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Let the space of all matrices over $\mathbb R$ of size $ n^2 $ , with the natural metric of $\mathbb R^{n^2 }$.

Prove that there exist a neighborhood of the identity matrix, such that all the matrices in that ball have a square root, i.e $$ \exists \varepsilon > 0\,:\forall X \in B\left( {I,\varepsilon } \right)\,\,\exists \,Y:\,Y^2 = X . $$

Hint: Consider the function $ F(X) = X^2 $ and use the Inverse Function Theorem

Help with this problem !! )=

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Let $n=1$. Then not all matrices (= numbers) in the neighborhood of $0$ have a square root. –  Christian Blatter Nov 18 '11 at 15:40
    
Wouldn't the following be a counterexample? Take any matrix with no square roots, and scale it (by a sufficiently small $\delta > 0$) so that the resulting matrix falls inside the ball. –  Srivatsan Nov 18 '11 at 15:42
    
Sorry!! The center of the ball is not the zero matrix, it´s the identity I –  August Nov 18 '11 at 19:41
    
Compute the derivative of $F$ at the identity and show that it is invertible... Since $F(I) = I$ the inverse function theorem then does the rest. –  t.b. Nov 18 '11 at 19:44
    
How can I don´t know how to calculate the derivate of a matrix D: –  August Nov 18 '11 at 19:47
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1 Answer

You want to solve the equation $(I+X)^2=I+Y$ for the $(n\times n)$-matrix $X$ when $Y$ is a given matrix near $0$. This amounts to

$$Z(X):=2X+X^2\ =\ Y\ ,\qquad (*)$$

an equation in the $n^2$ unknowns $X_{ik}$. When $Y=0$ then $X=0$ is a solution of this equation. According to the assumptions of the implicit function theorem we have to check whether the $(n^2\times n^2)$-matrix

$$J:=\Bigl({\partial(Z_{11},Z_{12},\ldots,Z_{nn})\over \partial(X_{11},X_{12},\ldots,X_{nn})}\Bigl)_{X=0}$$

is regular. As $Z_{ik}=2 X_{ik} +$ terms of degree $2$ in the $X_{ik}$, the matrix $J$ is $2$ times the $n^2\times n^2$ identity matrix; so it is certainly regular. It follows that the equation $(*)$ has a solution $X=F(Y)$ such that $F(0)=0$ and $F(\cdot)$ is a differentiable matrix valued function of (the matrix elements of) $Y$.

By the way, the square root $I+X$ of $I+Y$ for $\|Y\|<1$ is given by the binomial series

$$\sum_{k=0}^\infty\ {1/2 \choose k}\ Y^k\ .$$

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