problem involving norm

dannyboycurtis · #1 $Old$ November 18th, 2009, 06:02 PM

I need a little confirmation on this one, it seemed a bit too easy for me, which makes me think I assumed too much:
Let T be a linear operator on an inner product space V, suppose ||T(x)||=||x|| for all x in V. Prove T is injective.
Here is my proof:
let x,y exist in V such that T(x)=T(y).
So ||T(x)||=||x|| implies
||T(x)||=||T(y)|| implies
||x||=||y|| implies
x=y. (Its this implication Im suspicious of).
Any help please?

flyingsquirrel · #2 $Old$ November 19th, 2009, 03:18 AM

Quote:

Originally Posted by dannyboycurtis $View Post$

Here is my proof:
let x,y exist in V such that T(x)=T(y).
So ||T(x)||=||x|| implies
||T(x)||=||T(y)|| implies
||x||=||y|| implies
x=y.

If $x=-y$ then $\|x\| = \|y\|$ so $\|x\|=\|y\|$ does not imply $x=y$ .

As $T$ is a linear operator, $T(x)=T(y)$ implies $T(x-y)=T(x)-T(y)=0$ . Hence we have $\|T(x-y)\| = 0$ . I am sure you can take it from here.