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.

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。

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.

证明\(\mathbb{F}^\infty\)是无限维的。

Solution:

设\(e_1 = (1, 0, 0, …)\),\(e_2 = (0, 1, 0, …)\)…\(e_m = (0, 0, …【1前面共计m – 1个0】, 1, 0, …)\),对于向量组\(e_1, e_2, …, e_m\)张成的空间\(span(e_1, e_2, …, e_m)\),总存在向量\(e_{m + 1} = (0, 0, …【1前面共计m个0】, 1, 0, …) \notin span(e_1, e_2, …, e_m)\),由此可证明\(\mathbb{F}^\infty\)是无限维的。

证毕。

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:

函数\(f: R \rightarrow R\)称为偶函数,如果对所有\(x \in R\)均有\(f(-x) = f(x)\)。函数\(f: R \rightarrow R\)称为奇函数,如果对所有\(x \in R\)均有\(f(-x) = -f(x)\)。用\(U_e\)表示\(R\)上的实值偶函数的集合,用\(U_o\)表示\(R\)上实值奇函数的集合,证明\(R^R = u_e \oplus U_o\).

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++:

标记,目前没有完成这道题,想出证明后会进行完善。

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.

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两个子空间。

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.

证明V的两个子空间的并是B的子空间当且仅当其中一个子空间包含领一个子空间。

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 \(\mathbb{R}^{\mathbb{R}}\).