# Not Satisfied with Linear Algebra Theorem

1. Feb 15, 2013

### Vorde

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 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.

2. Feb 15, 2013

### chiro

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.

3. Feb 15, 2013

### Vorde

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?

4. Feb 15, 2013

### chiro

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).

5. Feb 15, 2013

### pwsnafu

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.

6. Feb 15, 2013

### Vorde

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.

7. Feb 15, 2013

### WannabeNewton

Let $T$ be injective (sorry I absolutely hate the term 1 - 1 god knows why it is even used). Because T is linear, $T(0) = 0$. This immediately implies this is the only vector for which this is true because if $\exists v\in V :T(v) = 0$, then the injectivity of $T$ implies that $v = 0$. Now let $T(v) = 0$ be true only for the zero element then if $T(v) = T(w)$ we have that $T(v - w) = 0\Rightarrow v = w$ thus $T$ 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. Feb 15, 2013

### Vorde

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. Feb 15, 2013

### lurflurf

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(u-v)=0
then we must have
u-v=0
so
u=v