How would I prove that an isometry is one to one?

1. Nov 12, 2006

wurth_skidder_23

How would I prove that an isometry is one to one?

General definition of isometry, A:

<Ax,Ay> = <x,y>

Where < , > is an inner product (scalar product, dot product, etc.)

How do I prove A has to be one to one for this to work?

2. Nov 12, 2006

StatusX

Note the distance between the images of two points will be the same as the distance between the points. So if two points map to the same image...

3. Nov 12, 2006

wurth_skidder_23

Is there a way to show that mathematically using the adjoint or something similar?

4. Nov 12, 2006

StatusX

To show what, that distances are preserved? What is distance in an inner product space?

5. Nov 12, 2006

wurth_skidder_23

distance between two vectors x, y in an inner product space is ||x-y||

6. Nov 12, 2006

slearch

Yes, and $$||x-y||^2$$ has an expression in terms of the inner product.

7. Nov 12, 2006

wurth_skidder_23

Okay, maybe I should state the whole problem:

Let X and Y be inner product spaces in R with inner products < , >_X and < , >_Y Suppose that T is in L(X,Y). Show that <x,y> = <Tx,Ty>_Y is an inner product on X if and only if T is one to one.

8. Nov 12, 2006

StatusX

The <- direction should be pretty easy, you just need to verify all the axioms of an inner product, and you'll need to use the linearity of T. I'm assuming your original question comes from proving the -> direction. It's been pretty much spelled out for you. Just use the definition of distance to show d(x,y)=d(Tx,Ty).