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Math Help - Showing that a C-infinity function exists

  1. #1
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    Showing that a C-infinity function exists

    I am running into some trouble proving this statement:

    There is a C^{\infty} square-root function in a neighborhood of the identity matrix I in Mat(n\times n,R).

    Basically I need to show that there is a C^{\infty} function f from a neighborhood U of I into Mat(n\times n,R), with f(I)= I, such that f(A)^{2}= A for all A\in U.

    I know that I can apply the inverse-function theorem with F(A)=A^{2}. But I am unsure where to begin.
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  2. #2
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    Quote Originally Posted by Newtonian27 View Post
    I am running into some trouble proving this statement:

    There is a C^{\infty} square-root function in a neighborhood of the identity matrix I in Mat(n\times n,R).

    Basically I need to show that there is a C^{\infty} function f from a neighborhood U of I into Mat(n\times n,R), with f(I)= I, such that f(A)^{2}= A for all A\in U.

    I know that I can apply the inverse-function theorem with F(A)=A^{2}. But I am unsure where to begin.
    One approach is to use the (binomial) power series expansion for (1+x)^{1/2}. In fact, if A = I+X, with \|X\|<1, then the series (I+X)^{1/2} = I+\frac12X-\frac18X^2+\frac1{16}X^3-\ldots converges to give a C^{\infty}-function of X.
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    Quote Originally Posted by Opalg View Post
    One approach is to use the (binomial) power series expansion for (1+x)^{1/2}. In fact, if A = I+X, with \|X\|<1, then the series (I+X)^{1/2} = I+\frac12X-\frac18X^2+\frac1{16}X^3-\ldots converges to give a C^{\infty}-function of X.

    That is great, so then all I would need to do is show that this converging series is in fact a C^{\infty}-function from U into Mat(n\times n, R)?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Newtonian27 View Post
    That is great, so then all I would need to do is show that this converging series is in fact a C^{\infty}-function from U into Mat(n\times n, R)?
    Although I like Opalg's solution considering it gives an explicit formula for \sqrt{A} I think you're suggestion is easier. Namely, consider S:\text{Mat}_n\left(\mathbb{R}\right)\to\text{Mat}  _n\left(\mathbb{R}\right) by A\mapsto A^2 one can check that S(A+H)-S(A)=AH+HA+o(H) so by the uniqueness of the derivative (in terms of it's linear approximation) we may conclude that \left(S'(A)\right)(H)=AH+HA and so in particular \left(S'(I)\right)(H)=2H which is invertible so that there exists some neighborhood U of I\in\text{Mat}_n\left(\mathbb{R}\right) such that S is invertible and has a C^1 inverse \text{Sq}. Can you show that it is now C^\infty?
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