Inverse functions in R^n

  • Thread starter GregA
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  • #1
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Main Question or Discussion Point

Sorry if this is the wrong place for my question, I'm having difficulty on a conceptual level getting my head round inverse functions and compositions of functions in R^n. I'm failing to understand my lecture notes as a result.

Suppose I have some function with domain R^n which maps to R^m given by f(x) = f[x1,x2,...,xn]T=[f1(x),f2(x),...,fm(x)]T it seems reasonable that you'd want to define f-1(x) such that f o f-1(x) = I, but is I an identity matrix?. I ask this because f(x) is a vector in R^m, I'd expect some other function g(f(x)) would also be a vector (as opposed to a matrix).
I'm clearly missing something. Can anyone throw me any hints or direct me to some online material that would help me (I have a book on the way in the post)
 
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Answers and Replies

  • #2
f(f^-1(x)) = x, not simply I.

You might be thinking of Ix which is x.
 
  • #3
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Your notation for I is incomplete. Your I is an identity on the image of f (a subset of Rm), while the alternative [itex]J = f^{-1}\circ f[/itex] is an identity on Rn.
Matrices represent linear transformations, not the result of a linear transformation. Ie., if g is a linear transformation from Rm into Rn, then g(x) is a vector in Rn while g can be represented by an nxm matrix.
 
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  • #4
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aha!...cheers guys! You're right, I was considering f acting on a vector instead of considering that f by itself is an mxn matrix :redface: :smile:

Things make sense again!
 

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