Chapter 1 Questions

Based on 1A. Preliminaries and Introduction,1B. What is a vector space?, and 1C. Subspaces.

1A

7.

Just algebra

10.

Each factor has to be scaled by the same amount. Lets take the first coordinates to start: $\frac{12-5i}{2-3i}=3+2i$ With this, we can see if this works for the rest of the coordinates.

For the second coordinate $(3+2i)(5+4i)=15+12i+10i-8=7+22i$ which does math

For the last coordinate, $(3+2i)(-6+7i)=-18+21i-12i-14=-32+9$

The third coordinate does not work as it is plus 9.

1B

1

Proof:

1. Firstly, notice that $v+(-v)=0$
2. Thus, the additive inverse of $-v$ is $v$.

4

Proof:

1. First, notice that we can subtract $v$ from each side to get the equation: $3x=w+(-v)⟹x=\frac{1}{3}(w+(-v))$
2. Now, as $V$ is closed under inverses we know $-v\in V$. Additionally as $w,-v\in V$ and $\frac{1}{3}\in \mathbb{F}$ we know through closure of addition and scaler multiplication $x=\frac{1}{3}(w+(-v))$ must also be in $V$.
3. To prove uniqueness, assume that $x^{\prime }$ and $x$ both satisfy the equation. Then:
4. $v+3x=v+3x^{\prime }=w$
5. subtracting $v$ we get: $v+3x-v=v+3x^{\prime }-v=w-v$
6. Thus, $3x=3x^{\prime }=w-v$
7. Now multiplying by the scaler $\frac{1}{3}\in \mathbb{R}$ we can see: $x=x^{\prime }=\frac{1}{3}(w-u)$
8. Thus $x=x^{\prime }$ and we have proved our $x$ is unique.

6.

Vector spaces have to obay the axioms for vector addition. However notice: $1=1+0=1+(\infty +-\infty )=(1+\infty )+(-\infty )=\infty +(-\infty )=0$ Which is clearly not true as $1\ne 2$.

8.

Proof:

1. Additive Commutativity:
   1. Let $v_1,v_2\in V_c$
   2. Then $v_1=(a.b)=a+bi\text{and}{v}_2=(c,d)=c+di$ for some $a,b,c,d\in V$
   3. Notice: $v_1+v_2=(a+bi)+(c+di)=(a+c)+(b+d)i$
   4. by the commutative property of addition we can see the following: $(a+c)+(b+d)i=(c+a)+(d+b)i=(c+di)+(a+bi)=v_2+v_i$
2. Additive Associativity:
   1. Let $v_1,v_2,v_3\in V_c$
   2. Then $v_1=(a.b)=a+bi\text{and}{v}_2=(c,d)=c+di\text{and}{v}_3=(x,y)=x+yi$
   3. Notice:
      1. $(v_1+v_2)+v_3=((a+bi)+(c+di))+(x+yi)$
      2. $=((a+c)+(b+d)i)+(x+yi)$
      3. $=(a+c+x)+(b+d+y)i$
      4. $=(a+bi)+((c+x)+(d+y)i)$
      5. $=(a+bi)+((c+di)+(x+yi))$
      6. $=v_1+(v_2+v_3)$
3. Additive identity:
   1. This exists as $0\in V$ thus $(0,0)\in V_c$
4. Additive inverses:
   1. Let $v\in V_c$
   2. Then $v=a+bi$ for some $a,b\in V$
   3. Notice, $(a+bi)+((-a)+(-b)i)=(a+(-a))+(b+(-b))i=0+0i$
   4. Thus we can see that the inverse of $v$ is $-v=(-a)+(-b)i$.
   5. Additionally, as $V$ is a vector space, therefore containing additive inverses for $a$ and $b$, it is clear $-a,-b\in V$. This means that $-v\in V_c$
5. Scaler multiplication associativity:
   1. Let $\lambda ,\alpha \in \mathbb{C}$ and $v\in V_c$
   2. Let the following definitions hold: $\lambda =a+bi\text{and}\alpha =c+di\text{and}v=x+yi$ For $a,b,c,d\in \mathbb{R}$ and $x,y\in V$. Remember $V$ is a real vector space
   3. Given the definition of Complex Scaler Multiplication, notice: $(\lambda \alpha )v=((a+bi)(c+di))(x+yi)=((ac-bd)+(ad+cb)i)(x+yi)$ $=((ac-bd)(x)+((ad+cb)(y))i)=(acx-bdx)+(ady+cby)i$
   4. Now from the other side: $\lambda (\alpha v)=(a+bi)((c+di)(x+yi))=(a+bi)((cx-dy)+(cy+dx)i)$ $=(cxa-dya)+$
   5. Skipping this because wtf
6. Multiplicative Identity:
   1. Let $(a+bi)=v\in V_c$ for $(a,b)\in V$
   2. Notice: $(1+0i)(a+bi)=a+bi$
   3. Thus scaler multiplication by $1\in \mathbb{C}$ holds
7. Distributive property
   1. $(\lambda +\alpha )v=\lambda v+\alpha v$ for $\lambda ,\alpha \in \mathbb{F}$ and $v\in V$
      1. Let $\lambda =(a+bi)$, $\alpha =(c+di)$, and $v=(x+yi)$ for $a,b,c,d\in \mathbb{R}$ and $x,y\in V$
      2. Notice: $(\lambda +\alpha )v=((a+c)+(b+d)i)v=(a+c)v+(b+d)vi$ $=av+cv+bvi+dvi=av+bvi+cv+dvi=(av+bvi)+(cv+dvi)$ $=(a+bi)v+(c+di)v=\lambda v+\alpha v$ #check Fr what is this problem???

1C

4.

Let $b\in \mathbb{R}$

Prove the set of functions over the interval $[0,1]$ such that ${\int }_0^{1}f=b$ is a subspace of ${\mathbb{R}}^{[0,1]}⟺$ $b=0$

$\Rightarrow$

1. Assume that the set of functions over the interval $[0,1]$ such that ${\int }_0^{1}f=b$ is a subspace of ${\mathbb{R}}^{[0,1]}$
2. Then, notice that the $0$ function must be in the subspace. Thus ${\int }_0^{1}0=0$ $\Leftarrow$
3. Assume that $b=0$
   1. Then the identity is in the subspace as ${\int }_0^{1}0=0$
   2. For additive closure, assume $f,g$ are in the subspace. This means ${\int }_0^{1}f=0$ and ${\int }_0^{1}g=0$.
   3. Notice, the sum $(f+g)$ must also be in the subspace as ${\int }_0^{1}f+g={\int }_0^{1}f+{\int }_0^{1}g=0+0=0$.
   4. Scaler Multiplication is also closed, as for any $f$ in the subspace and $\lambda \in \mathbb{R}$, $\lambda f$ is also in the subspace as : ${\int }_0^1\lambda f=\lambda {\int }_0^{1}f=\lambda 0=0$

5.

No, it is not. This is because it is not closed under scaler multiplication as we are talking about a complex vector space, for example $(3+2i)(1,5)\notin R^2$

7.

No, here is why.

1. This does meet 0 as it is closed under addition and $u+(-u)=0\in U$
2. This is closed under addition
3. This is not closed under scaler multiplication. Lets say you have the subset {-1,0,1}. This is closed under addition and additive inverses. However $5(1,0)=(5,0)\notin U$.

8.

1. You can take the two subsets $\{(x,0)\in {\mathbb{R}}^2\mid x\in \mathbb{R}\}and\{(0,y)\in {\mathbb{R}}^2\mid y\in \mathbb{R}\}$
2. Their union is a subset of ${\mathbb{R}}^2$ and closed under scaler multiplcation. However, it is not closed under addition as $(1,0)+(0,1)=(1,1)$ is not in the union. A better way to say this is one of the coordinates is 0.

11.

Proof: 1. As 0 is in every subspace, it is in the intersection of all subspaces 2. For additive closure, suppose $a,b$ are in every subspace of $V$. As they are subspaces they must also be closed under addition, which means $a+b$ must also be in every subspace, and therefor the intersection 3. For closure under scaler multiplication, the same logic follows. As every individual vector space is closed under scaler multiplication, so must the intersection. This means that if any element $u$ is in the intersection, all of its scaler multiples are also within the intersection as they are also in all the subspaces.

12

Let $U,W$ be subspaces of $V$

$U\cup W$ is a subspace of $V$ $⟺$ $U\subset W$ or $W\subset U$ $\Rightarrow$

1. Lets consider the contrapositive: $U\not\subset W$ and $W\not\subset U$ $\to$ $U\cup W$ is not a subspace of $V$
   1. This means there is an $u\in U/W$ and an element $w\in W/U$. Note, $u\ne w\ne 0$
   2. This means that $u,w\in U\cup W$.
      1. For the sake of contradiction, assume $U\cup W$ is a subspace. Then it would have additive closure, and as $u,w\in U\cup W$ it must mean that $u+w\in U\cup W$
      2. This means that either $u+w\in U$ or $u+w\in W$.
      3. W.L.O.G assume $u+w\in U$. Now notice that all the following summands are in $U$: $(u+w)+(-u)$
      4. However, notice that $(u+w)+(-u)=w$ which is clearly not in $U$. Thus we have reached a contradiction as $U$ is a subspace with additive closure. $\Leftarrow$
2. Assume $U\subset W$ or $W\subset U$
3. Then, if $U\subset W$ the union is just $W$ which is defined as a subspace. Similarly, if $W\subset U$ the union is just $U$ which is defined to be a subspace.