Are inverse functions in R^n an identity matrix?

Click For Summary
SUMMARY

The discussion centers on the concept of inverse functions and identity matrices in the context of functions mapping from R^n to R^m. The user initially struggles with the notation and understanding that the identity matrix I represents an identity on the image of the function f, while the composition of f and its inverse f^-1 yields an identity on R^n. The clarification provided emphasizes that matrices represent linear transformations, and the output of these transformations is a vector, not a matrix. This distinction resolves the user's confusion regarding the relationship between functions and their inverses.

PREREQUISITES
  • Understanding of linear transformations in R^n and R^m
  • Familiarity with function composition and inverse functions
  • Knowledge of matrix representation of linear transformations
  • Basic concepts of identity matrices and their properties
NEXT STEPS
  • Study the properties of inverse functions in linear algebra
  • Learn about matrix representations of linear transformations
  • Explore the concept of identity matrices in different dimensions
  • Research function composition and its implications in R^n and R^m
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as anyone seeking to deepen their understanding of inverse functions and their applications in vector spaces.

GregA
Messages
210
Reaction score
0
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)
 
Last edited:
Physics news on Phys.org
f(f^-1(x)) = x, not simply I.

You might be thinking of Ix which is x.
 
Your notation for I is incomplete. Your I is an identity on the image of f (a subset of Rm), while the alternative J = f^{-1}\circ f 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.
 
Last edited:
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!
 

Similar threads

Undergrad On inverse function theorem in Spivak's CoM
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad Ambiguity of the term "indefinite integral"
  • · Replies 11 ·
Replies
11
Views
2K
Undergrad On (real) entire functions and the identity theorem
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Inverse Functions: x=f(y) and X=f^-1(y)
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Open interval or Closed interval in defining convex function
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Doubt about theorem in Calculus on Manifolds
  • · Replies 9 ·
Replies
9
Views
3K
Undergrad What does "the sequence of functions has limit in R" mean?
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad Second derivative of chained function
  • · Replies 14 ·
Replies
14
Views
2K
Undergrad A correct definition of sequential right continuity of a function
  • · Replies 2 ·
Replies
2
Views
2K
High School How do I invert this exponential function?
  • · Replies 3 ·
Replies
3
Views
4K