I'm trying to prove that a linear map is injective

[catcherintherye]

hello, I've been reading some proofs and in keep finding this same argument tyo prove that a linear map is injective viz, we suppose that t(a,c) = 0 and then we deduce that a,c = 0,0. is it the case that the only way a linear map could be non injective is if it took two elements to zero? i.e. t injective iff ker(t) not zero?

 

[HallsofIvy]

First, it's easy to prove that any linear map takes the 0 vector to 0. If a linear maps takes some non-zero vector to 0 also, then it clearly is not injective.

On the other hand, suppose the linear map T takes two vectors, u, v, with [itex]u\ne v[/itex], into the same thing. Then T(u- v)= T(u)- T(v)= 0 so T takes the non-zero vector u-v into 0.

However, your final statement is exactly reversed: t is injective if and only if ker(t) is {0}.

 

[tacman]

Yes, that is correct. The definition of injectivity for a linear map is that if t(a) = t(b), then a = b. In other words, if two elements in the domain map to the same element in the co-domain, then they must be the same element. This means that the only way for a linear map to be non-injective is if there exists two distinct elements in the domain that map to the same element in the co-domain, which is exactly the definition of the kernel (ker(t)). Therefore, a linear map is injective if and only if its kernel is zero.