Example for a linear transformation with $\dim(\ker(f))=1$

Question

My answer:

$f: \Bbb{Z} \longrightarrow \Bbb{Z}$ defined by $f(x) = x$

Is that wrong? :D

Answer 1

Let $T : \mathbb{R}^{2} \longrightarrow \mathbb{R}^{2}$ an Linear Transformation, defined by : $$T(x,y) = (x-y, x-y)$$.

$\text{dim}(Ker(T)) = 1$

Answer 2

The simplest example is the zero transformation of the real numbers into itself. So, $f: \Bbb{R} \to \Bbb{R} $ defined by $f(x) =0$. Its kernel is all of $\Bbb{R} $, and is hence one dimensional.

What you wrote doesn't make sense because to talk about linear transformations, the domain and target space need to be vector spaces over a field, but I don't see how the set of integers, $\Bbb{Z}, $ can be made into a vector space.

Answer 3

You just need a transformation that maps a line through origin (1-dimensional vector space) to zero vector, for e.g. $f:\mathbb R^3\rightarrow \mathbb R^3$ defined as $f(x,y,z)=(x,y,0)$ maps z-axis to zero vector.