(1) Suppose $a$ and $b$ are real numbers, not both $0$. Find real numbers $c$ and $d$ such that
$$ \frac{1}{a+bi}=c+di\tag{1.1} $$
Solution Let $\alpha=a+bi$. Then $\frac{1}{\alpha}=\frac{1}{a+bi}$ is its multiplicative inverse. Multiplying both sides of $1.1$ by $\alpha=a+bi$, we get
$$ 1+0i=1=(a+bi)\frac{1}{a+bi}=(a+bi)(c+di)=ac-bd+(ad+bc)i $$
where the last equality follows from the defintion of multiplication of complex numbers. This gives the two equations
$$ ac-bd=1\\ bc+ad=0 $$
Solving for $c$ and $d$, we get
$$ \begin{bmatrix}a&-b&1\\b&a&0\end{bmatrix}\rightarrow\begin{bmatrix}a&-b&1\\0&a+\frac{(-b)(-b)}{a}&0+\frac{(1)(-b)}{a}\end{bmatrix}\rightarrow\begin{bmatrix}a&-b&1\\0&a+\frac{b^2}{a}&-\frac{b}{a}\end{bmatrix} $$
$$ \rightarrow\begin{bmatrix}1&-\frac{b}{a}&\frac{1}{a}\\0&\frac{a^2+b^2}{a}&-\frac{b}{a}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{1}{a}-\frac{b}{a}\frac{b}{a^2+b^2}\\0&\frac{a^2+b^2}{a}&-\frac{b}{a}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{a^2+b^2}{a(a^2+b^2)}-\frac{b^2}{a(a^2+b^2)}\\0&1&-\frac{b}{a}\frac{a}{a^2+b^2}\end{bmatrix} $$
$$ \rightarrow\begin{bmatrix}1&0&\frac{a^2}{a(a^2+b^2)}\\0&1&-\frac{b}{a^2+b^2}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{a}{a^2+b^2}\\0&1&-\frac{b}{a^2+b^2}\end{bmatrix} $$
(2) Show that
$$ \frac{-1+\sqrt{3}i}{2} $$
is a cube root of $1$ (meaning that its cube equals $1$).
Solution Note that
$$\begin{align*} \Big(\frac{-1+\sqrt{3}i}{2}\Big)^2 &= \Big(-\frac{1}{2}+\frac{\sqrt{3}i}{2}\Big)^2 \\\\ &= \frac{1}{4}+\frac{3i^2}{4}-2\Big(\frac{1}{2}\frac{\sqrt{3}i}{2}\Big)\\\\ &= \frac{1}{4}-\frac{3}{4}-\frac{\sqrt{3}i}{2}\\\\ &= -\frac{1}{2}-\frac{\sqrt{3}i}{2} \end{align*}$$
Hence
$$\begin{align*} \Big(\frac{-1+\sqrt{3}i}{2}\Big)^3 &= \Big(\frac{-1+\sqrt{3}i}{2}\Big)^2\Big(\frac{-1+\sqrt{3}i}{2}\Big)\\\\ &= \Big(-\frac{1}{2}-\frac{\sqrt{3}i}{2}\Big)\Big(-\frac{1}{2}+\frac{\sqrt{3}i}{2}\Big)\\\\ &= \frac{1}{4}-\frac{3i^2}{4}-\frac{\sqrt{3}i}{4}+\frac{\sqrt{3}i}{4}\\\\ &= \frac{1}{4}+\frac{3}{4}+0\\\\ &= 1 \end{align*}$$
(3) Prove that $−(−v)=v$ for every $v\in V$.
Solution For any vector $v$ in the vector space $V$, there is an additive inverse $-v\in V$ such that $v+(-v)=0$. In particular, $-v$ has an additive inverse $-(-v)$:
$$ -v+(-(-v))=0 $$
But the commutativity of addition gives
$$ -v+v=v+(-v)=0 $$
So we have shown that $-(-v)$ and $v$ are both additive inverses of $-v$. But the text proved the uniqueness of the additive inverse. Hence it must be that $-(-v)=v$.
(4) Prove that if $a\in \mathbb{F}$, $v\in V$, and $av=0$, then $a=0$ or $v=0$.
Solution Suppose $a\neq0$ and $v\neq0$. Then $v=1\cdot v=\big(\frac{1}{a}\cdot a\big)\cdot v=\frac{1}{a}\cdot(a\cdot v)=\frac{1}{a}\cdot0=0$, a contradiction.
(5) For each of the following subsets of $\mathbb{F}^3$, determine whether it is a subspace of $\mathbb{F}^3$:
(a) $U=\{(x_1,x_2,x_3)\in \mathbb{F}^3:x_1+2x_2+3x_3=0\}$
(b) $U=\{(x_1,x_2,x_3)\in \mathbb{F}^3:x_1+2x_2+3x_3=4\}$
(c) $U=\{(x_1,x_2,x_3)\in \mathbb{F}^3:x_1x_2x_3=0\}$
(d) $U=\{(x_1,x_2,x_3)\in \mathbb{F}^3:x_1=5x_3\}$
Solution (a) $U$ is a subspace because it
- contains the additive identity $0$: $0+2\cdot0+3\cdot0=0$
- is closed under addition: if $x,y\in U$, then $(x_1+y_1)+2(x_2+y_2)+3(x_3+y_3)=(x_1+2x_2+3x_3)+(y_1+2y_2+3y_3)=0+0=0$. Hence $x+y\in U$.
- is closed under scalar multiplication: fix any $a\in \mathbb{F}$ and any $x\in U$. Then $ax_1+2(ax_2)+3(ax_3)=a(x_1+2x_2+3x_3)=a\cdot0=0$. Hence $ax\in U$.
(b) $U$ is not a subspace because it doesn’t contain the additive identity: $0+2\cdot0+3\cdot0=0\neq4$. Hence $0\notin U$.
(c) $U$ isn’t closed under addition: let $x=(1,1,0)$ and $y=(0,0,1)$. Then $x,y\in U$ but $x+y=(1,1,1)\notin U$.
(d) $U$ is a subspace because it
- contains the additive identity $0$: $0=5\cdot0\implies 0\in U$
- is closed under addition: if $x,y\in U$, then $(x_1+y_1)=(5x_3+5y_2)=5(x_3+y_3)$. Hence $x+y\in U$.
- is closed under scalar multiplication: fix any $a\in \mathbb{F}$ and any $x\in U$. Then $ax_1=a(5x_3)=(a5)x_3=(5a)x_3=5(ax_3)$. Hence $ax\in U$.
(6) Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $−u\in U$ whenever $u\in U$), but $U$ is not a subspace of $\mathbb{R}^2$.
Solution Let $\mathbb{Z}\subset\mathbb{R}$ denote the set of integers and let $U=\mathbb{Z}^2$. Then
- $U$ is closed under addition: for any $x,y\in\mathbb{Z}^2$, $x+y\in\mathbb{Z}^2$
- $U$ is closed under taking additive inverses
But $U$ is not closed under scalar multiplication: $\frac{1}{2}\cdot(3,4)=\big(\frac{3}{2},2\big)\notin\mathbb{Z}^2=U$.
(7) Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^2$.
Solution Define
$$ U=\{(x,y)\in\mathbb{R}^2:x=y\}\cup\{(x,y)\in\mathbb{R}^2:x=-y\} $$
Then for any $a,x\in\mathbb{R}$, we have $a(x,x)=(ax,ax)\in U$ and $a(x,-x)=(ax,-ax)\in U$. Hence $U$ is closed under scalar multiplication.
But for any $x\in\mathbb{R}$ with $x\neq0$, we have $(x,x)\in U$ and $(x,-x)\in U$ but $(x,x)+(x,-x)=(2x,0)\notin U$. Hence $U$ is not closed under addition.
(8) Prove that the intersection of any collection of subspaces of $V$ is a subspace of $V$.
Solution Let $\{U_i\}$ be a collection of subspaces of $V$. Then $\cap_iU_i$
- contains the additive identity: $0\in U_i$ $\forall i\implies0\in\cap_iU_i$
- is closed under addition: let $x,y\in\cap_iU_i$. Then $x,y\in U_i$ $\forall i$ hence $x+y\in U_i$ $\forall i$ hence $x+y\in\cap_iU_i$.
- is closed under scalar multiplication: fix any $a\in \mathbb{F}$ and any $x\in\cap_iU_i$. Then $x\in U_i$ $\forall i$ hence $ax\in U_i$ $\forall i$ hence $ax\in\cap_iU_i$.
(9) Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.
Solution Let $U,W$ be subspaces of $V$.
Suppose $U\subset W$. Then $U\cup W=W$ is a subspace of $V$.
Let $U\cup W$ be a subspace of $V$ and suppose $U\not\subseteq W$ and $W\not\subseteq U$. Fix $x\in U\smallsetminus W$ and $y\in W\smallsetminus U$. Then $x+y\notin U$. For if $x+y\in U$, then $y=(x+y)-x\in U$. Similarly, $x+y\notin W$ else $x=(x+y)-y\in W$. Hence $x\in U\subset U\cup W$ and $y\in W\subset U\cup W$ but $x+y\notin U,W\implies$ $x+y\notin U\cup W$. Contradiction hence it must be that $U\subseteq W$ or $W\subseteq U$.
(10) Suppose that $U$ is a subspace of $V$. What is $U+U$?
Solution By definition, we have $U+U=\{u+u’:u,u’\in U\}$. But $U$ is closed under addition so $u+u’\in U$ $\forall u,u’\in U$. Hence we’re not adding anything to $U$ and $U+U=U$.
Alternate proof: Since $U\subset U$ and $U+U$ is the smallest subspace containing $U$, we have $U+U\subset U$; on the other hand, $U\subset U+U$ is clear. Hence, $U+U=U$.
(11) Is the operation of addition on the subspaces of $V$ commutative? Associative?
Solution Yes. Let $U_1,U_2,U_3$ be subspaces of $V$. Then commutativity holds from the commutativity of vector addition:
$$\begin{align*} U_1+U_2 &= \{u_1+u_2:u_1\in U_1,u_2\in U_2\}\\ &= \{u_2+u_1:u_2\in U_2,u_1\in U_1\}\\ &= U_2+U_1 \end{align*}$$
And associativity holds from the associativity of vector addition:
$$\begin{align*} (U_1+U_2)+U_3 &= \{u_1+u_2:u_1\in U_1,u_2\in U_2\}+U_3\\ &= \{(u_1+u_2)+u_3:u_1\in U_1,u_2\in U_2,u_3\in U_3\}\\ &= \{u_1+(u_2+u_3):u_1\in U_1,u_2\in U_2,u_3\in U_3\}\\ &= U_1+\{u_2+u_3:u_2\in U_2,u_3\in U_3\}\\ &= U_1+(U_2+U_3) \end{align*}$$
(12) Does the operation of addition on the subspaces of $V$ have an additive identity? Which subspaces have additive inverses?
Solution The subspace $\{0\}$ is the additive identity on subspace addition: let $U$ be a subspace of $V$ and note that
$$ U+\{0\}=\{u+0:u\in U\}=\{u:u\in U\}=U $$
Only the subspace $\{0\}$ has an additive inverse: suppose that $U$ is a subspace and its additive inverse is $W$. That is $U+W=\{0\}$. Since $0\in U$, we have $0+w=0$ for all $w\in W$. Hence $W=\{0\}$. Hence $\{0\}=U+\{0\}$. But this occurs if and only if $U=\{0\}$.
(13) Prove or give a counterexample: if $U_1$, $U_2$, $W$ are subspaces of $V$ such that
$$ U_1+W=U_2+W $$
then $U_1=U_2$.
Counterexample Let $U_1,U_2\subset W$ such that $U_1\neq U_2$. Then $U_1+W=\{u_1+w:u_1\in U_1\subset W,w\in W\}=W$ since $W$ is closed under addition. Similarly $U_2+W=W$. Hence $U_2+W=W=U_1+W$.
A more concrete counterexample: define $W\equiv V\equiv\mathbb{R}^2$ and
$$ U_1\equiv\{(x,y)\in\mathbb{R}^2: x=y\} \quad\quad\text{and}\quad\quad U_2\equiv\{(x,y)\in\mathbb{R}^2: x=-y\} $$
In Wayne’s Example 1.1, we showed that $U_1$ and $U_2$ are subspaces of $V=\mathbb{R}^2$. Note that $U_1+W=\mathbb{R}^2=U_2+W$ but $U_1\neq U_2$.
(14) Suppose $U$ is the subspace of $\mathscr{P}(\mathbb{F})$ consisting of all polynomials $p$ of the form
$$ p(z)=az^2+bz^5 $$
where $a,b\in \mathbb{F}$. Find a subspace $W$ of $\mathscr{P}(\mathbb{F})$ such that $\mathscr{P}(\mathbb{F})=U\oplus W$.
Proof In the notes for ch.2, we verify that $\mathscr{P}(\mathbb{F})$ (the set of all polynomials with coefficients in $\mathbb{F}$) is a subspace of $\mathbb{F}^\mathbb{F}$.
Next let’s show that $U=\{p\in\mathscr{P}(\mathbb{F}): p(z)=az^2+bz^5,a,b\in \mathbb{F}\}$ is a subspace of $\mathscr{P}(\mathbb{F})$.
- $0_{\mathscr{P}(\mathbb{F})}(z)\equiv0\equiv0z^2+0z^5$. Then $0_{\mathscr{P}(\mathbb{F})}\in U$ and $0_{\mathscr{P}(\mathbb{F})}$ is the additive identity since $p(z)+0_{\mathscr{P}(\mathbb{F})}=p(z)$ for any $p\in U$.
- Closed under addition: define $p_1(z)\equiv az^2+bz^5$ and $p_2(z)\equiv cz^2+dz^5$. Then $(p_1+p_2)(z)\equiv p_1(z)+p_2(z)=(a+c)z^2+(b+d)z^5$. Hence $p_1+p_2\in U$.
- Closed under scalar multiplication: for $c\in \mathbb{F}$, $p(z)\equiv az^2+b^5$, and $(cp)(z)\equiv c(az^2+bz^5)=caz^2+cbz^5$. Then $cp\in U$.
Define $W\equiv\{w(z)\in\mathscr{P}(\mathbb{F}):w(z)=w_0+w_1z+w_3z^3+w_4z^4+\sum_{i=6}^{\infty}w_iz^i,w_i\in\mathbb{F}\}$. We can similarly show that $W$ is a subspace of $\mathscr{P}(\mathbb{F})$. By definition, we have
$$ U+W=\{u+w:u\in U,w\in W\} $$
Then clearly $U+W\subset\mathscr{P}(\mathbb{F})$ and $\mathscr{P}(\mathbb{F})\subset U+W$. Hence $\mathscr{P}(\mathbb{F})=U+W$.
To show $U+W$ is a direct sum, the text proved that it’s sufficient to show that $U\cap W=\{0\}$. This follows from the definitions of $U$ and $W$. Hence $\mathscr{P}(\mathbb{F})=U\oplus W$.
$\blacksquare$
(15) Prove or give a counterexample: if $U_1$, $U_2$, $W$ are subspaces of $V$ such that
$$ V=U_1\oplus W\quad\quad\text{and}\quad\quad V=U_2\oplus W $$
then $U_1=U_2$.
Counterexample Define $V\equiv\mathbb{R}^2$, $W\equiv\{(y,0)\in\mathbb{R}^2:y\in\mathbb{R}\}$, $U_1\equiv\{(x,x)\in\mathbb{R}^2:x\in\mathbb{R}\}$, and $U_2\equiv\{(x,-x)\in\mathbb{R}^2:x\in\mathbb{R}\}$. We showed in Wayne’s Example 1.1 that $U_1$ and $U_2$ are subspaces of $V=\mathbb{R}^2$. Similarly $W$ is a subspace of $V=\mathbb{R}^2$. Also note that
$$\begin{align*} U_1+W&=\{(x+y,x)\in\mathbb{R}^2:x,y\in\mathbb{R}\}\\ U_2+W&=\{(x+y,-x)\in\mathbb{R}^2:x,y\in\mathbb{R}\}\\ \end{align*}$$
Let’s show that $U_i+W=\mathbb{R}^2$ for $i=1,2$. This will imply that we can write $U_i\oplus W$. For any $(a,b)\in\mathbb{R}^2$, we want to find $x,y$ such
$$ x+y=a\\ x=b $$
$$ \begin{bmatrix}1&1&a\\1&0&b\end{bmatrix}\rightarrow\begin{bmatrix}1&0&b\\1&1&a\end{bmatrix}\rightarrow\begin{bmatrix}1&0&b\\0&1&a-b\end{bmatrix} $$
And similarly for $U_2+W$:
$$ x+y=a\\ -x=b $$
$$ \begin{bmatrix}1&1&a\\-1&0&b\end{bmatrix}\rightarrow\begin{bmatrix}-1&0&b\\1&1&a\end{bmatrix}\rightarrow\begin{bmatrix}-1&0&b\\0&1&a+b\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-b\\0&1&a+b\end{bmatrix} $$
Hence $U_1\oplus W=V=U_2\oplus W$ but $U_1\neq U_2$.