Is T a One-to-One Linear Operator?

[H12504106]

Homework Statement

Let T be a linear operator on a finite dimensional vector space V. Suppose ||T(x)|| = ||x|| for all x in V, prove that T is one to one.

Homework Equations

||T(x)||^2 = <T(x),T(x)>
||x||^2 = <x,x>

The Attempt at a Solution

Suppose T(x) = T(y) x, y in V
Then ||T(x)|| = ||T(y)||
and so <T(x),T(x)> = <T(y),T(y)>
By assumption, we have <x,x> = <y,y>
But then i can't proceed on to show that x=y

 

[Mark44]

Homework Statement

Let T be a linear operator on a finite dimensional vector space V. Suppose ||T(x)|| = ||x|| for all x in V, prove that T is one to one.

Homework Equations

||T(x)||^2 = <T(x),T(x)>
||x||^2 = <x,x>

The Attempt at a Solution

Suppose T(x) = T(y) x, y in V
Then ||T(x)|| = ||T(y)||
and so <T(x),T(x)> = <T(y),T(y)>
By assumption, we have <x,x> = <y,y>
But then i can't proceed on to show that x=y

One form of the definition for one-to-one is this:
For all x and y in V, T(x) = T(y) ==> x = y.
You can do a proof by contradiction by assuming that for some x and y in V, T(x) = T(y) and x $\neq$ y.

You need to use the fact that T is a linear operator.

 

[spamiam]

If you've seen the right theorem, proving that $\ker(T) = \{0\}$ might be a slightly easier way to do this.