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1) Why is true that there is an open ball in $M_n$ centered at the identity matrix and a continuous function $f$ defined on this open ball s.t. $f(M)^2=M$ for $M$ in the ball?

2) Extending the question a bit: Would I be right in thinking that we cannot do the same for any arbitrary matrix $M\in M_n$ because considering $n=1$ this clearly fails for the negative numbers?

3) What about higher powers? The case $n=1$ clearly works for all odd powers. But not for the even ones. But perhaps there is a catch for the odd powers too?

Thanks.

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    @DavideGiraudo: thanks!2011-11-23

1 Answers 1

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Let $p\geq 2$. We can write $(1-x)^{\frac 1p}=\sum_{n=0}^{+\infty}a_nx^n$ as a power series of radius of convergence $1$. Put $f(M):=\sum_{n\geq 1}^{+\infty}a_n(I-M)^n$. This series is normally convergent on $B\left(I,\frac 12\right)$ and using Cauchy product we can see that $f(M)^p= I-(I-M)=M$ for each $M\in B\left(I,\frac 12\right)$.
As @Joel showed, it doesn't work if we substitute $I$ by an other point.