# problem involving norm

• Nov 18th 2009, 05:02 PM
dannyboycurtis
problem involving norm
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).
• Nov 19th 2009, 02:18 AM
flyingsquirrel
Quote:

Originally Posted by dannyboycurtis
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 \$\displaystyle x=-y\$ then \$\displaystyle \|x\| = \|y\|\$ so \$\displaystyle \|x\|=\|y\|\$ does not imply \$\displaystyle x=y\$.

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