4
$\begingroup$

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.

  • 1
    You have a formula of $\sqrt{1-x}$ written as a power series $\sum_n a_nx^n$. Consider this formal power series for matrices $f(M)=\sum_n a_nM^n$ and put $g(M)=f(M-I)$. For the second question, indeed, you will have problems at $0$ for example, already for $n=1$.2011-11-22
  • 0
    $$A = \left(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\right)$$ is not a square, cube or any $n^{\text{th}}$ power for $n \ge 2$. Indeed, assume there exists $n \ge 2$ and $B$ such that $B^n = A$. We have $B^{2n} = 0$ so $B$ is nilpotent. But because it's a $2\times2$ matrix, we have $B^2 = 0$. So $A = B^2 B^{n-2} = 0$, which is a clear contradiction.2011-11-23
  • 0
    @DavideGiraudo: Thanks, but how do I show continuity and the open ball?2011-11-23
  • 0
    @JoelCohen: Thanks for the example.2011-11-23
  • 0
    Let $2r$ the radius of convergence of $\sum_n a_nx^n$. Then the series is normally convergent on the set $|x|$g\colon M\mapsto f(I-M)$ is continuous on the ball $B(I,r)$, and the Cauchy product shows that we have $g(M)^2=I-(I-M)=M$ on this ball. For question 3, you want to do again the question 1 and 2 substituting $2$ by an integer $p$, right? – 2011-11-23
  • 0
    @DavideGiraudo: thanks!2011-11-23

1 Answers 1

1

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.