Section 1.4 Homework

1. Find a subset Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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. 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha \in \mathbb {R} } we have 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)\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 $(1,-1)\in C$, 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: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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}} , Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (1,1)\in A} for all of these $A$ by $i$ then you get $(i,i)\notin A$.

3. Find a subset Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle Q\subset \mathbb {R} } that is closed under addition but not scalar multiplication by real scalars.

Solution:

Here them using Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\sqrt {2}}={\sqrt {2}}\cdot 1\notin Q} . Another possible answer is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $1\in Q$ 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \mathbb {C} } . Define $V^{*}$ as the complex vector space whose additive structure is that of $V$ but where complex scalar multiplication is given by Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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) $(\alpha +\beta )*x={\overline {\alpha +\beta }}x+({\bar {\alpha }}+{\bar {\beta }})x={\bar {\alpha }}x+{\bar {\beta }}x=\alpha *x+\beta *x$ Therefore $V^{*}$ is a complex vector space.

7. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha \in \mathbb {R} } . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (1)x=x} , $\alpha (\beta x)=(\alpha \beta )x$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P_{n}-P_{n-1}} and does not form a subspace.

Solution:

First off, the space is equal to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P_{n}-P_{n-1}} because the polynomials that have exactly degree $n$ are in Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P_{n}} but not in $P_{n-1}$ and there are no other polynomials in Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(t):\mathbb {F} \to \mathbb {F} } , show that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x,y\in V} and $\alpha \in \mathbb {F}$. 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle q\in P_{n}} . Then $x+y=p(t)f(t)+q(t)f(t)=(p(t)+q(t))f(t)$. But since Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p(t)+q(t)} is just another polynomial in Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P_{n}} , then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 $\alpha x\in V$. Therefore 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} 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 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^*} is a complex vector space.

8. Let 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 = \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} .