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 />
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What do you get after applying the Inverse Function Theorem on ##f(X) = X^2##?
