# Section 1.4

1. Find a subset ${\displaystyle C\subset \mathbb {F} ^{2}}$ that is closed under scalar multiplication but not under addition of vectors.

Solution:
There are many possible answers that work. Here is one of them.
${\displaystyle \mathbb {F} ^{2}=\mathbb {R} ^{2}}$ with ${\displaystyle C=\{(x,y):x^{2}=y^{2}\}}$. Then if ${\displaystyle (x,y)\in C}$ and ${\displaystyle \alpha \in \mathbb {R} }$ we have

${\displaystyle \alpha (x,y)=(\alpha x,\alpha y)\in C}$ because ${\displaystyle x^{2}=y^{2}\Rightarrow (\alpha x)^{2}=\alpha ^{2}x^{2}=\alpha ^{2}y^{2}=(\alpha y)^{2}}$ so that ${\displaystyle C}$ is closed under scalar multiplication. However, ${\displaystyle (1,1)\in C}$ and ${\displaystyle (1,-1)\in C}$, but ${\displaystyle (1,1)+(1,-1)=(2,0)\notin C}$ so that ${\displaystyle C}$ is not closed under addition of vectors.

2. Find a subset ${\displaystyle A\subset \mathbb {C} ^{2}}$ that is closed under vector addition but not under multiplication by complex number.

Solution:
Many possible answers again. Here are a few:

${\displaystyle A=\mathbb {N} ^{2}}$, ${\displaystyle A=\mathbb {Z} ^{2}}$, ${\displaystyle A=\mathbb {Q} ^{2}}$, ${\displaystyle A=\mathbb {R} ^{2}}$, all are closed under addition, but if you multiply ${\displaystyle (1,1)\in A}$ for all of these ${\displaystyle A}$ by ${\displaystyle i}$ then you get ${\displaystyle (i,i)\notin A}$.

3. Find a subset ${\displaystyle Q\subset \mathbb {R} }$ that is closed under addition but not scalar multiplication by real scalars.

Solution:
Here them using ${\displaystyle Q}$ can be a hint. If you let ${\displaystyle Q=\mathbb {Q} }$, the rational numbers, then they will be closed under addition, but not scalar multiplication. That is because ${\displaystyle 1\in Q}$ and ${\displaystyle {\sqrt {2}}\in \mathbb {R} }$, but ${\displaystyle {\sqrt {2}}={\sqrt {2}}\cdot 1\notin Q}$.

Another possible answer is ${\displaystyle Q=\{x\in \mathbb {R} :x>0\}}$. Then this will be closed under addition since the sum of two positive numbers is still positive, but ${\displaystyle 1\in Q}$ and ${\displaystyle -1\in \mathbb {R} }$ and ${\displaystyle -1=-1\cdot 1\notin Q}$.

6. Let ${\displaystyle V}$ be a complex vector space i.e., a vector space where the scalars are ${\displaystyle \mathbb {C} }$. Define ${\displaystyle V^{*}}$ as the complex vector space whose additive structure is that of ${\displaystyle V}$ but where complex scalar multiplication is given by ${\displaystyle \lambda *x={\bar {\lambda }}x}$. Show that ${\displaystyle V^{*}}$ is a complex vector space.

Solution:
To begin we don’t need to show any of the first four axioms are true as they only involve addition of vectors and since ${\displaystyle V^{*}}$ has the same additive structure as ${\displaystyle V}$ and ${\displaystyle V}$ is a vector space, the first four axioms will still be true. For the remaining four properties we simply check that they will hold.

5) ${\displaystyle 1*x={\bar {1}}x=1x=x}$
6) ${\displaystyle \alpha *(\beta *x)=\alpha *({\bar {\beta }}x)={\bar {\alpha }}{\bar {\beta }}x={\overline {\alpha \beta }}x=(\alpha \beta )*x}$
7) ${\displaystyle \alpha *(x+y)={\bar {\alpha }}(x+y)={\bar {\alpha }}x+{\bar {\alpha }}y=\alpha *x+\alpha *y}$
8) ${\displaystyle (\alpha +\beta )*x={\overline {\alpha +\beta }}x+({\bar {\alpha }}+{\bar {\beta }})x={\bar {\alpha }}x+{\bar {\beta }}x=\alpha *x+\beta *x}$
Therefore ${\displaystyle V^{*}}$ is a complex vector space.

7. Let ${\displaystyle P_{n}}$ be the set of polynomials in ${\displaystyle \mathbb {F} [t]}$ of degree ${\displaystyle \leq n}$.
(a) Show that ${\displaystyle P_{n}}$ is a vector space.

Solution:
Suppose ${\displaystyle x,y\in P_{n}}$ and ${\displaystyle \alpha \in \mathbb {R} }$. Then ${\displaystyle x=a_{n}t^{n}+\cdots +a_{0}}$ and ${\displaystyle y=b_{n}t^{n}+\cdots +b_{0}}$. So that ${\displaystyle x+y=(a_{n}+b_{n})t^{n}+(a_{0}+b_{0})\in P_{n}}$. Also ${\displaystyle \alpha x=(\alpha a_{n})t^{n}+\cdots \alpha a_{0}\in P_{n}}$ so that ${\displaystyle P_{n}}$ is closed under addition and scalar multiplication. Also ${\displaystyle x+y=(a_{n}+b_{n})t^{n}+(a_{0}+b_{0})=(b_{n}+a_{n})t^{n}+(b_{0}+a_{0})=y+x}$ so addition is commutative. Similarly for associative. The zero vector in ${\displaystyle P_{n}}$ is the zero polynomial with all coefficients equal to 0. The additive inverse of ${\displaystyle x}$ is ${\displaystyle (-x)=(-a_{n})t^{n}+(-a_{0})}$. Also, ${\displaystyle (1)x=x}$, ${\displaystyle \alpha (\beta x)=(\alpha \beta )x}$, ${\displaystyle (\alpha +\beta )x=\alpha x+\beta x}$ and ${\displaystyle \alpha (x+y)=\alpha x+\alpha y}$ just by writing each of them out. Therefore ${\displaystyle P_{n}}$ is a vector space.

(b) Show that the space of polynomials of degree ${\displaystyle n\geq 1}$ is ${\displaystyle P_{n}-P_{n-1}}$ and does not form a subspace.

Solution:
First off, the space is equal to ${\displaystyle P_{n}-P_{n-1}}$ because the polynomials that have exactly degree ${\displaystyle n}$ are in ${\displaystyle P_{n}}$ but not in ${\displaystyle P_{n-1}}$ and there are no other polynomials in ${\displaystyle P_{n}}$ that aren’t also in ${\displaystyle P_{n-1}}$. Now this set does not form a subspace because it is not closed under addition. ${\displaystyle p(t)=t^{n}}$ and ${\displaystyle q(t)=-t^{n}+1}$ are both polynomials of degree exactly ${\displaystyle n}$. However, ${\displaystyle p+q=1}$ is a polynomial of degree 0 not ${\displaystyle n}$.

(c) If ${\displaystyle f(t):\mathbb {F} \to \mathbb {F} }$, show that ${\displaystyle V=\{p(t)f(t):p\in P_{n}\}}$ is a subspace of ${\displaystyle Func(\mathbb {F} ,\mathbb {F} )}$.

Solution:
To show it is a subspace, we only need to check that it is closed under addition and scalar multiplication. Let ${\displaystyle x,y\in V}$ and ${\displaystyle \alpha \in \mathbb {F} }$. Then ${\displaystyle x=p(t)f(t)}$ for some polynomial ${\displaystyle p\in P_{n}}$ and ${\displaystyle y=q(t)f(t)}$ for some polynomial ${\displaystyle q\in P_{n}}$. Then ${\displaystyle x+y=p(t)f(t)+q(t)f(t)=(p(t)+q(t))f(t)}$. But since ${\displaystyle p(t)+q(t)}$ is just another polynomial in ${\displaystyle P_{n}}$, then ${\displaystyle x+y}$ is exactly of the form polynomial times ${\displaystyle f(t)}$. Thus ${\displaystyle x+y\in V}$. Also ${\displaystyle \alpha x=\alpha (p(t)f(t))=(\alpha p(t))f(t)}$ which is again of the form polynomial times ${\displaystyle f(t)}$ so ${\displaystyle \alpha x\in V}$. Therefore ${\displaystyle V}$ is a subspace. 8) ${\displaystyle (\alpha +\beta )*x={\overline {\alpha +\beta }}x+({\bar {\alpha }}+{\bar {\beta }})x={\bar {\alpha }}x+{\bar {\beta }}x=\alpha *x+\beta *x}$
Therefore ${\displaystyle V^{*}}$ is a complex vector space.

8. Let ${\displaystyle V=\mathbb {C} ^{\times }=\mathbb {C} -\{0\}}$. Define addition on ${\displaystyle V}$ by ${\displaystyle x\boxplus y=xy}$. Define scalar multiplication by ${\displaystyle \alpha \boxdot x=e^{\alpha }x}$.
(a) Show that if we use ${\displaystyle 0_{V}=1}$ and ${\displaystyle -x=x^{-1}}$, then the first four axioms for a vector space are satisfied.

Solution:
1) ${\displaystyle x\boxplus y=xy=yx=y\boxplus x}$ so addition is commutative.

2) ${\displaystyle (x\boxplus y)\boxplus z=(xy)\boxplus z=(xy)z=xyz=x(yz)=x(y\boxplus z)=x\boxplus (y\boxplus z)}$ and addition is associative.
3) ${\displaystyle x\boxplus 0_{V}=x\boxplus 1=1x=x}$
4) ${\displaystyle x\boxplus -x=x\boxplus x^{-1}=xx^{-1}=1=0_{V}}$

(b) Which of the scalar multiplication properties do not hold?

Solution:
The only property that won’t hold is associativity of scalar multiplication.

${\displaystyle \alpha \boxdot (\beta \boxdot x)=\alpha (e^{\beta }x)=e^{\alpha }e^{\beta }x}$ which is not the same as ${\displaystyle (\alpha \beta )\boxdot x=e^{\alpha \beta }x}$.