Section 1.4 Homework

1. Find a subset $C\subset \mathbb {F} ^{2}$ 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.
$\mathbb {F} ^{2}=\mathbb {R} ^{2}$ with $C=\{(x,y):x^{2}=y^{2}\}$. Then if $(x,y)\in C$ and $\alpha \in \mathbb {R}$ we have
$\alpha (x,y)=(\alpha x,\alpha y)\in C$ because $x^{2}=y^{2}\Rightarrow (\alpha x)^{2}=\alpha ^{2}x^{2}=\alpha ^{2}y^{2}=(\alpha y)^{2}$ so that $C$ is closed under scalar multiplication. However, $(1,1)\in C$ and $(1,-1)\in C$, but $(1,1)+(1,-1)=(2,0)\notin C$ so that $C$ is not closed under addition of vectors.

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

Solution Many possible answers again. Here are a few:
$A=\mathbb {N} ^{2}$, $A=\mathbb {Z} ^{2}$, $A=\mathbb {Q} ^{2}$, $A=\mathbb {R} ^{2}$, all are closed under addition, but if you multiply $(1,1)\in A$ for all of these $A$ by $i$ then you get $(i,i)\notin A$.

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

Solution Here them using $Q$ can be a hint. If you let $Q=\mathbb {Q}$, the rational numbers, then they will be closed under addition, but not scalar multiplication. That is because $1\in Q$ and ${\sqrt {2}}\in \mathbb {R}$, but ${\sqrt {2}}={\sqrt {2}}\cdot 1\notin Q$.
Another possible answer is $Q=\{x\in \mathbb {R} :x>0\}$. Then this will be closed under addition since the sum of two positive numbers is still positive, but $1\in Q$ and $-1\in \mathbb {R}$ and $-1=-1\cdot 1\notin Q$.

6. Let $V$ be a complex vector space i.e., a vector space where the scalars are $\mathbb {C}$. Define $V^{*}$ as the complex vector space whose additive structure is that of $V$ but where complex scalar multiplication is given by $\lambda *x={\bar {\lambda }}x$. Show that $V^{*}$ 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 $V^{*}$ has the same additive structure as $V$ and $V$ 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) $1*x={\bar {1}}x=1x=x$
6) $\alpha *(\beta *x)=\alpha *({\bar {\beta }}x)={\bar {\alpha }}{\bar {\beta }}x={\overline {\alpha \beta }}x=(\alpha \beta )*x$
7) $\alpha *(x+y)={\bar {\alpha }}(x+y)={\bar {\alpha }}x+{\bar {\alpha }}y=\alpha *x+\alpha *y$
8) $(\alpha +\beta )*x={\overline {\alpha +\beta }}x+({\bar {\alpha }}+{\bar {\beta }})x={\bar {\alpha }}x+{\bar {\beta }}x=\alpha *x+\beta *x$
Therefore $V^{*}$ is a complex vector space.

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

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

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

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

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

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

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

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

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

Solution The only property that won’t hold is associativity of scalar multiplication.
$\alpha \boxdot (\beta \boxdot x)=\alpha (e^{\beta }x)=e^{\alpha }e^{\beta }x$ which is not the same as $(\alpha \beta )\boxdot x=e^{\alpha \beta }x$.