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Show that there exists a neighborhood $U$ of $I$ in $\mbox{Mat}(2 \times 2)$ such that for all $B\in U$ there are at least two solutions in $\mbox{Mat}(2 \times 2)$ for the equation $A^2 = B$.

I have this question for homework and I would be happy to get a hint. I think it uses the implicit function theorem.

Rodrigo de Azevedo
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  • You're actually looking for the inverse theorem (which is closely related to the implicit function theorem). – Mark Saving Dec 27 '21 at 21:22
  • Hint: prove that $f(A)=A^2$ has regular Jacobian at the identity matrix. For this you can use a curve of the form $\gamma(t) = I + tC$. – Laz Dec 27 '21 at 23:51
  • When $B$ is close to $I_2$, it is nonsingular. In this case, there are explicit formulae for its square roots. See some of the answers to https://math.stackexchange.com/q/59384 – user1551 Dec 28 '21 at 07:53

2 Answers2

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If $A^2 = B,$ then $(-A)^2 = B.$ This only fails when $A=-A,$ so $A=0,$ but $0$ is "far" from $I.$

Now it comes down to showing that there is such an $A.$ The soft solution is that the map $x\to x^2$ is an open map in a neighborhood $V$ of identity, so its image is onto a neighborhood $U$ of identity.

The "hard" solution is to expand $(I+x)^{\frac12}$ in a power series (by the binomial theorem) and note that the series converges for $\rho(x)$ small. The "even harder" solution is to use the Jordan canonical form to note that $B$ is conjugate to either a diagonal matrix (in which case the solution is obvious) or to an upper triangular matrix, in which case a solution is easy to get.

Igor Rivin
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In the (sufficiently small) neighborhood of $I$ matrices have two (not necessarily distinct) non-zero eigenvalues. It means that any such matrix, say $B$, has (so called Jordan Form) representation as $B = C*D*C^{-1}$ where $D$ is diagonal matrix. Now, $D$ has a root , it is just diagonal matrix with roots of $D$, let's denote it by $D^{\frac{1}{2}}$. Then if $A=C*D^{\frac{1}{2}}*C^{-1}$ one checks that $A^2=B$. To get second solution change signs for the "root" matrix diagonals.

Salcio
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