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Not Satisfied with Linear Algebra Theorem 
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#1
Feb1513, 07:03 PM

P: 784

Hello,
I am just not satisfied with the following theorem (I don't know it's name): Let T:R^n > R^m be a linear transformation. Then T is onetoone if and only if the equation T(x) = 0 has only the trivial solution. The "proof" involves saying that if T is not onetoone, then there are two different vectors U and V such that T(U)=T(V)= some vector B. And since T is linear it follows that T(UV) = T(U)T(V) = B  B = 0. It then concludes by saying "hence there are nontrivial solutions to T(X)= 0. So, either the two conditions in the theorem are both true or they are both false." I just don't see how that proved the theorem in any way, perhaps because I don't fully understand which two conditions it is talking about. Could anyone help me here? Thank you. 


#2
Feb1513, 08:17 PM

P: 4,573

Hey Vorde.
A 11 would imply that T(UV) = T(U)  T(V) != 0 for U != V. If however this were false then it would imply that for some U != V that T(U) = T(V) proving that the mapping is not 11. But then you have to take into account the trivial solution (i.e. the zero vector) as a special case where T(U) = 0 for U = 0. The formal proof for 11ness is to show that for U != V then T(U) != T(V) for all U and V in the domain of the mapping. 


#3
Feb1513, 08:57 PM

P: 784

Hey Chiro,
Right, so I understand that part. What I don't understand is how it is enough to show that there aren't any nonzero vectors that map to the zerovector to know that the mapping is onetoone everywhere. Why couldn't the zero vector be the only vector that maps to the zero vector but still have non 11ness elsewhere? 


#4
Feb1513, 10:38 PM

P: 4,573

Not Satisfied with Linear Algebra Theorem
We know that the zero vector always maps to the zero vector, but we also know that if everything is 11, then it means that only the zero vector maps to the zero vector and everything else maps to some other vector (that isn't the zero vector).



#5
Feb1513, 10:41 PM

Sci Advisor
P: 827

Suppose T(u) = T(v). Then T(u)  T(v) = 0. But be linearity, T(u)  T(v) = T(uv). Then only vector that maps to 0 is the zero vector. Hence, u  v = 0, so u = v, proving T was 1to1. 


#6
Feb1513, 10:44 PM

P: 784

Okay, but that only works if you posit that the transformation is 11, and the theorem doesn't start with the assumption that the transformation is 11. I understand that the zero vector will always map to itself, just not why that says anything about the rest of the transformation.
Ah, wait, I think I might see it now. 


#7
Feb1513, 10:48 PM

C. Spirit
Sci Advisor
Thanks
P: 5,585

Let [itex]T[/itex] be injective (sorry I absolutely hate the term 1  1 god knows why it is even used). Because T is linear, [itex]T(0) = 0[/itex]. This immediately implies this is the only vector for which this is true because if [itex]\exists v\in V :T(v) = 0[/itex], then the injectivity of [itex]T[/itex] implies that [itex]v = 0[/itex]. Now let [itex]T(v) = 0[/itex] be true only for the zero element then if [itex]T(v) = T(w)[/itex] we have that [itex]T(v  w) = 0\Rightarrow v = w[/itex] thus [itex]T[/itex] is injective. This is essentially what the proof you quoted is saying but the quoted proof is more concise.
EDIT: Seems like people responded while I was typing this up but I guess I'll leave it here anyways =D. 


#8
Feb1513, 10:53 PM

P: 784

Okay, thank you to all who have been helping.
What was really bothering me was that I didn't see why the trivial solution had to be the only solution for the zero vector. But I just went back and thought about it and now I can see that this must be the case. That assumed, I can follow the rest of the theorem. Once again, thank you all. 


#9
Feb1513, 11:01 PM

HW Helper
P: 2,264

T is one to one if
whenever T(u)=T(v) we must have u=v now suppose T(x)=0 only when x=0 If T(u)=T(v) by linearity T(uv)=0 then we must have uv=0 so u=v 


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