1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Linear Transformation and Proving Norms

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose T : V --> W is a linear transformation and one-to-one. Show, if ||.|| is a norm on W, then ||x|| =||T(x)|| is a norm on V.
    (V and W are vector spaces)

    2. Relevant equations

    T is linear, so T(x+y)= T(x) + T(y) and T(ax)= aT(x)
    T is one-to-one, so T(x)=T(y) implies that x=y.
    ||.|| is a norm, so ||v||=0 iff v=0 and is always greater than or equal to 0;
    ||v+w|| is less than or equal to ||v||+||w||

    3. The attempt at a solution

    I know since T is one-one, then ker(T)={0} and since T is linear, then T(0)=0. I tried using the properties of linear transformations to prove the three properties (listed above) of a norm, and I think it can be solved in this way, but I haven't been able to figure it out.
  2. jcsd
  3. Feb 25, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    I think it can be solved that way as well. Why don't you try it? To be clear, I'd suggest you write ||x|| to indicate the given norm on W and ||x||'=||T(x)|| to indicate the other norm. So just start with the first property you need to prove. Is ||x||'>=0 with ||x||'=0 only if x=0?
    Last edited: Feb 25, 2010
  4. Feb 25, 2010 #3
    Go through each one and apply the properties that you are given.
    For the last one, the triangle inequality, take ||x+y|| = ||T(x+y)|| = ||T(x) + T(y)|| (by linearity). ||.|| is a norm on w, so apply the triangle inequality here.
  5. Feb 25, 2010 #4
    So, for the first property, I know that ker(T)={0} so then
    ||x||' = ||Tx|| = 0 iff x=0, but how do you know that ||Tx||>=0? Is it just because ||x|| is a norm?

    For the second, I know
    ||cx||' = ||T(cx)|| = ||cT(x)|| = |c|*||Tx|| = |c|*||x||', so this holds.

    For the triangle inequality, I know that
    ||x+y||' = ||T(x+y)|| = ||Tx+Ty||, but how is this <= to ||Tx||+||Ty||? Can you use the fact that ||x+y||<=||x||+||y||? I mean, I know ||x|| is a norm, which is given, but does that mean I can use the triangle inequality from that to show that ||x||' is a norm?
  6. Feb 25, 2010 #5


    User Avatar
    Science Advisor
    Homework Helper

    I think the answers to both your questions are, yes, we know these things because we are given that ||x|| is a norm on W. The only property where there is really much of anything to prove is the first one, ||x||'=0 means ||T(x)||=0 which is only true if T(x)=0, which is only true if x=0 since T is one to one. Since ker(T)={0} as you've already said.
  7. Feb 25, 2010 #6
    Okay, thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook