Inner Product of a Linear Transformation

[gysush]

Homework Statement

Let V be a vector space over a field F = R or C. Let W be an inner product space over F. w/ inner product <*,*>. If T: V->W is linear, prove <x,y>' = <T(x),T(y)> defines an inner product on V if and only if T is one-to-one

Homework Equations

What we know, W is an inner product space, so it satisfies for x,y,z in W and c in F the properties of inner product space:
<x + z, y> = <x,y> + <z,y>
<cx,y>=c<x,y>
<x,y>= conjugate<y,x>
<x,x> > 0 if x does not = 0

T linear, thus for x,y in V and c in F
T(x+cy)=T(x) + cT(y)

Forward phase:
If <x,y>' = <T(x),T(y)> is an inner product then it satisfies the same requirements for an inner product space mentioned above.

Want to show T is one-to-one. => by def. if T: V-> W is one-to-one then for x,y in V T(x),T(y) in W...T(x)=T(y) => x=y...or equivalently the contra-positive.
Also, T is one-to-one iff Ker(T) = {0}

By def. an inner product on V is a function that assigns x,y in V/F to a scalar in F denoted by <x,y>

3. Attempt

We want to show that if T(x)=T(y) then x=y...
consider <T(x),T(y)> = <T(x),T(x)> which is > 0 unless T(x) = 0.
Now, I am failing to see how this helps at all. If <T,T> is > 0...than how does this give us any information about x and y? If T(x)=0...then this argument is the Ker argument...

So...Consider all x in V s.t. T(x)=0...we want to show x=0.

then...<T(x),T(y)> = < 0 , T(y)> = 0 = <T(y),0>

Again..I do not see how this gives us any info about x and y.

Any starting hints?

 

[phyzguy]

Try forming T(x-y), and then form <T(x-y),T(x-y)>. I think this will give you the answer.

 

[JSuarez]

If $\left\langle Tx,Ty\right\rangle$ is an inner product, what can you say about $\left\langle Tx,Tx\right\rangle$? What does this tell you about $ker\left( T\right)$?

 

[gysush]

I think I see what I was missing. I was trying to only manipulate <T(x),T(y)>.
Here is what I think now...

Let T(x)=T(y) and consider <x-y,x-y>' = <T(x-y),T(x-y)> = <T(x),T(x-y)> - <T(y),T(x-y)> Then, <x-y,x-y>' = <T(x),T(x-y)> - <T(x),T(x-y)> = 0
Then, <a,a>' = 0 implies a=0 => x-y = 0 => x=y

Then for the reverse direction Let T(x)=T(y) => x=y...consider <x + z, y>' = <T(x+z),T(y)>
then <T(x+z),T(y)> = <T(x),T(y)> + <T(z),T(y)> since W is inner product space
which equals <x,y>' + <z,y>'

Then similarly for the other the other conditions.