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

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

 

 

类似问题:

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。

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 \(R^2\).

 

 

类似问题:

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

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 \(\pm\frac{1}{\sqrt{2}}(i + 1)\).