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Inverse function theorem over matrices
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[QUOTE="micromass, post: 4767688, member: 205308"] I don't know why you use / here. It's better to just write "such that". (sorry, nitpicky) 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 [tex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2[/tex] The Jacobian of that is just the derivative [tex]J(x):\mathbb{R}\rightarrow \mathbb{R}:h\rightarrow 2xh[/tex] This suggests that in the general case we have a map [tex]f:\mathbb{R}^{n^2}\rightarrow \mathbb{R}^{n^2}:X\rightarrow X^2[/tex] and that the Jacobian would be [tex]J(X):\mathbb{R}^{n^2}\rightarrow \mathbb{R}^{n^2}:H\rightarrow 2XH[/tex] 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 [tex]\lim_{h\rightarrow 0} \frac{f(x+h) - f(x) - A(x)h}{\|h\|} = 0[/tex] So you must check that this holds for ##A(X) = 2X##. Thus [tex]\lim_{H\rightarrow 0} \frac{(X+H)^2 -X^2 - 2XH}{\|H\|} = 0[/tex] If you try to caluclate this, you will find that this is [b]not[/b] true. So ##A(X) = 2X## is not the right Jacobian. Can you make a small modification that will provide the right Jacobian? What do you get after applying the Inverse Function Theorem on ##f(X) = X^2##? [/QUOTE]
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