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Prove there exist non-empty open sets U and V of $n \times n$ matrices over $\mathbb{R}$ such that for every matrix $A \in U$ there exists exactly one matrix $B \in V$ such that $B^4 = A$.

I've tried to approach this problem in several ways, using characteristic polynomials, Jordan Canonical Form and facts from calculus but failed to get anything useful.

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  • $\begingroup$ I think you can take $U$ and $V$ both to be the set of matrices with $n$ distinct positive eigenvalues. $\endgroup$
    – Theo Bendit
    Jun 16, 2019 at 8:55

2 Answers 2

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Hint: take $U$ to be a sufficiently small neighbourhood of the identity. Recall the inverse function theorem.

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  • $\begingroup$ i think the implicit function theorem might be more directly applicable, no? I know both are equivalent so it doesn't technically matter which we use... but it seems like implicit function theorem's hypothesis are very directly satisfied. $\endgroup$
    – peek-a-boo
    Jun 16, 2019 at 10:25
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    $\begingroup$ @peek-a-boo Depends on how you think about it. I fixed the identity and use the inverse function theorem because the derivative of matrix fourth power at identity is invertible, so is a local diffeomorphism and be done with it. You take two copies and compute the derivative, etc. $\endgroup$
    – user10354138
    Jun 16, 2019 at 10:33
  • $\begingroup$ perhaps, it's just that with implicit function theorem, the proof was immediate in my head. but regardless, this is a very good hint! usually people learn (multivariable) calculus and linear algebra as separate subjects so it might not even occur to them that uniting both subjects gives an elegant answer $\endgroup$
    – peek-a-boo
    Jun 16, 2019 at 10:39
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Using the implicit function theorem, we can obtain a general result.

Let $B_0\in M_n(\mathbb{R})$, $spectrum(B_0)=(\lambda_i)$, ${B_0}^4=A_0$, $f:X\rightarrow X^4$.

$\textbf{Proposition}$. If, for every $(i,j)$, $\lambda_i^3+\lambda_j^3+\lambda_i\lambda_j^2+\lambda_i^2\lambda_j\not= 0$, then there are open neighborhoods $V,U$ of $B_0,A_0$ s.t. $f$ is a diffeomorphism from $V$ onto $U$.

$\textbf{Proof}$. It suffices to show that the derivative $Df_{B_0}:H\in M_n\mapsto HB_0^3+B_0HB_0^2+B_0^2HB_0+B_0^3H$ is one to one.

If we stack row by row the matrices into vectors, then

$Df_{B_0}=I\otimes {B_0^3}^T+B_0\otimes {B_0^2}^T+B_0^2\otimes B_0^T+B_0^3\otimes I$. cf.

https://en.wikipedia.org/wiki/Kronecker_product

Since the $4$ above linear applications pairwise commute, $spectrum(Df_{B_0})=(\lambda_i^3+\lambda_j^3+\lambda_i\lambda_j^2+\lambda_i^2\lambda_j)_{i,j}$ and we are done.

Note that, when $B_0=I$, all considered eigenvalues are equal to $4$.

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