10.23 Vector Spaces

Check whether the following sets with the defined operation are vector space or not.
1. Let \(V=\left\lbrace\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^2:x\geq0,y\geq0\right\rbrace\) over \(\mathbb{F}=\R\) with
  
  \(\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}w\\ z\end{pmatrix}=\begin{pmatrix}x+w\\ y+z\end{pmatrix}\) and \(c\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}cx\\ cy\end{pmatrix}\)
  
  Not a vector space because we don't have additive inverses
  
  For example \(\begin{pmatrix}1\\ 1\end{pmatrix}\in V,-\begin{pmatrix}1\\ 1\end{pmatrix}=\begin{pmatrix}-1\\ -1\end{pmatrix}\notin V\) 2. Let \(V=\left\lbrace\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^2:xy\geq0\right\rbrace\) over \(\mathbb{F}=\R\) with same operations as (a)
  
  We should have \(v+w\in V\) for all \(v,w\in V\). NO!!
  
  For example \(\begin{pmatrix}1\\ 2\end{pmatrix}+\begin{pmatrix}-2\\ -2\end{pmatrix}=\begin{pmatrix}-1\\ 1\end{pmatrix}\rightsquigarrow\left(-1\right)\cdot1=-1<0\) 3. Let \(V=M_n(\R)\) over \(\mathbb{F}=\R\) with \(A+B=\frac12(AB+BA)\) and \(a\cdot A=0\)
  
  Not a vector space since we have no multiplicative identity
  
  Exercise: check with \(a\cdot B=0\) 4. Let \(X\) be an arbitrary set and \(P(x)=\{A:A\subseteq X\}\)
  
  Consider \(V=P(x)\) over \(\mathbb{F}=\Z_2\) with \(A+B=(A\setminus B)\cup (B\setminus A)\) and \(\begin{cases}0\cdot A=\emptyset\\ 1\cdot A=A\end{cases}\)
  1. Commutativity
    
    \(A+B=(A\setminus B)\cup(B\setminus A)=(B\setminus A)\cup(A\setminus B)=B+A,\checkmark\) 2. Associativity
    1. \((A+B)+C=...=(A\cap B^{c}\cap C^{c})\cup(A^{c}\cap B\cap C^{c})\cup(A^{c}\cap B^{c}\cap C)\cup(A\cap B\cap C)=A+(B+C)\)
    2. \(0\cdot(1\cdot A)=0\cdot A=\emptyset=(0\cdot1)\cdot A\)
      
      \(0\cdot(0\cdot A)=0\cdot\emptyset=\emptyset=(0\cdot0)\cdot A\)
      
      \(1\cdot(1\cdot A)=1\cdot A=A=(1\cdot1)\cdot A\) 3. Additive identity
    Note that \(A+\emptyset=\left(A\setminus\emptyset\right)\cup\left(\emptyset\setminus A\right)=A\cup\emptyset=A\), thus \(\empty\) is additive identity 4. Multiplicative identity
    
    Note that \(1\cdot A=A\) by definition 5. Additive inverse
    
    \(A+A=\left(A\setminus A\right)\cup\left(A\setminus A\right)=\emptyset\cup\emptyset=\emptyset\) 6. Distributivity
    
    Exercise 5. Let \(V=\emptyset\) over any field \(\mathbb{F}\) with any operations
  Not a vector space because we have no additive identity
  
  \(\nexists 0\in V\) because there is nothing in \(V=\empty\)
  
  Exercise: check that all other properties are valid.
In the definition of a vector space, show that (the additive inverse condition) \(c_1\) can be replaced with (the condition that \(0\cdot v=0\)) \(c_2\)

We have to show that \(c_1\Leftrightarrow c_2\)

\(\Rightarrow\)) Suppose we have \(c_1\) (so we have a vector space as normally defined)

Since \(0\cdot v=(0+0)\cdot v=0\cdot v+0\cdot v\) and since \(c_1\) holds, \(\exists -(0\cdot v)\in V\)

Then \(-0\cdot v+0\cdot v=-0\cdot v+0\cdot v+0\cdot v\Rightarrow0\cdot v=0\)

\(\Leftarrow\)) Suppose we have \(c_2\), note that \(0=0\cdot v=(1+(-1))\cdot v\) which \(1+(-1)\) is in field not vector space thus we don't use \(c_1\)

Then we have \(1\cdot v+(-1)\cdot v=v+(-1)\cdot v=0\)

Then \((-1)\cdot v\) is an additive inverse for \(v\)
Consider the vector space \(\R^3=\{(a,b,c):a,b,c\in \R\}\)
1. Is it possible to write the vector \((0,0,0)\in \R^3\) in the form \(0=a(2,2,2)+b(0,0,3)+c(0,1,1)\) with \(a,b,c\in\R\) not all zero?
  
  We have \(\begin{cases}2a=0\\ 2a+c=0\\ 2a+3b+c=0\end{cases}\Rightarrow\begin{cases}a=0\\ c=0\\ b=0\end{cases}\).
  
  Thus it is impossible. 2. Is it possible to write any vector \(v=(v_1,v_2,v_3)\in \R^3\) in the form \(v=a(2,2,2)+b(0,0,3)+c(0,1,1)\) in a unique way?
  
  We have \(\begin{cases}2a=v_1\\ 2a+c=v_2\\ 2a+3b+c=v_3\end{cases}\Rightarrow\begin{cases}a=\frac{v_1}{2}\\ c=v_2-v_1\\ b=\frac{v_3-v_2}{3}\end{cases}\).
  
  Thus we can.
  
  Then we need to prove it is unique
  
  Suppose we have \(v=a(2,2,2)+b(0,0,3)+c(0,1,1)\) and \(v=\tilde a(2,2,2)+\tilde b(0,0,3)+\tilde c(0,1,1)\)
  
  We minus them and get \(0=\left(a-\tilde{a})(2,2,2\right)+(b-\tilde{b})\left(0,0,3\right)+(c-\tilde{c})(0,1,1)\)
  
  By exercise a, we can prove it is unique since \(a=\tilde a,b=\tilde b,c=\tilde c\) 3. Is it possible to write \(u=(-1,2,-1)\) in the form \(u=a(1,2,3)+b(1,0,2)\) for \(a,b\in\mathbb{R}\)?
  
  We have \(\begin{cases}a+b=-1\\ 2a=2\\ 3a+2b=-1\end{cases}\Rightarrow\begin{cases}a=1\\ b=-2\end{cases}\). Yes 4. Is it possible to write any vector \(v=(v_1,v_2,v_3)\in \R^3\) in the form \(v=a(1,2,3)+b(1,0,2)+c(-1,2,-1)\) in a unique way?
  
  No, we can take \(v=\tilde a(1,2,3)+\tilde b(1,0,2)+\tilde c(-1,2,-1)\) and get \(0=(a-\tilde a)(1,2,3)+(\tilde b-b)(1,0,2)+(\tilde c-c)(-1,2,-1)\)
  
  We can take \(\begin{cases}a-\tilde{a}=1\\ b-\tilde{b{}}=-2\\ c-\tilde{c}=-1\end{cases}\), so it is not unique
Consider the real vector space \(V = \{p \in \mathbb{R}[x] : \deg(p) \leq 2\}\) and the following polynomials in \(V\): \(p_1(x) = 2 - x + 3x^2\), \(p_2(x) = 1 + 4x + 2x^2\) and \(p_3(x) = 8 - 10x + x^2\).

Can we write the polynomial \(0 \in V\) in the form \(0 = a p_1(x) + b p_2(x) + c p_3(x)\) with \(a, b, c \in \mathbb{R}\) not all zero?

We need:
\(\begin{aligned}0 &= 2a + b + 8c \quad &\leftarrow \quad 1 \\ 0 &= -a + 4b - 10c \quad &\leftarrow \quad x \\ 0 &= 3a + 2b + c \quad &\leftarrow \quad x^2\end{aligned}\)

\(\implies \begin{cases} b = -2a - 8c \\ a = 4b - 10c \\ c = -3a - 2b\end{cases}\)\(\implies \begin{cases} b = -2a - 8c \\ a = -8a - 32c - 10c \\ c = -3a + 4a + 16c\end{cases}\)\(\implies \begin{cases} b = -2a - 8c \\ 9a = -42c \\ -15c = a\end{cases}\)\(\implies\begin{cases}b=-2a-8c\\ a=-\frac{42}{9}c=-\frac{14}{3}c\\ a=-15c\end{cases}\)\(\implies\begin{cases}b=0\\ a=c=0\\ a=c=0\end{cases}\)

So can we write any polynomial \(p \in V\) in the form \(p(x) = a p_1(x) + b p_2(x) + c p_3(x)\) in a unique way? (exercise)

\(p(x)=\tilde{a}p_1(x)+\tilde{b}p_2(x)+\tilde{c}p_3(x)\)

Then we have \(0=(a-\tilde{a})p_1(x)+(b-\tilde{b})p_2(x)+(c-\tilde{c})p_3(x)\)

We need:
\(\begin{aligned}0 & =2\left(a-\tilde{a}\right)+(b-\tilde{b})+8(c-\tilde{c}) & \leftarrow\quad1\\ 0 & =-\left(a-\tilde{a}\right)+4(b-\tilde{b})-10(c-\tilde{c})\quad & \leftarrow\quad x\\ 0 & =3\left(a-\tilde{a}\right)+2(b-\tilde{b})+(c-\tilde{c})\quad & \leftarrow\quad x^2\end{aligned}\)

By previous exercise, is unique

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