Not Satisfied with Linear Algebra Theorem

by Vorde
Tags: algebra, linear, satisfied, theorem
Vorde is offline
Feb15-13, 07:03 PM
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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 one-to-one if and only if the equation T(x) = 0 has only the trivial solution.

The "proof" involves saying that if T is not one-to-one, 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(U-V) = 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.
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chiro is offline
Feb15-13, 08:17 PM
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Hey Vorde.

A 1-1 would imply that T(U-V) = 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 1-1.

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 1-1-ness is to show that for U != V then T(U) != T(V) for all U and V in the domain of the mapping.
Vorde is offline
Feb15-13, 08:57 PM
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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 non-zero vectors that map to the zero-vector to know that the mapping is one-to-one everywhere.

Why couldn't the zero vector be the only vector that maps to the zero vector but still have non 1-1-ness elsewhere?

chiro is offline
Feb15-13, 10:38 PM
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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 1-1, 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).
pwsnafu is offline
Feb15-13, 10:41 PM
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Quote Quote by Vorde View Post
Take two vectors, u and v.
Suppose T(u) = T(v).
Then T(u) - T(v) = 0.
But be linearity, T(u) - T(v) = T(u-v).
Then only vector that maps to 0 is the zero vector.
Hence, u - v = 0, so u = v, proving T was 1-to-1.
Vorde is offline
Feb15-13, 10:44 PM
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Okay, but that only works if you posit that the transformation is 1-1, and the theorem doesn't start with the assumption that the transformation is 1-1. 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.
WannabeNewton is offline
Feb15-13, 10:48 PM
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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.
Vorde is offline
Feb15-13, 10:53 PM
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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.
lurflurf is offline
Feb15-13, 11:01 PM
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T is one to one if
we must have

now suppose
only when

by linearity
then we must have

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