kernel

Aug 2009
639
2
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 2009
1,176
412
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 \(\displaystyle T\) be a linear map.

Then \(\displaystyle T\) is injective
\(\displaystyle \Leftrightarrow (Tx = Ty \Rightarrow x=y) \)
\(\displaystyle \Leftrightarrow (Tx-Ty = 0 \Rightarrow x=y) \)
\(\displaystyle \Leftrightarrow (T(x-y) = 0 \Rightarrow x = y) \)
\(\displaystyle \Leftrightarrow (T(x-y) = 0 \Rightarrow x-y = 0) \)
\(\displaystyle \Leftrightarrow (Tu = 0 \Rightarrow u =0) \)
\(\displaystyle \Leftrightarrow kerT = \{0\}\).
 
  • Like
Reactions: HallsofIvy
Aug 2009
639
2
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 2009
1,176
412
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 \(\displaystyle ab^{-1}\)). So I'll use group notation.

Is your question basically,

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

The answer is `no'.

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

The same holds for linear maps, etc.
 
  • Like
Reactions: alexandrabel90
Aug 2009
639
2
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 2009
1,176
412
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 \(\displaystyle +\) with \(\displaystyle *\) and \(\displaystyle T\) with \(\displaystyle \phi\).

So it starts,

\(\displaystyle \phi\) is injective
\(\displaystyle \Leftrightarrow (g\phi = h\phi \Rightarrow g=h)\)
\(\displaystyle \Leftrightarrow ((g\phi)(h\phi)^{-1} = 1 \Rightarrow g=h)\)
\(\displaystyle \vdots\)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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".
 
  • Like
Reactions: alexandrabel90