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Show it is differentiable

  1. Nov 21, 2008 #1
    Let B(V,V) be the set of bounded linear transformations from V to V. Let U be the set of invertible elements of B(V,V) and define the map [tex]^{-1}[/tex]: [tex] U\rightarrow U[/tex] by [tex]^{-1}(T)=T^{-1}[/tex]
    Show that the map [tex]^{-1}[/tex] is differentiable at each [tex]T \in U[/tex].
     
  2. jcsd
  3. Nov 22, 2008 #2
    For us to be able to help you, you would first have to show your work. Moreover it wouldn't hurt to say what V is. It surely has some topology?
     
  4. Nov 22, 2008 #3
    The only information I have about V is that V is a Banach space. Also, I know that [tex]^{-1}: U\rightarrow U[/tex] is continuous.
    I tried to use the definition of derivative. So the map [tex]^{-1}[/tex] would have a derivative [tex](^{-1})'(T)[/tex] at each T in U if for T,S in U, [tex]S^{-1}=T^{-1}+m(S,T)(S-T)[/tex] and
    [tex]lim_{S\rightarrow T} m(S,T)= (^{-1})'(T)[/tex].
    Then [tex]m(S,T)=(S^{-1} -T^{-1})(S-T)^{-1}[/tex]. So
    [tex](^{-1})'(T)=lim_{S\rightarrow T} m(S,T)=lim_{S\rightarrow T}(S^{-1} -T^{-1})(S-T)^{-1}= lim_{S\rightarrow T}-S^{-1}(S-T)T^{-1}(S-T)^{-1}[/tex]
    but what is this limit? or how do I know it exists?
    By the way, a part b of the question says: show that [tex](^{-1})'(T)(U)=T^{-1}UT^{-1}[/tex].
     
  5. Nov 22, 2008 #4

    Dick

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    You want a linear function A_x such that lim |(x+h)^(-1)-x^(-1)-A_x(h)|/|h| -> 0 as |h|->0. (x+h)^(-1)=(x(1+x^(-1)h))^(-1)=(1+x^(-1)h)^(-1)*x^(-1). Now use the good old reliable 1/(1+S)=1-S+S^2-... Can you see what A_x(h) is?
     
  6. Nov 22, 2008 #5
    Hi Dick,
    Can I see what A_x(h) is?
    No, I can't.
    I get the part of (x+h)^(-1)=(x(1+x^(-1)h))^(-1)=(1+x^(-1)h)^(-1)*x^(-1).
    But then you say: Now use the good old reliable 1/(1+S)=1-S+S^2-...
    where do I use that? You mean use that on (1+x^(-1)h)^(-1)?
    but then how do I get to A_(x)?
    HEEELP
     
  7. Nov 22, 2008 #6

    Dick

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    Sure. (1+x^(-1)h)^(-1)=1-x^(-1)h+(x^(-1)h)^2- ... The linear function A_x(h) has to be the part of (x+h)^(-1)-x^(-1) that is linear in h (i.e. first power). Parts that don't have an h in them have to cancel and higher powers of h will go to zero as h->0.
     
  8. Nov 22, 2008 #7
    Let' see,
    the term linear in h is -x^(-1)hx^(-1).
    Is [tex](^{-1})'(T)=T^{-1}T^{-1}[/tex]?
    but then [tex](^{-1})'(T)(U)=U^{-1}T^{-1}U^{-1}T^{-1}[/tex] and part b of the exercise says show [tex](^{-1})'(T)(U)=T^{-1}UT^{-1}[/tex]
    ???
     
  9. Nov 22, 2008 #8

    Dick

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    Good! The linear term is A_x(h)=-x^(-1)hx^(-1). That's your derivative. In the notation of the second part, you've shown:
    [tex]
    (^{-1})'(T)(U)=(-1)*T^{-1}UT^{-1}
    [/tex]
    I think the exercise answer is missing that (-1) factor.
     
  10. Nov 22, 2008 #9
    aahhhhh?
    now I'm really confused.
    Can you please write the same in terms of [tex]T\in U[/tex].
    I mean, for each [tex]T\in U[/tex], [tex](^{-1})'(T)=[/tex] ?
     
  11. Nov 22, 2008 #10

    Dick

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    You are confused, and I'm not surprised. W=(^-1)'(T) is a linear function from B(V,V) to B(V,V) defined for each T in U. To say what W is you have to say how it acts on an A in B(V,V). W(A)=(-1)*T^(-1)AT^(-1). Try relating this to ordinary derivatives on R. If f'(x)=m, then |f(x+h)-f(x)-m*h|/|h| goes to zero as h goes to zero (everything ordinary numbers). If f(x)=1/x, then the derivative m at x=a is -1/a^2. So m*h=(-1)(1/a)*h*(1/a). Do you see how that relates to your problem? The fact linear transformations don't commute makes it a little different. But only a little.
     
  12. Nov 23, 2008 #11
    Hi Dick,
    First of all, thank you for your help.
    I see how your example relates to the problem.
    It just seems weird to me that if you can define (-1)(T) as (-1)(T) =T^(-1) (without using an aditional element A), you say that we can't do the same for the derivative. That is, to give the derivative at T without using some additional element A.
    I'm not sure about the following:
    Wouldn't the derivative of T be (-1)'(T)= -T^(-1)T^(-1) because if (-1)'(T)(U)= -T^(-1)UT^(-1) then (-1)'(T)= (-1)'(T)(I)= -T^(-1)IT^(-1)= -T^(-1)T^(-1).

    Perhaps the derivative can be expressed as (-1)'(T)= -(*T^(-1))(T^(-1)) where (*w)(v)=v*w. So for example
    (-1)'(T)(U)(V)= -(*T^(-1))(T^(-1))UV= -(*T^(-1))(T^(-1)UV) =-T^(-1)UVT^(-1).
     
  13. Nov 23, 2008 #12

    Dick

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    I'm not really sure where you are going there. But I'm pretty sure (^(-1))'(T)(U)=-T^(-1)UT^(-1). We got that out of the linearized form of (^(-1)) and being careful about order. For derivatives in R, you can be careless about the ordering. Here you can't. (^(-1))'(T) is supposed to be a linear operator from B(V,V) to B(V,V), so it has to act on something. -T^(-1)UT^(-1) is not the same as -T^(-1)T(-1)U. If U=I, it is, but that's not true in general. The derivative here is an operator.
     
    Last edited: Nov 23, 2008
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