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## Proof of the Day: The Kernel’s Injective Fried Chicken

Let $T : V\longrightarrow W$ be a linear map between vector spaces. Show that T is injective if and only if $ker T = \{0\}$.

Proof:

Suppose first that T is injective. Then $Tx = 0 = T0$ implies $x = 0$, so $ker(T) = \{ 0\}$.

Now suppose $ker(T) = \{ 0\}$. Then $Tx = Ty$ implies $T(x - y) = 0$ which implies $x - y = 0$, and $x = y$. So, T is injective.

$\Box$

## Proof of the Day: Homomorphisms and Abelian Groups

Let $f:G \rightarrow H$ be a homomorphism of Abelian groups.  Show that $f(-x)=-f(x)$ for each $x\epsilon G$.

Proof:

Since f is a homomorphism, and since $x\epsilon G$, an Abelian group, then:

$f(x) = f(-x)+0$

$= f(-x) + (f(x) - f(x))$

$= (f(-x) + f(x)) - f(x)$ (by associativity)

$= f(-x+x)-f(x)$ (f is a homomorphism)

$= f(0)-f(x)$

$= -f(x)$

The last line is true because $f(0)=0$ (but that’s another proof).

$\Box$

## Proof of the Day (POD): Rings, Cancellation, and Zero Divisors

I’m going to try and start a new feature here, the “proof of the Day”.  It’s based on the crossfit idea of the Workout of the Day where they post a new workout everyday.  In this case, I’ll try (and maybe some of my colleagues here will also) and post a new proof (not EVERY day, but often) on a particular topic, without much but the proof itself.

While I’ll also try and post on these topics generally in other articles, the POD’s will consist of simply interesting, important, or fun proofs for their own sake.  Proofs are pretty.

A ring has the cancellation property if and only if it has no zero divisors.

Proof:
Part I.

Suppose R is a ring that has the cancellation property.  Then for $a,b,c\epsilon R$, $ab=ac$ implies $b=c$.  Now, let $a\epsilon R$ such that $a\neq 0$.  Further, suppose a is a zero divisor.  Then, there exists a $b\epsilon R$ such that $b\neq 0$, but $ab=0$. Then, $ab=a0$.  But, since R has the cancellation property, then this implies that $b=0$. This is a contradiction.  Therefore, if R has the cancellation property, R has no zero divisors.

Part II.

Suppose R has no zero divisors.  Let $a,b,c\epsilon R$, and let $ab=ac$, where $a\neq 0$.  Then, $ab-ac=0$.  This implies that $a(b-c)=0$.  Since $a\neq 0$, then $(b-c)$ must equal 0, which implies $b=c$.  So, R has the left cancellation property.

Similarly, if $ba=ca$, and $a\neq 0$, then $(b-c)a=0$ which implies $b-c=0$, so $b=c$.  So, R also has the right cancellation property.  Therefore, if R has no zero divisors, then it has the cancellation property.

$\Box$

## Busy-ness! And a love for Maple…

The quarter is off to a running start! I wish I could post more here, but… Whew!  Hopefully this post will make up for my absence!

So here’s a little potpourri for you:

## Basic Math: Dumb as a Monkey

New study shows that even college students can perform as well as monkeys on an arithmetic test.

The results indicate that monkeys perform approximate mental addition in a manner that is remarkably similar to the performance of the college students. These findings support the argument that humans and nonhuman primates share a cognitive system for nonverbal arithmetic, which likely reflects an evolutionary link in their cognitive abilities.

## Let’s Study Linear Algebra

For anyone interested in getting a jump ahead for MTH 444/544: Advance Linear/Multilinear Algebra, we’ll be meeting at 10 a.m. on Monday, 17 December in the Neuberger Atrium. Read the first four sections of the text (up through page 33) and be ready to discuss all the homework problems. That is – take a look at them and either do them or at least be sure that you can do them on the fly. There’s only 26 total for the four sections and some of those are duplications (the same kind of problem, just with different parameters – several induction proofs, for example…)

Here’s Part I

And Part II