Showing that a C-infinity function exists

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Apr 11th 2011, 09:18 AM
Newtonian27

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.
Apr 11th 2011, 10:47 AM
Opalg

Quote:

Originally Posted by Newtonian27

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.
Apr 11th 2011, 11:32 AM
Newtonian27

Quote:

Originally Posted by Opalg

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 $Mat(n\times n, R)$ ?
Apr 11th 2011, 02:14 PM
Drexel28

Quote:

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 $C^{\infty}$ -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 $C^\infty$ ?