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

Part 1, additive identity:

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

Part 2, closed under addition:

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}.

Part 1, additive identity（加法单位元）:

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

Part 2, closed under addition（加法封闭性）:

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 $$R^{[0, 1]}$$.

## What is the square root of i?

Solution:

Let $$z = a + bi$$ be the complex number which is a square root of i, that is

$$z^{2} = (a + bi)^{2} = a^{2} – b^{2} + 2abi = i$$

Equating real and imaginary parts, we have

$$\begin{cases} a^{2} – b^{2} & = 0,\\ 2ab & = 1.\end{cases}$$

The two real solution to this pair of equations are

$$\begin{cases} a = \frac{1}{\sqrt{2}},\\ b = \frac{1}{\sqrt{2}},\end{cases}$$

and

$$\begin{cases} a = -\frac{1}{\sqrt{2}},\\ b = -\frac{1}{\sqrt{2}}.\end{cases}$$

The two square root of i therefor are $$\pm\frac{1}{\sqrt{2}}(i + 1)$$.