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Does there locally exist a square root of the matrix \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} in Mat(2,2)? I'm not sure how to approach this problem. We want to use the Inverse function theorem for the squaring map near \begin{pmatrix} 1 & 2 \\ -2 & -1 \end{pmatrix} whose derivative is $AH+HA$. We want this to be invertible. It seems since $H$ must be a small matrix that this can never be invertible. Am I going about it the right way? Can someone help me finish?

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up vote 1 down vote accepted

No. What you should concern about is not the invertibility/non-invertibility of the matrix $AH+HA$, but the invertibility/non-invertibility of the linear map $H\mapsto AH+HA$. In other words, the square map $A\mapsto A^2$ has a local inverse at $A=\begin{pmatrix}1&2\\-2&1\end{pmatrix}$ if the equation $AH+HA=0$ has only the trivial solution $H=0$.

Edit: Note that what the inverse function theorem gives is a sufficient condition rather than a necessary condition for the existence of a local inverse. So, if the derivative mapping $H\mapsto AH+HA$ turns out to be singular, all you can say is merely that the inverse function theorem doesn't apply. To prove that the square map has no local inverse at $A$, you must use other methods. For example, you may find two sequences of matrices $\{B_n\}$ and $\{C_n\}$ such that $\lim_{n\to\infty}B_n=\lim_{n\to\infty}C_n=A$, but $B_n\not=C_n$ and $B_n^2=C_n^2$ for every $n$. Alternatively, you may find a sequence of matrices $\{S_n\}$ such that $\lim_{n\to\infty}S_n=A^2$, but $S_n$ has no (real) square root. To this end you may consider $S_n=\begin{pmatrix}-3&1/n\\0&-3\end{pmatrix}$.

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