# kernel

• May 31st 2010, 01:49 AM
alexandrabel90
kernel
i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.

how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity?
• May 31st 2010, 01:59 AM
Swlabr
Quote:

Originally Posted by alexandrabel90
i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.

how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity?

What are we "in" here? Linear maps? Group homomorphisms?

The proof is pretty much the same in every context through.

Let $T$ be a linear map.

Then $T$ is injective
$\Leftrightarrow (Tx = Ty \Rightarrow x=y)$
$\Leftrightarrow (Tx-Ty = 0 \Rightarrow x=y)$
$\Leftrightarrow (T(x-y) = 0 \Rightarrow x = y)$
$\Leftrightarrow (T(x-y) = 0 \Rightarrow x-y = 0)$
$\Leftrightarrow (Tu = 0 \Rightarrow u =0)$
$\Leftrightarrow kerT = \{0\}$.
• May 31st 2010, 02:11 AM
alexandrabel90
im studying group homomorphisms now.

in this case, if it is a function from A to B, where the identity of B =0, then the element in A has to be 0 since u=0 (from your above statement) right?

what if the element in A that get mapped to 0 in B is not 0, wont u be not equals to 0? in such case, Tu=0 but u wont be zero?

thanks!
• May 31st 2010, 02:26 AM
Swlabr
Quote:

Originally Posted by alexandrabel90
im studying group homomorphisms now.

in this case, if it is a function from A to B, where the identity of B =0, then the element in A has to be 0 since u=0 (from your above statement) right?

what if the element in A that get mapped to 0 in B is not 0, wont u be not equals to 0? in such case, Tu=0 but u wont be zero?

thanks!

My above proof can be quite easily changed to look at group homomorphisms (just change a+b to ab, and a-b to $ab^{-1}$). So I'll use group notation.

Is it possible that $1 \not\in ket(\phi)$?

Let $1 \phi = a$. Then $a^2 = (1\phi)^2 = (1^2)\phi = 1\phi = a$. Thus, $a^2=a$ ( $a$ is an idempotent). Then, $a^2 = a \Rightarrow a^2a^{-1} = aa^{-1} \Rightarrow a=1$. Thus, $1_G\phi = 1_H$ for $\phi: G\rightarrow H$ a group homomorphism.

The same holds for linear maps, etc.
• May 31st 2010, 02:38 AM
alexandrabel90
for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

becos im not sure when it should be under + and when it should be under x.

thanks
• May 31st 2010, 02:42 AM
Swlabr
Quote:

Originally Posted by alexandrabel90
for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

becos im not sure when it should be under + and when it should be under x.

thanks

Just go through the proof I gave for linear maps and replace $+$ with $*$ and $T$ with $\phi$.

So it starts,

$\phi$ is injective
$\Leftrightarrow (g\phi = h\phi \Rightarrow g=h)$
$\Leftrightarrow ((g\phi)(h\phi)^{-1} = 1 \Rightarrow g=h)$
$\vdots$
• May 31st 2010, 05:19 AM
HallsofIvy
Quote:

Originally Posted by alexandrabel90
for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

becos im not sure when it should be under + and when it should be under x.

thanks

You can use any symbol you like for the operation!

It is a general convention that \abelian (commutative) groups are written "additively" and non-abelian groups are written "multiplicatively".