I have some confusion about injective linear maps and can't get help anywhere. My book says that:
Let $T$ be a linear map $T:V \rightarrow W$ then $T$ is injective if and only if $kerl(T) = 0$. That means two different elements under $T$ should have two different images.
Now this a confusing part for me:
When we write $T(V) = W$ then we mean that we multiply input vectors $v \in V$ with some matrix(transformation matrix) then we get an output-vector/image $(w \in W)$.
$(1):$ What I understood is that the kernel of the transformation matrix should be $0$ but what happens if the kernel of $T$(which I assume is a matrix) is $0$ but the input vector-vector ($v \in V$) are linearly dependent? Is that still an injective linear map?
$(2)$ Suppose I'm given a linear map $T:V \rightarrow W$ where
${ \{v_1 = (1, 2, -2), v_2 = (3, 6, -4), v_3 =(0,1,a^2-1)}\} \in V$
and
$\{w_1 = (2,-a^2,-1) ,w_2 = (4, 6-2a, 2), w_3 = (0, 3, 2)\} \in W$
I was asked for which values of $a$ there is at least one injective linear map such that $T(v_1) = w_1 ,T(v_2) = w_2,T(v_3) = w_3$
and for which values there is a unique injective linear map.
Here I'm not given the transformation matrix $T$ but only the input-vector and images, so how do i determine when there is at least one injective linear map?
I only need hint/(if possible steps with explanation if possible) and not full solution.
Recall the definitions: an injective map is a map that does not map any two different elements to the same element. This is in general. For the specific case of linear maps, you can prove that it is equivalent to not mapping any non-zero element to $0$, which in turn is equivalent to $\ker(T)=\{0\}$.
If you write $T(V)=W$, then you actually mean that the map $T$ is surjective, since for every $w \in W$ there is $v \in V$ such that $T(v)=w$. If you don't intend $T$ to be surjective, you should just write $T \colon V \to W$ (or $T(V) \subseteq W$).
(1) your question is not clear: being injective is a property of the map, it does not depend on the input vectors you choose. But if the map is injective, then linearly independent vectors in $V$ are mapped to linearly independent vectors in $W$ (can you see why this is so?). Linearly dependent vectors are always mapped to linearly dependent vectors, even if $T$ is not injective (can you see why?)
(2) if linearly independent vectors are mapped to linearly dependent vectors, then there can be no injective map that does it, as we have seen in (1). If on the other hand all the $v_i$ are linearly independent, then they form a basis (you know why?), so their images (if they are linearly independent) determine a unique injective linear map. If the $v_i$ are linearly dependent, then you have to see whether it is possible to define a map with the required properties and whether it is injective or not. It is difficult to say more, not knowing which facts you can use.