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Prove adjoint invertible

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Let V be a finite-dimensional inner product space, and let T be a linear operator on V. Prove that if T is invertible, then T* is invertible and (T*)^-1 = (T^-1)*

    2. Relevant equations
    As shown above.
    <T(x),y> = <x,T*(y)>

    3. The attempt at a solution
    Well, I figure you only need to show that the equation holds, that shows that T* is invertible, since its inverse exists.
    Now, I try to do something with the inner products:

    <(T^-1)(x),y> = <x,(T^-1)*(y)>

    I’m not sure how to “flip” inverse and the star.

    Thanks for your help! =)
  2. jcsd
  3. May 7, 2007 #2


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    Start with <T*(T-1)*x,y> and see what you end up with.
  4. May 8, 2007 #3
    <T*(T-1)*x,y> = <T*(x), T-1(y)> = <(T-1)*x,T(y)> = <x, (T-1)(T)y> = <x,y>

    so T*(T-1)* = I, so (T-1)* is the inverse of T*, hence (T-1)* = (T*)-1)

    thanks! =)
  5. May 8, 2007 #4


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    Step 1 is to deduce that <T*(T-1)*x,y> = <x,y>. Once you get that, step 2 is exactly what you did:

    T*(T-1)* = I, so (T-1)* is the inverse of T*, hence (T-1)* = (T*)-1

    But you appear to have done step 1 wrong. You made the right conclusion, but all your steps look invalid. For example, you first line is of the form <A*B*x,y> = <A*x,By>. Why is this wrong in general? Well if it were always true, we'd get:

    <A*B*x,y> = <A*x,By>
    <A*B*x,y> = <B*A*x,y>
    <(A*B* - B*A*)x,y> = 0 (for all y)
    A*B* - B*A* = 0
    AB = BA (in general, i.e. for all A and B)

    But matrix multiplication is not commutative in general, so this is wrong. Try again.
  6. Aug 21, 2010 #5
    then how is?
  7. Aug 22, 2010 #6


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    Then do it being very careful about order.

    <A*B*x,y> = <B*x,Ay>= <x, BAy>.

    <B*A*x,y>= <A*x, By>= <x, ABy>.
  8. Aug 22, 2010 #7
    Understand but, In first prove (T* is invertivel) but how sure that <T*(T-1)*x,y> = <x,y> , if i dont know that (TT-1) is I, because is just that want will prove
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