LADR Exercises 1B

Problem 1

Prove that $-(-v)=v$ for every $v\in V$.

Proof. We wish to show that $v$ is the additive inverse of $(-v)$. Thus, we need to show that $(-v)+v=0$. We have:

$$\begin{array}{r}(-v)+v=(-1)v+1v=(-1+1)v=0v=0\end{array}$$

as desired.

Problem 2

Suppose $a\in \mathbb{F},v\in V$, and $av=0$. Prove that $a=0$ or $v=0$.

Proof.

Case 1: Suppose $a=0$. By the property of scalar multiplication in a vector space, we have $av=0\cdot v=0$.

Case 2: Suppose $a\ne 0$. Since $a\in \mathbb{F}$, $a$ will have a multiplicative inverse such that $a^{-1}a=1$. Then we have:

$$\begin{array}{rl}av& =0\\ a^{-1}(av)& =a^{-1}\cdot 0\\ (a^{-1}a)v& =0\\ 1v& =0\\ v& =0\end{array}$$

Therefore, in the case that $a\ne 0$, we must have $v=0$.

Problem 3

Suppose $v,w\in V$. Explain why there exists a unique $x\in V$ such that $v+3x=w$.

Proof. Suppose that we have $x=\frac{1}{3}(w-v)$. Then we have:

$$\begin{array}{rl}v+3x& =v+3\left(\frac{1}{3}(w-v)\right)\\ & =v+3\left(\frac{1}{3}\right)\cdot (w-v)\\ & =v+(w-v)\\ & =w\end{array}$$

which proves existence. To show uniqueness, suppose that there is some $y\in V$ such that $v+3y=w$. Then

$$\begin{array}{rl}v+3y& =v+3x\\ 3y& =3x\\ y& =x\end{array}$$

proving uniqueness.

Problem 5

The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in the definition of a vector space. Which one?

The empty set fails because it does not have an additive identity; there is no element $0\in V$ such that $v+0=v$ for $v\in V$.

Why does the multiplicative identity not fail? There are technically no counterexamples for which $1v=v$ fails, so the property is satisfied vacuously. However, the additive identity requires a specific zero element in the set.

Problem 6

Let $\infty$ and $-\infty$ denote two distinct objects, neither of which is in $\mathbb{R}$. Define an addition and scalar multiplication on $\mathbb{R}\cup \{\infty ,-\infty \}$ as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for $t\in \mathbb{R}$ define

$$t\infty =\{\begin{array}{ll}-\infty & \text{ if }t<0\\ 0& \text{ if }t=0\\ \infty & \text{ if }t>0\end{array}\text{and}t(-\infty )=\{\begin{array}{ll}\infty & \text{ if }t<0\\ 0& \text{ if }t=0\\ -\infty & \text{ if }t>0\end{array}$$

and

$$\begin{array}{rl}t+\infty & =\infty +t=\infty +\infty =\infty \\ t+(-\infty )& =(-\infty )+t=(-\infty )+(-\infty )=-\infty \\ \infty +(-\infty )& =(-\infty )+(\infty )=0\end{array}$$

With these operations of addition and scalar multiplication, is $\mathbb{R}\cup \{\infty ,-\infty \}$ a vector space over $\mathbb{R}$? Explain.

$\mathbb{R}\cup \{\infty ,-\infty \}$ is not a vector space over $\mathbb{R}$. If it were a vector space over $\mathbb{R}$, the distributive property would give us:

$$\begin{array}{rl}\infty & =(2+(-1))\infty \\ & =2\infty +(-1)\infty \\ & =\infty +(-\infty )\\ & =0\end{array}$$

$\infty =0$ is a contradiction because $\infty$ and $0$ are distinct elements by definition, since the zero vector must be unique.

Problem 7

Suppose $S$ is a non-empty set. Let $V^S$ denote the set of functions from $S$ to $V$. Define a natural addition and scalar multiplication on $V^S$, and show that $V^S$ is a vector space with these definitions.

$V$ is a vector space over $\mathbb{F}$, and $V^S$ is the set of functions from $S$ to $V$. This means each element of $V^S$ is a function that assigns each element of $S$ to a vector in $V$.

Definitions:

• Addition: For two functions, $f,g\in V^S$, we define their sum $(f+g)(s)=f(s)+g(s)$ for all $s\in S$. Here, $f(s)+g(s)$ is the usual vector addition in $V$.
• Scalar multiplication: For any scalar $\alpha \in \mathbb{F}$, and any function $f\in V^S$, we define scalar multiplication $(\alpha f)\in V^S$ by $(\alpha f)(s)=\alpha \cdot f(s)$ for all $s\in S$. Here, $\alpha \cdot f(s)$ is the usual scalar multiplication in $V$.

With the above properties, we can show that $V^S$ is a vector space by verifying the properties of vector spaces:

Commutativity: We have

$$(f+g)(s)=f(s)+g(s)=g(s)+f(s)=(g+f)(s)$$

Therefore, $f+g=g+f$.

Associativity of Addition: We have

$$((f+g)+h)(s)=(f+g)(s)+h(s)=(f(s)+g(s))+h(s)$$

Since $f(s),g(s),h(s)$ are elements of $V$, we can use associativity in $V$ to have

$$(f(s)+g(s))+h(s)=f(s)+(g(s)+h(s))=(f+(g+h))(s)$$

Therefore, $(f+g)+h=f+(g+h)$.

Associativity of Multiplication: For all $a,b\in \mathbb{F}$ and $f\in V^S$, we have

$$((ab)f)(s)=(ab)\cdot f(s)$$

We can then use associativity in $V$ to get

$$(ab)\cdot f(s)=a\cdot (b\cdot f(s))=(a(bf))(s)$$

Therefore, $(ab)f=a(bf)$.

Additive Identity: Define a zero function $0\in V^S$ by $0(s)=0_V$ for all $s\in S$, where $0_V$ is the zero vector in $V$. Then, for any $f\in V^S$, we have

$$(f+0)(s)=f(s)+0(s)=f(s)+0_V=f(s)$$

Therefore $f+0=f$.

Additive Inverse: For any $f\in V^S$, define $-f\in V^S$ by $(-f)(s)=-f(s)$ for all $s\in S$. Then, we have

$$(f+(-f))(s)=f(s)+(-f(s))=0_V$$

Therefore, $f+(-f)=0$, showing that $-f$ is the additive inverse of $f$ in $V^S$.

Multiplicative Identity: For the scalar $1\in \mathbb{F}$ and any $f\in V^S$, we have $(1f)(s)=1\cdot f(s)=f(s)$. Therefore, $1f=f$.

Distribute Property 1: For all $a\in \mathbb{F}$ and $f,g\in V^S$, we have

$$(a(f+g))(s)=a\cdot (f+g)(s)=a\cdot (f(s)+g(s))$$

Then, using the distributive property in $V$, we have

$$a\cdot (f(s)+g(s))=a\cdot f(s)+a\cdot g(s)=(af)(s)+(ag)(s)$$

Therefore, we have $a(f+g)=af+ag$.

Distributive Property 2: For all $a,b\in \mathbb{F}$ and $f\in V^S$, we have

$$((a+b)f)(s)=(a+b)\cdot f(s)$$

By the distributive property in $V$, we have

$$(a+b)\cdot f(s)=a\cdot f(s)+b\cdot f(s)=(af)(s)+(bf)(s)$$

Therefore, we have $(a+b)f=af+bf$.

Problem 8

Suppose $V$ is a real vector space.

• The complexification of $V$, defined by $V_{\mathbb{C}}$, equals $V\times V$. An element of $V_{\mathbb{C}}$ is an ordered pair $(u,v)$, where $u,v\in V$, but we write this as $u+iv$.
• Addition on $V_{\mathbb{C}}$ is defined by

$$(u_1+i{v}_1)+(u_2+i{v}_2)=(u_1+u_2)+i(v_1+v_2)$$

for all $u_1,v_1,u_2,v_2\in V$.

• Complex scalar multiplication on $V$ is defined by

$$(a+bi)(u+iv)=(au-bv)+i(av+bu)$$

for all $a,b\in \mathbb{R}$ and all $u,v\in V$.

Prove that with the definitions of addition and scalar multiplication as above, $V_{\mathbb{C}}$ is a complex vector space.

Think of $V$ as a subset of $V_{\mathbb{C}}$ by identifying $u\in V$ with $u+i0$. The construction of $V_{\mathbb{C}}$ from $V$ can then be thought of as generalizing the construction of ${\mathbb{C}}^n$ from ${\mathbb{R}}^n$ .

The complexification $V_{\mathbb{C}}$ of a real vector space $V$ is a complex vector space; so, we’re just showing that $V_{\mathbb{C}}$ behaves like ${\mathbb{C}}^n$ as a vector space over $\mathbb{C}$.

Thus, this is essentially just showing commutativity, associativity, additive and multiplicative identities, additive inverse, and distributive properties for complex numbers. Much of this is done in LADR Exercises 1A.