- #1
gysush
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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?