Let $P\in R^{n\times n}$ be an orthogonal matrix. I want to ask whether or not there exists some vector $x\in R^n$ containing no zero entries such that $Px$ also contains no zero entries.
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1$\begingroup$ yes : the set of vectors with no zero entry is an open in Zariski topology, so is its images by P, and it preimage by P, so the intersection of these three sets is open, and since $\mathbf R$ is infinite, is non-empty. $\endgroup$– few_repsCommented Aug 29, 2016 at 2:03
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4$\begingroup$ True but more advanced than necessary, and also hiding the key point that $P$, being orthogonal, is invertible. Hence the set of $x$ such that $x$ or $Px$ does have at least one zero entry is the union of $2n$ subspaces of positive codimension. This can't be all of ${\bf R}^n$ because $\bf R$ is infinite. (For example, any hyperplane intersects the "rational normal curve" $\{(1,t,t^2,\ldots,t^{n-1})\}$ in fewer than $n$ points.) $\endgroup$– Noam D. ElkiesCommented Aug 29, 2016 at 2:08
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1 Answer
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Instead of Zariski topology consider usual topology. Every open ball will contain a vector with all entries non-zero. (I leave this to you to verify). As $P$ is invertible it is a homeomorphism. Image of open ball will contain open ball. QED
