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Let

  • $X$ be compact Hausdorff topological space,
  • $C(X)$ denote the algebra of complex-valued continuous functions on $X$,
  • $b\in \mathbb{C}^m$,
  • $\mathbf{A}\in C(X)^{m\times n}$,
  • for all $x\in X$, $b\in \textrm{range}(\mathbf{A}(x))$.

Question: Does there exist an $\mathbf{x}\in C(X)^{n\times 1}$ such that for all $x\in X$, $\mathbf{A}(x) \mathbf{x}(x)=b$?

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  • $\begingroup$ I am not sure what you are after. Do you want to show that the solution depends continuously on the matrix? Are you especially interested in the case where $A$ gets singular? $\endgroup$
    – Dirk
    Commented Dec 4, 2014 at 16:08

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Let $$A(x)=\pmatrix{x&1\cr 0&x},$$ for $x\in [-1,1]$ and $$b=\pmatrix{1\cr 0}.$$ Then $b$ is in the range of $A(x)$ for every $x$, but for $x\neq 0$, we have $$A^{-1}(x)b=\pmatrix{1/x\cr 0},$$ so there is no continuous solution.

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  • $\begingroup$ Many thanks for the illuminating example. $\endgroup$
    – Amol Sasane
    Commented Dec 4, 2014 at 18:16

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