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

[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?

 

[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...

 

[wurth_skidder_23]

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

 

[StatusX]

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

 

[wurth_skidder_23]

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

 

[slearch]

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

 

[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.

 

[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).