Let $X$ be a compact Hausdorff topological space, and $C(X)$ denote the ring of complex-valued, continuous functions on $X$. Let $A$ be a matrix with entries from $C(X)$ of size $m\times n$ and $b\in \mathbb{C}^{m\times 1}$. Suppose that for each $x\in X$, the equation $A(x) v=b$ has a solution $v\in \mathbb{C}^n$. Then does there exists an $V\in C(X)^{n\times 1}$ such that $AV=b$?
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$\begingroup$ I do not completely understand your notation. Does this have something to do with selection theorems? See for example en.wikipedia.org/wiki/Michael_selection_theorem $\endgroup$– András BátkaiDec 15, 2015 at 10:32
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$\begingroup$ Selection theory might indeed be relevant, although know nothing about its machinery---e.g. how one would go about checking the "lowerhemicontinuity" of a multivalued map---do you know of a good reference for this? About the notation: sorry---I meant if $R$ is a ring, then $R^{m\times n}$ denotes the set of matrices with $m$ rows and $n$ columns and having entries from $R$, and my notation $A(x)$ means the matrix obtained by evaluating each of the entries of $A\in C(X)^{m\times n}$ at $x\in X$. Does this help in clarification? $\endgroup$– PernillaDec 15, 2015 at 10:42
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$\begingroup$ A quick remark: the answer is "yes" by Cramer's rule if $A(x)$ is nonsingular for all $x$. $\endgroup$– Paul SiegelDec 15, 2015 at 12:23
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No. Take $n=m=2$, $A(x)=\pmatrix{1&0\\x&x^2}$, $x\in \mathbb{R}$, $b=\pmatrix{1\\0}$. For $x\ne 0$ the only solution of $AV=b$ is $V(x)=\pmatrix{1\\-1/x}$, it has discontinuity at 0 for any choice of $V(0)$.
answered Dec 15, 2015 at 12:32
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$\begingroup$ Thank you. From your example, I see that some additional condition is needed. What if: $\exists \delta>0$ such that $\forall x\in X$ and $\forall y\in \mathbb{C}^{n\times 1}$, $\|(A(x))^* y\|\geq \delta \|Py\|$? Here all norms are the usual Euclidean norms on $\mathbb{C}^{\cdot\times 1}$ and $P$ denotes the orthogonal projection onto span($b$). Your example does not satisfy this uniformity condition. $\endgroup$– PernillaDec 15, 2015 at 13:21
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$\begingroup$ It is possible. On first glance it looks similar to Michael's selection theorem condition. $\endgroup$ Dec 15, 2015 at 13:42
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$\begingroup$ Okay, thanks for this. Will try to check this. $\endgroup$– PernillaDec 15, 2015 at 13:49
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$\begingroup$ You can always obtain the solution by applying the Moore-Penrose inverse of $A(x)$ to $b$. Hence, the answer will be yes in any situation where this depends continuously on $x$. The above example shows that this is not always the case but it is so if the $A(x)$ are all either surjective or injective since one then has simple explicit formulae for it. $\endgroup$– dalryDec 15, 2015 at 16:57