1. Doubt about Linear transformation

Hi (excuse me about my english).

I have a doubt about this linear transformation: f(x) = x - 1

There is a theorem that says that if the kernel of the transformation is 0; then the transformation is inyective:

if the ker (f(x)) = 0

But in this linear transformation the ker (f) = { x = 1 } ; because 1 - 1 = 0

So the transformation could not be linear, what am I doing wrong?.

Thank you in advance

2. Re: Doubt about Linear transformation

Originally Posted by supertren
Hi (excuse me about my english).
I have a doubt about this linear transformation: f(x) = x - 1
There is a theorem that says that if the kernel of the transformation is 0; then the transformation is injective:
But in this linear transformation the ker (f) = { x = 1 } ; because 1 - 1 = 0
So the transformation could not be linear, what am I doing wrong?.
The theorem is: If $\mathcal{F}$ is a linear transformation is an injection then $\mathcal{Ker(F)}=\{0\}$
It does not say: If $\mathcal{Ker(F)}=\{0\}$ then $\mathcal{F}$ is a injective linear transformation.

3. Re: Doubt about Linear transformation

Originally Posted by Plato
The theorem is: If $\mathcal{F}$ is a linear transformation is an injection then $\mathcal{Ker(F)}=\{0\}$
It does not say: If $\mathcal{Ker(F)}=\{0\}$ then $\mathcal{F}$ is a injective linear transformation.
The theorem is in both directions:

If $\mathcal{F}$ is a linear transformation is an injection then $\mathcal{Ker(F)}=\{0\}$; and If $\mathcal{Ker(F)}=\{0\}$ then $\mathcal{F}$ is a injective linear transformation

4. Re: Doubt about Linear transformation

$f(x)=x-1$, $f:\mathfrak{R}\longrightarrow \mathfrak{R}$,

is not a linear transformation because $f(x+y)\neq f(x)+f(y)$

5. Re: Doubt about Linear transformation

Originally Posted by Idea
$f(x)=x-1$, $f:\mathfrak{R}\longrightarrow \mathfrak{R}$,

is not a linear transformation because $f(x+y)\neq f(x)+f(y)$
OMG, you are right.

Thanks.