I have this linear Transformation –
$ T : \mathbb{P}^2 → \mathbb{R}^2 $
in which T is defined by : $T(ax^2+bx+c)$ = $(a+c,2b)$
If we try to find the kernel of this Ker(T) = $0$,
so we set $(a+c,2b)$ = $0$ to get the values of b = $0$, and c = -a.
If we put this back into the quadratic polynomial we have:
$ax^2+bx+c$ = $ax^2+ 0x+(-a)$ = $ax^2-a$
So T($ax^2-a$) = $0$, and we can choose infinitely many values of 'a', so that means we get many of these polynomial vector types being sent to the $0$ vector in the output space.
So i can choose 2 values of 'a' to show that we have more than one input that maps to output.
This shows that this is not One-To-One.
BUT i am having difficulty with the basic definition of when it is one-to-one.
I read that only ONE vector can go to the zero vector in the output space.
But does that mean it can be any one vector?
OR
Since this is a linear transformation, is the only vector allowed be the Zero vector in the Input space and it goes to the Zero vector in the output space.
So my understanding of One-to-One is not that clear when we look at it from the perspective of the Kernel.
Input on this is greatly appreciated!
Best Answer
Try proving: A linear function is one to one if and only if it's kernel is a trivial vector space(contains only zero).
Hints: $\impliedby$ take two distinct vectors and try finding the difference in their image.