Derivatives of inverses

  • Thread starter Benny
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  • #1
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Hi, I'm having trouble getting started on the following question and would like some help.

Q. Let U be an open subset of R^n and V be an open subset of R^m. Assume that [itex]f:U \to V[/itex] is a C^1 function with a C^1 inverse [itex]g:V \to U[/itex]. (Thus (f o g) is the identity function [itex]1_V :V \to V[/itex] and (g o f) is the identity function [itex]1_U :U \to U[/itex].)

a) Show that the matrix [itex]Df\left( {\mathop {x_0 }\limits^ \to } \right)[/itex] is invertible with inverse [itex]Dg\left( {f\left( {\mathop {x_0 }\limits^ \to } \right)} \right)[/itex] for each x_0 in U.

I'm thinking that I might need to use matrix multiplication for example AA^-1 = I or something like that. I know that the derivative matrices for f and g evaluated at x_0 are defined since they are both C^1. Those are just some rough ideas and I don't know where to begin. Is there something obvious that I should I work with to get started? Any help would be good thanks.
 
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Answers and Replies

  • #2
HallsofIvy
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Just apply the chain rule to (f o g)(x)= x.
 
  • #3
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I considered it this way.

By the chain rule (g o f)'(x) = Dg(f(x))Df(x)

(g o f) is the identity function which is (I think) linear. So I write (g o f)(x) = Ix. The derivative of a linear function is just the matrix.

So I = Dg(f(x))Df(x). If that takes care of the inverse part then is there anything else I need to do in order to show that Df(x) is invertible? If CD = I then does it automatically follow that C is invertible?
 
  • #4
0rthodontist
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Well, you should also show that DC = I... but since Df is not nxn I don't really know what you're doing.
 
  • #5
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Ok I see your point.

If I use (f o g)(x) = x and I differentiate both sides I obtain Df(g(x))Dg(x) = 0 for a fixed vector x. Differentiating (g o f)(x) = x I obtain Dg(f(x))D(g(x)) = 0 for a fixed vector x. So Df(g(x))Dg(x) = Dg(f(x))Df(x). That doesn't appear to be going anywhere. :confused:
 
  • #6
0rthodontist
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No, it is not true in general that Df(g(x))Dg(x) = Dg(f(x))Df(x) because they are matrices of different dimensions.
 
  • #7
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The next part of the question says to conclude that n = m. But since I can't use that in this question, how would I even show that Df(x_0) is invertible with inverse Dg(f(x_0))?

To show the given result for this question I would need to arrive at Dg(f_0))Df(x_0) = I = Df(x_0)Dg(f(x_0)). But this doesn't even make sense if the matrices aren't square.

For example its like saying AB = I = BA implies A^-1 = B even though A and B aren't square matrices. I just don't see how this question is supposed to be done.

The definition that I have seen for invertible matrices require the matrices themselves to be square. So the statement "AB = I = BA implies A^-1 = B" only makes sense when A and B are square. In the case of my question Df(x_0) is not square (or at least I can't assume it is). So I don't know what to do.

To clarify my question. It consist of two parts.

a) The question in my original post,
b) Conclude that n = m. (a result which I can't really assume in part 'a').
 
  • #8
0rthodontist
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Oh, my bad...

From what you were saying before you have
f o g(x) = x
g o f(x) = x
Df(g(x))g(x) = I
Dg(f(x))f(x) = I
with I the same in both because they are both Dx and x has the same dimensions in both
So you really can conclude, as you said
Df(g(x))Dg(x) = Dg(f(x))Df(x)
Therefore DfDg and DgDf have the same dimensions, and n = m, and the inverse of Df(x) is Dg(f(x)).
 
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  • #9
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Hmm but f o g: V -> V where x is not in V so (f o g)(x) doesn't make sense. I can't assume that n = m for part (a) (the question in my original post). But then how can I possibly show that Df(x_0) is invertible?
 
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  • #10
0rthodontist
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O.K. Well, I have screwed this up twice in one thread, let's see if this is another strike.

Df(g(x))Dg(x) = Im
Dg(f(x))Df(x) = In

Now if n > m then dim(ker(Df(x))) is greater than 0 for every x, so (Df(x))v = 0 for some nonzero v, contradicting the second equation.
If m > n then dim(ker(Dg(x))) is greater than 0 for every x, so (Dg(x))v = 0 for some nonzero v, contradicting the first equation.
So m = n.
 
  • #11
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Thanks for your response, it made things more clear. I got down to the two equations involving I_m and I_n as you did. But I just said that if n is not equal to m then one of the equations doesn't make sense. Your argument is much stronger. I'm weak on the linear algebra (can't remember quite a lot of stuff from last year) so I'll need to go over your working. Thanks again.
 

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