## 403/503 – Homework 2

February 15, 2010

This homework is due Wednesday, February 24.

Solve the following problems from Chapter 3 of the textbook: 2, 4, 5, 7, 9, 10, 26.

In addition, solve the following problem:

Let $T:V\to W$ be a linear map between finite dimensional vector spaces over ${\mathbb F}={\mathbb R}.$ For $S\subseteq W,$ we denote by $T^{-1}[S]$ the preimage of $S$ under $T.$ This is the set of all vectors $v\in V$ such that $Tv\in S.$

Recall that a subset $C$ of a vector space $U$ over ${\mathbb R}$ is convex iff whenever $u_1,u_2\in C$ and $\alpha\in[0,1],$ then also $\alpha u_1+(1-\alpha)u_2\in C.$ Show that $T^{-1}[S]$ is convex iff $S\cap{\rm ran}(T)$ is convex. Is it necessary that $T$ is linear for this to hold, or does it suffice that $T$ is continuous?

Assume that $S$ is a subspace of $W,$ and show that $T^{-1}[S]$ is a subspace of $V.$

## 187 – Quiz 3

February 15, 2010

Here is quiz 3.

Problem 1 requests a counterexample to the statement: An integer $x$ is positive if and only if $x+1$ is positive.

To find a counterexample to a statement of the form “an integer $x$ has property $A$ if and only if it has property $B,$ we need an integer $x$ such that one of the properties is true of $x$ while the other is false. Note that if $x$ is positive, i.e., $x>0,$ then $x+1>1>0,$ i.e., $x+1$ is also positive. This means that the only way the statement is going to fail is if we have that $x+1$ is positive yet $x$ is not.

We can now easily see that $x=0$ is a counterexample. (And, in fact, it is the only counterexample.)

Problem 2 asks to show that $C\subseteq D,$ where $C=\{x\in{\mathbb Z}\colon x|12\}$ and $D=\{x\in{\mathbb Z}\colon x|60\}.$

By definition, $C\subseteq D$ iff any element of $C$ is also an element of $D.$ Let us then consider an element $x$ of $C.$ By definition of $C,$ this means that $x$ is an integer, and $x|12.$ We need to show that $x\in D,$ i.e., that $x\in{\mathbb Z}$ and that $x|60.$ The first requirement is automatic, so we only need to verify that $x|60.$

To see this, we use that $x|12.$ By definition, this means that there is an integer, let us call it $a,$ such that $xa=12.$ To show that $x|60,$ we need (again, by definition) that there is an integer $b$ such that $xb=60.$ Now, since $xa=12,$ then $5(xa)=5\times 12=60,$ or $x(5a)=60.$ We immediately see that we can take $b=5a.$ This is an integer, and $xb=60,$ as needed.

Problem 3 asks for the cardinality of the set $2^{2^{\{1,3,7,11\}}}.$

Recall that for any set $A,$ then $|2^A|=2^{|A|},$ as shown in lecture. In particular, if $A=\{1,3,7,11\}$, then $|2^A|=2^4=16.$ Now let $A'=2^{\{1,3,7,11\}}=2^A.$ Then $|2^{A'}|=2^{|A'|}=2^{16}=65536.$