Showing that a C-infinity function exists

I am running into some trouble proving this statement:

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

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

I know that I can apply the inverse-function theorem with $\displaystyle F(A)=A^{2}$. But I am unsure where to begin.

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Originally Posted by Newtonian27

I am running into some trouble proving this statement:

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

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

I know that I can apply the inverse-function theorem with $\displaystyle F(A)=A^{2}$. But I am unsure where to begin.

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One approach is to use the (binomial) power series expansion for $\displaystyle (1+x)^{1/2}$. In fact, if $\displaystyle A = I+X$, with $\displaystyle \|X\|<1$, then the series $\displaystyle (I+X)^{1/2} = I+\frac12X-\frac18X^2+\frac1{16}X^3-\ldots$ converges to give a $\displaystyle C^{\infty}$-function of X.

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Originally Posted by Opalg

One approach is to use the (binomial) power series expansion for $\displaystyle (1+x)^{1/2}$. In fact, if $\displaystyle A = I+X$, with $\displaystyle \|X\|<1$, then the series $\displaystyle (I+X)^{1/2} = I+\frac12X-\frac18X^2+\frac1{16}X^3-\ldots$ converges to give a $\displaystyle C^{\infty}$-function of X.

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That is great, so then all I would need to do is show that this converging series is in fact a $\displaystyle C^{\infty}$-function from $\displaystyle U$ into $\displaystyle Mat(n\times n, R)$?

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Originally Posted by Newtonian27

That is great, so then all I would need to do is show that this converging series is in fact a $\displaystyle C^{\infty}$-function from $\displaystyle U$ into $\displaystyle Mat(n\times n, R)$?

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Although I like Opalg's solution considering it gives an explicit formula for $\displaystyle \sqrt{A}$ I think you're suggestion is easier. Namely, consider $\displaystyle S:\text{Mat}_n\left(\mathbb{R}\right)\to\text{Mat} _n\left(\mathbb{R}\right)$ by $\displaystyle A\mapsto A^2$ one can check that $\displaystyle S(A+H)-S(A)=AH+HA+o(H)$ so by the uniqueness of the derivative (in terms of it's linear approximation) we may conclude that $\displaystyle \left(S'(A)\right)(H)=AH+HA$ and so in particular $\displaystyle \left(S'(I)\right)(H)=2H$ which is invertible so that there exists some neighborhood $\displaystyle U$ of $\displaystyle I\in\text{Mat}_n\left(\mathbb{R}\right)$ such that $\displaystyle S$ is invertible and has a $\displaystyle C^1$ inverse $\displaystyle \text{Sq}$. Can you show that it is now $\displaystyle C^\infty$?