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. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \mathbb {F} ^{2}=\mathbb {R} ^{2}} with $C=\{(x,y):x^{2}=y^{2}\}$. Then if Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x,y)\in C} and $\alpha \in \mathbb {R}$ we have $\alpha (x,y)=(\alpha x,\alpha y)\in C$ because Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (1,1)\in C} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (1,-1)\in C} , but Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (1,1)+(1,-1)=(2,0)\notin C} so that $C$ is not closed under addition of vectors.

2. Find a subset Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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: $A=\mathbb {N} ^{2}$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A=\mathbb {Z} ^{2}} , $A=\mathbb {Q} ^{2}$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle Q=\mathbb {Q} } , the rational numbers, then they will be closed under addition, but not scalar multiplication. That is because Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1\in Q} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1\in Q} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -1\in \mathbb {R} } and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha *(\beta *x)=\alpha *({\bar {\beta }}x)={\bar {\alpha }}{\bar {\beta }}x={\overline {\alpha \beta }}x=(\alpha \beta )*x} 7) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha *(x+y)={\bar {\alpha }}(x+y)={\bar {\alpha }}x+{\bar {\alpha }}y=\alpha *x+\alpha *y} 8) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \mathbb {F} [t]} of degree Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \leq n} .
(a) Show that $P_{n}$ is a vector space.

Solution:

Suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x+y=(a_{n}+b_{n})t^{n}+(a_{0}+b_{0})\in P_{n}} . Also Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $P_{n}$ is the zero polynomial with all coefficients equal to 0. The additive inverse of $x$ is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (-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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p(t)=t^{n}} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle q(t)=-t^{n}+1} are both polynomials of degree exactly $n$. However, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha \in \mathbb {F} } . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=p(t)f(t)} for some polynomial Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p\in P_{n}} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=q(t)f(t)} for some polynomial $q\in P_{n}$. Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(t)} . Thus Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(t)} so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha x\in V} . Therefore $V$ is a subspace. 8) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \boxplus y = xy} . Define scalar multiplication by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha \boxdot x = e^{\alpha}x} .
(a) Show that if we use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0_V = 1} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -x = x^{-1}} , then the first four axioms for a vector space are satisfied.

Solution:

1) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\boxplus y = xy = yx = y \boxplus x} so addition is commutative. 2) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \boxplus 0_V = x \boxplus 1 = 1x = x} 4) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha \boxdot (\beta \boxdot x) = \alpha (e^\beta x) = e^\alpha e^\beta x} which is not the same as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\alpha \beta) \boxdot x = e^{\alpha \beta} x} .