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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.

$\Box$