 2:

The key to the answer of this question is that for $$a, b \in \mathbb{C}$$, we can’t deduce $$a = b$$ from $$a^3 = b^3$$. Therefor, if we take

$$u, w \in$$ {$$(a, b, c) \in \mathbb{C}^3 : a^3 = b^3$$},

and assume $$u = (u_1, u_2, u_3), v = (v_1, v_2, v_3)$$, we can’t deduce $$(u_1 + v_1)^3 = (u_2 + v_2)^3$$ from $$u_1^3 = u_2^3 and v_1^3 = v_2^3$$.

Thus {$$(a, b, c) \in \mathbb{C}^3 : a^3 = b^3$$} isn’t closed under addition.

i.e. {$$(a, b, c) \in \mathbb{C}^3 : a^3 = b^3$$} is a subspace of $$\mathbb{C}^3$$.

Prove that the set of continuous real-valued functions on the interval [0, 1] is a subspace of R^[0, 1]

Note Before Proof: 证明一个集合是另外一个集合的子空间，只要证明这个集合具有加法单位元0、加法封闭性、标量乘法封闭性即可。

Prove:

Let V = {$$f | f: [0, 1] \rightarrow R$$ such that f is continuous}.

Take $$f_{0} = 0 (\forall x \in [0, 1])$$. Clearly $$f_{0}$$ is continuous and $$f_{0} \in V$$.

Take $$f, g \in V$$.

For each $$\epsilon > 0$$ and for each $$x \in [0, 1]$$ there exists a $$\delta > 0$$. Such that if $$|x_{1} – x_{2}| < \delta$$, then $$|f(x_{1}) – f(x_{2})| < \frac{\epsilon}{2}$$, and $$|g(x_{1}) – g(x_{2})| < \frac{\epsilon}{2}$$.

Since $$f + g = (f + g)(x) = f(x) + g(x)$$, we have

\begin{aligned} & |(f + g)(x_{1}) – (f + g)(x_{2})|\\= & |[f(x_{1}) – f(x_{2})] + [g(x_{1}) – g(x_{2})]|\\ \le & |[f(x_{1}) – f(x_{2})]| + |[g(x_{1}) – g(x_{2})]|\\< & \epsilon.\end{aligned}

i.e. $$f + g$$ is continuous at all $$x \in [0, 1]$$.

Therefor, $$f + g \in V$$.

Part 3, closed under scalar multiplication（标量乘法封闭性）:

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

Assume a= 0, then

$$(af)(x) = a \cdot f(x) = 0, x \in [0, 1]$$.

Clearly, $$af$$ is a continuous real-valued function on the interval [0, 1].

Assume $$a \neq 0$$, for each $$\epsilon > 0$$ and for each $$x \in [0, 1]$$, there exists a $$\delta > 0$$ such that if

$$|x_{1} – x_{2}| < \delta$$,

then

$$|f(x_{1}) – f(x_{2})| < \frac{\epsilon}{a}$$.

Now we have

$$|(af)(x_{1}) – (af)(x_{2})| = |a[f(x_{1}) – f(x_{2})]| = |a|\cdot|f(x_{1}) – f(x_{2})| < \epsilon$$.

Therefor, $$af \in V$$.

Thus V is a subspace of $$R^{[0, 1]}$$.