Inverse function theorem over matrices

Click For Summary
SUMMARY

The discussion centers on the application of the Inverse Function Theorem to the function f: M_{n×n} → M_{n×n}, defined as f(X) = X^2. Participants explore the validity of the theorem at the identity matrix and the calculation of the Jacobian for this function. The Jacobian is initially proposed as J(X): ℝ^{n^2} → ℝ^{n^2}: H → 2XH, but it is concluded that this formulation does not satisfy the necessary limit condition for the Jacobian. The discussion also touches on proving the existence of a unique matrix X such that X^2 = Y when Y is near the identity matrix.

PREREQUISITES
  • Understanding of the Inverse Function Theorem in multivariable calculus
  • Familiarity with Jacobian matrices and their computation
  • Knowledge of matrix operations and properties
  • Basic concepts of linear algebra, particularly regarding matrix spaces
NEXT STEPS
  • Study the derivation and properties of Jacobians for matrix functions
  • Research the implications of the Inverse Function Theorem in higher dimensions
  • Explore the uniqueness of solutions in matrix equations, particularly X^2 = Y
  • Examine the relationship between matrix multiplication and vector spaces
USEFUL FOR

Mathematicians, students of advanced calculus, and anyone interested in the application of the Inverse Function Theorem to matrix functions and their properties.

brunob
Messages
15
Reaction score
0

Homework Statement


I have a function f:M_{n×n} \to M_{n×n} / f(X) = X^2.

The questions
Is valid the inverse function theorem for the identity matrix? It talks about the Jacobian at the identity, but I have no idea how get a Jacobian of that function. Can I see the matrices as vectors and redefine the function as f:R^{n^2} \to R^{n^2} / f(x) = x^2 using a new dot product that represents the matrix multiplication?

Also, how can I prove that if a matrix Y[/itex] is near to the identity then \exists ! X / X^2 = Y ?<br /> <br /> Thanks!
 
Physics news on Phys.org
brunob said:

Homework Statement


I have a function f:M_{n×n} \to M_{n×n} / f(X) = X^2.

I don't know why you use / here. It's better to just write "such that". (sorry, nitpicky)

The questions
Is valid the inverse function theorem for the identity matrix? It talks about the Jacobian at the identity, but I have no idea how get a Jacobian of that function. Can I see the matrices as vectors and redefine the function as f:R^{n^2} \to R^{n^2} / f(x) = x^2 using a new dot product that represents the matrix multiplication?

The dot product has nothing to do with things here. You need to find the Jacobian somehow of squaring matrices. Now, if we look to the one-dimensional case, we have the map

f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2

The Jacobian of that is just the derivative
J(x):\mathbb{R}\rightarrow \mathbb{R}:h\rightarrow 2xh

This suggests that in the general case we have a map

f:\mathbb{R}^{n^2}\rightarrow \mathbb{R}^{n^2}:X\rightarrow X^2

and that the Jacobian would be
J(X):\mathbb{R}^{n^2}\rightarrow \mathbb{R}^{n^2}<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\rightarrow 2XH
Of course, you must check this first. Given a multivariable map ##f## and a matrix ##A(x)## at every point, you have that this is the Jacobian if

\lim_{h\rightarrow 0} \frac{f(x+h) - f(x) - A(x)h}{\|h\|} = 0

So you must check that this holds for ##A(X) = 2X##. Thus

\lim_{H\rightarrow 0} \frac{(X+H)^2 -X^2 - 2XH}{\|H\|} = 0

If you try to caluclate this, you will find that this is not true. So ##A(X) = 2X## is not the right Jacobian. Can you make a small modification that will provide the right Jacobian?

Also, how can I prove that if a matrix Y[/itex] is near to the identity then \exists ! X / X^2 = Y ?<br />
<br /> <br /> What do you get after applying the Inverse Function Theorem on ##f(X) = X^2##?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
Replies
8
Views
2K