Ok I understand the definitions but am still struggling to decide if T is injective or not, for the kernel(T) obviously we know this equals the null space(T) so if I find one vector in R4 such that Tv=the zero vector does this mean it is injective.
eg. if I choose the vector
[1]
[0]
[0]
[-1]
To show that is injective you have to show that the only vector such that is ; that's not what you're doing. Anyway, this idea is often used to show that a linear application is not injective. For if you find a non-zero vector such that then contains both 0 and hence and can't be injective.
EDIT: by the way, for
No! is true for any linear application , even if is not injective (this result comes from ). A linear application is injective if . You are saying that a linear application is injective if . These two sentences are very different and only the former is correct.
Now, how can one show that ? Well, the only thing we have to do is solve for . Let .
hence is injective.