Linear Transformation and Proving Norms

[cassiew]

Homework Statement

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)

Homework 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;
||cv||=c||v||
||v+w|| is less than or equal to ||v||+||w||

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.

 

[Dick]

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?

 

[VeeEight]

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.

 

[cassiew]

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?

 

[Dick]

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.

 

[cassiew]

Okay, thanks!