Differentiability of Inverse Map in Bounded Linear Transformations

[symbol0]

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 ^{-1}: U\rightarrow U by ^{-1}(T)=T^{-1}
Show that the map ^{-1} is differentiable at each T \in U.

 

[Pere Callahan]

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?

 

[symbol0]

The only information I have about V is that V is a Banach space. Also, I know that ^{-1}: U\rightarrow U is continuous.
I tried to use the definition of derivative. So the map ^{-1} would have a derivative (^{-1})'(T) at each T in U if for T,S in U, S^{-1}=T^{-1}+m(S,T)(S-T) and
lim_{S\rightarrow T} m(S,T)= (^{-1})'(T).
Then m(S,T)=(S^{-1} -T^{-1})(S-T)^{-1}. So
(^{-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}
but what is this limit? or how do I know it exists?
By the way, a part b of the question says: show that (^{-1})'(T)(U)=T^{-1}UT^{-1}.

 

[Dick]

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?

 

[symbol0]

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

 

[Dick]

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.

 

[symbol0]

Let' see,
the term linear in h is -x^(-1)hx^(-1).
Is (^{-1})'(T)=T^{-1}T^{-1}?
but then (^{-1})'(T)(U)=U^{-1}T^{-1}U^{-1}T^{-1} and part b of the exercise says show (^{-1})'(T)(U)=T^{-1}UT^{-1}
?

 

[Dick]

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:
<br /> (^{-1})&#039;(T)(U)=(-1)*T^{-1}UT^{-1}<br />
I think the exercise answer is missing that (-1) factor.

 

[symbol0]

aahhhhh?
now I'm really confused.
Can you please write the same in terms of T\in U.
I mean, for each T\in U, (^{-1})&#039;(T)= ?

 

[Dick]

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.

 

[symbol0]

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).

 

[Dick]

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.