Thread: Proof Linear Transformation / 1 - 1

Hi everyone, I was given a set of proofs that I must understand but the explanation I have found after much research is way to complicated for an into to linear Algebra course, it explains it for an Injection what I need is it proved for a 1-1, after reading online about injection I foudn its the same thing, but this proof mentions things like homomorphism which I have no clue what it is and not in curriculum

So Please if anyone can Please please explain this proof in simpler terms I would more then appreciate it!!!

so this is the question


This is what I Found online

Pf:
Assume
Then if for if then which means which means or thus 1-1

Now for opposite, assume T is 1-1, then since (0 vector always maps to 0 vectors under linear transforms) then

wow that was fast seems like a very neat and simple solution

much much appreciated

it's ironic that people seem to have no trouble with the concept of "vector space", but more trouble with the concept of "abelian group".

the first 4 rules (axioms) of a vector space usually go like this:

V1) there is a binary operation, called vector addition, which sends every pair of vectors (u,v) to a vector u+v.

V2) vector addition is associative: for all u,v,w in V: (u+v)+w = u+(v+w)

V3) vector addition is commutative: for all u,v in V: u+v = v+u

V4) there is a vector (usually called the 0-vector) 0_V with the property that v+0_V = 0_V+v = v, for all v in V.

V5) for every vector v, there is another vector, -v (usually called "negative v") with v+(-v) = 0_V.

there are more "rules" vector spaces follow, but these 5 say that (V,+) is an abelian group.

the analogue for a linear transformation of a vector space, is an abelian group homomorphism. now this might SOUND scary, but all it means is:

f(x+y) = f(x) + f(y) (abelian group homomorphisms are like linear maps except there's no scalar multiplication to worry about).

it turns out that for linear transformations, we don't need to worry about the "scalar multiplication part" of a linear map: we never need to use the fact that: T(av) = aT(v).

the reason for this is: if T(v) = 0, then T(av) = a0 = 0, and if T(av) = 0, then aT(v) = 0, so unless a = 0, T(v) must have been 0 in the first place (if a = 0, then av = 0, but if T preserves vector addition:

T(0) + T(0) = T(0+0) = T(0), so subtracting T(0) from both sides gives T(0) = 0: if T is a linear map, T(0) is ALWAYS 0, so 0 is ALWAYS in any kernel).

so we don't get any "new elements" of the kernel of T by multiplying v in Ker(T) by a, so we only need to look at the vector addition (that is, consider V as just an abelian group).

jakncoke's answer is the same as what you posted, you are just unfamiliar with that terminology.

Deveno your reply is greatly appreciated, I learned something new and interesting thank you for that