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Homework Help: Inner Product of a Linear Transformation

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant 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>
    <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?
  2. jcsd
  3. Mar 30, 2010 #2


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    Science Advisor

    Try forming T(x-y), and then form <T(x-y),T(x-y)>. I think this will give you the answer.
  4. Mar 30, 2010 #3
    If [itex]\left\langle Tx,Ty\right\rangle[/itex] is an inner product, what can you say about [itex]\left\langle Tx,Tx\right\rangle[/itex]? What does this tell you about [itex]ker\left( T\right)[/itex]?
  5. Mar 30, 2010 #4
    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.
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