Homework 2
999027873 Yue Shi
-
(45 pts) Check whether the following sets \(V\) with the defined operations are vector spaces over \(\mathbb{R}\) or not. If they are vector spaces, prove the properties. If not, what property fails?
-
\(V = \mathbb{R}_{>0}, \mathbb{F} = \mathbb{Q}\) with \(a \oplus b = a \cdot b ,\,\,\,\, \frac{n}{m}\odot a = \sqrt[m]{a^{n}}\)
-
The sum is associative and commutative \(\checkmark\)
\(a\oplus\left(b\oplus c\right)=a\oplus\left(b\cdot c\right)=a\cdot\left(b\cdot c\right )=\left(a\cdot b\right)\cdot c=\left(a\oplus b\right)\oplus c,\checkmark\)
\(a\oplus b =a\cdot b =b\cdot a =b\oplus a,\checkmark\) 2. There exists a neutral element \(\checkmark\)
The neutral element is \(1\), that is \(1\oplus a=1\cdot a =a,\checkmark\) 3. There exists a additive inverse element \(\checkmark\)
Suppose \(b\) is the additive inverse of \(a\), then \(a\oplus b=a\cdot b=0_{V}\Rightarrow a=b^{-1}\) satisfied \(\checkmark\) 4. \(1\cdot v=v,\forall v\in V,\checkmark\)
\(\frac{1}{1}\odot a=\sqrt[1]{a^{1}}=a,\checkmark\) 5. \(\lambda\cdot(v+w)=\lambda\cdot v+\lambda\cdot w,\forall\lambda\in\mathbb{F},\forall v,w\in V,\checkmark\)
\(\frac{n}{m}\odot(v\oplus w)=\frac{n}{m}\odot(v\cdot w)=\frac{n}{m}\odot(v\cdot w )=\sqrt[m]{(v\cdot w)^{n}}=\sqrt[m]{v^{n}}\cdot\sqrt[m]{w^{n}}=\left(\frac{n}{m}\odot v\right)\oplus\left(\frac{n}{m}\odot w\right),\checkmark\) 6. \(\left(\lambda+\mu\right)\cdot v=\lambda\cdot v+\mu\cdot v,\forall\lambda,\mu\in\mathbb{F} ,\forall v\in V,\checkmark\)
\((\frac{n}{m}+\frac{b}{a})\odot a=(\frac{an+bm}{am})\odot a=\sqrt[am]{a^{an+bm}}= \sqrt[am]{a^{an}\cdot a^{bm}}=\sqrt[am]{a^{an}}\cdot\sqrt[am]{a^{bm}}=\left(\frac{an}{am} \odot a\right)\oplus\left(\frac{bm}{am}\odot a\right)=\left(\frac{n}{m}\odot a\right )\oplus\left(\frac{b}{a}\odot a\right)\checkmark\) 7. \(\lambda\cdot\left(\mu\cdot v\right)=\left(\lambda\cdot\mu\right)\cdot v,\forall\lambda ,\mu\in\mathbb{F},\forall v\in V,\checkmark\)
\(\frac{n}{m}\odot(\frac{b}{a}\odot a)=\frac{n}{m}\odot\sqrt[a]{a^{b}}=\sqrt[m]{(\sqrt[a]{a^{b}})^{n}} =\sqrt[m]{(\sqrt[a]{a})^{bn}}=\sqrt[am]{a^{bn}}=\frac{bn}{am}\odot a=\left(\frac{n}{m} \cdot\frac{b}{a}\right)\odot a,\checkmark\)
Thus this is a vector space 2. \(V = \mathbb{R}^2, \mathbb{F} = \mathbb{R}\) with \((a_{1}, a_{2}) \oplus (b_{1}, b_{2}) = (a_{1}+ a_{2}- 2, b_{1}+ b_{2}- 3) ,\,\,\, \alpha \odot (a, b) = \alpha (a - 2, b - 3) + (2, 3)\)
-
The sum is not associative and commutative
\((a_{1},a_{2})\oplus((b_{1},b_{2})\oplus(c_{1},c_{2}))=\left(a_{1},a_{2}\right)\oplus \left(b_{1}+b_{2}-2,c_{1}+c_{2}-3\right)=\left(a_{1}+a_{2}-2,b_{1}+b_{2}+c_{1}+c_{2} -5\right)\)
\(\left((a_{1},a_{2}\right)\oplus\left(b_{1},b_{2})\right)\oplus(c_{1},c_{2})=(a_{1} +a_{2}-2,b_{1}+b_{2}-3)\oplus(c_{1},c_{2})=\left(a_{1}+a_{2}+b_{1}+b_{2}-5,c_{1}+ c_{2}-3\right)\)
Thus it's not associative
\((a_{1},a_{2})\oplus(b_{1},b_{2})=(a_{1}+a_{2}-2,b_{1}+b_{2}-3)\neq\left(b_{1}+b_{2} -2,a_{1}+a_{2}-3\right)=(b_{1},b_{2})\oplus(a_{1},a_{2})\)
Thus it's not commutative 2. There doesn't exist a neutral element
Assume \((b_1,b_2)\) is neutral element, then:
\((a_{1},a_{2})\oplus(b_{1},b_{2})=(a_{1}+a_{2}-2,b_{1}+b_{2}-3)=\left(a_{1},a_{2} \right)\)
But it's impossible 3. There doesn't exist a additive inverse element
Assume \((b_1,b_2)\) is additive inverse, then:
\((a_{1},a_{2})\oplus(b_{1},b_{2})=(a_{1}+a_{2}-2,b_{1}+b_{2}-3)=\left(0,0\right)\)
But it's impossible
Thus this is not a vector space 3. \(V = \mathrm{M}_n(\mathbb{R}), \mathbb{F} = \mathbb{R}\) with \(A \oplus B = 2A + 2B, \,\,\, c \odot A = c A^{T}\)
Except commutativity of addition and distributivity over vector addition, others are all invalid
For example:
The sum is not associative
\(A\oplus (B\oplus C)=A\oplus (2B+2C)=2A+4B+4C\)
\(\left(A\oplus\left.B\right)\oplus C=\left(2A+2B\right.\right)\oplus C=4A+4B+2C\)
Thus it's not vector space
-
-
-
(25 pts) Let \(V\) be an arbitrary real vector space. Prove that either \(V\) has one element or \(V\) has infinitely many elements.
We know \(0_V\in V\), then if \(0_V\) is the only element in \(V\), then \(V\) has one element, we are done
If not, we need to prove \(V\) has infinitely many elements.
Since \(\{0_V\}\subset V\), then \(\exists 0\neq v_1\in V\), then \(\lambda v_1\in V\) where \(\lambda =\mathbb{R}\)
Since \(|\R|\) is infinite, then \(\lambda v_1\) is infinite, then there are infinitely many elements
-
(30 pts) Are the following subsets \(S\) of the vector space \(V\) subspaces? Justify
-
\(S = S_1 \cap S_2\) for \(S_{1} = \{(x, y, z) \in \mathbb{R}^{3} : x + y + z = 0\}\) and \(S_2 = \{(x, y, z) \in \mathbb{R}^3 : 2x - y + 3z = 0\}\) where \(V = \mathbb{R}^3\).
Take \((x,y,z)\in S\), then \((x,y,z)\in S_1\cap S_2\)
Thus \(x+y+z=0\) and \(2x-y+3z=0\), then \(3x+4z=0\Rightarrow x=-\frac{4}{3}z,y=\frac{1}{3}z\)
Then \(S=\{(-4z,z,3z)\in\mathbb{R}^{3}:z\in\mathbb{R}\}\)
Take \((-4z_{1},z_{1},3z_{1}),(-4z_{2},z_{2},3z_{2})\in S\), then \((-4z_{1},z_{1},3z_{1})+(-4z_{2},z_{2},3z_{2})=(-4\left(z_{1}+z_{2}),z_{1}+z_{2}, 3\left(z_{1}+z_{2}\right)\right)\in S\)
Take \(\lambda\in\mathbb{R},(-4z,z,3z)\in S\), then \(\lambda\cdot\left(-4z,z,3z\right)=\left(-4\lambda z,\lambda z,3\lambda z\right)\in W\)
Thus this is a subspace 2. \(S = \{A \in \mathrm{M}_n(\mathbb{R}) : A^T = A\}\) where \(V = \mathrm{M}_n(\mathbb{R})\).
Take \(A,B\in S\), then \((A+B)^T=A^T+B^T=A+B\), thus \(A+B\in S\)
Take \(\lambda \in \R,A\in S\), then \((\lambda A)^{T}=\lambda A^{T}=\lambda A\), thus \(\lambda A\in S\)
Thus this is a subspace 3. \(S = \{p(x) \in \mathbb{R}[x] : p(x) \text{ has only even powers of } x\}\) where \(V = \mathbb{R}[x]\).
Take \(p(x),q(x)\in S\), then \(p(x)+q(x)\) also has only even powers of \(x\)
Take \(\lambda \in \R,p(x)\in S\), then \(\lambda p(x)\) also has only even powers of \(x\)
Thus this is a subspace
-