Sorry for the confusion from earlier. I tried to fix the thread. The old version can be found below.

For $6$-dimensional subspaces $V$ of the space $\mathbb{R}^{3\times 3}$ of real three-times-three matrices, consider its intersection with $O(3)$.

  • Is it generically true that $V \cap O(3)$ is non-empty (i.e. the set of such $V$ is an open subset of $\mathrm{Gr}(3, 6)$)? Can one say how many points the intersection generically includes?
  • Is it possible to characterize the $6$-dimensional subspaces $V$ such that $V \cap O(3) = \emptyset$? (For example, if $V$ consists only of singular matrices, then $V \cap O(3) = \emptyset$; conversely, does every $V$ with $V \cap O(3) = \emptyset$ consist only of singular matrices?)

Old thread Let $V$ be a $6$-dimensional subspace of the space $\mathbb{R}^{3\times 3}$ of real three-times-three matrices. Is the intersection of $V$ with $O(3) \subset \mathbb{R}^{3 \times 3}$ always non-empty? If not, how can we characterize those subspaces $V$ such that $V \cap O(3) = \emptyset$? \Edit: Anton made the trivial comment below that one can take matrices with first (or some other) column equal to zero. However, how can we characterize such subspaces?

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    $\begingroup$ Take the subspace of matrices with first column equal to zero. $\endgroup$ – user1688 Jan 26 '16 at 10:36
  • $\begingroup$ You should say what you mean by 'characterize'. The set of 6-dimensional subspaces of $\mathbb{R}^{3\times 3}$ is a compact manifold $\mathrm{Gr}(3,6)$ of dimension $18$, and it's clear (because of the examples already given) that the set of 6-dimensional spaces that don't meet $\mathrm{O}(3)$ is a (nonempty) open set $U\subset \mathrm{Gr}(3,6)$, so you are asking for a 'characterization' of an open subset in $\mathrm{Gr}(3,6)$. It can't be by equations, but maybe by some kind of inequality that defines the boundary of the set $U$ in $\mathrm{Gr}(3,6)$. Is that the sort of answer you seek? $\endgroup$ – Robert Bryant Jan 26 '16 at 12:30
  • $\begingroup$ This would be the kind of answer I would be interested in. Is $U$ a dense subset? Probably one could characterize its complement by equations? I edited the post above. $\endgroup$ – Matthias Ludewig Jan 26 '16 at 12:54
  • $\begingroup$ While I can't correct it now, in my comment above, wherever I wrote $\mathrm{Gr}(3,6)$, I should have written $\mathrm{Gr}(6,9)$, i.e., the Grassmannian of $6$-dimensional subspaces of a $9$-dimensional subspace. $\endgroup$ – Robert Bryant Jan 26 '16 at 17:51

In general the answer is no. For example, the matrices whose first column is a column of zeroes form a $6-$dimensional subspace not containing any ortogonal matrix, because it does not contain any invertible matrix.

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    $\begingroup$ In addition, since $O(3)$ is compact and does not contain zero, any subspace close enough to this given subspace still does not contain any orthogonal matrix. So the intersection is not generically non-empty. $\endgroup$ – YCor Jan 26 '16 at 17:25

The answer to your first question is 'yes, generically, the intersection is finite and transverse'. More precisely, the set $U$ of 6-dimensional subspaces that intersect $\mathrm{SO}(3)$ transversely (and hence in a finite number of points), is a dense open set in $\mathrm{Gr}(6,9)$. (However, $U$ is not connected; the number of intersection points can be (and are) different for different components of $U$.) For example, the $6$-dimensional subspace $V_+$ consisting of upper triangular matrices intersects $\mathrm{SO}(3)$ transversely in the $4$ points that consist of the diagonal elements of $\mathrm{SO}(3)$. Thus every $6$-dimensional subspace sufficiently near $V_+$ also intersects $\mathrm{SO}(3)$ in $4$ points and transversely.

As for your parenthetical question, the answer to that is 'no, there exist $6$-dimensional subspaces $W$ that do not meet $\mathrm{SO}(3)$ and for which the generic element in $W$ is invertible'. For example, the generic $W$ that is sufficiently near the $W_0$ consisting of the matrices with zero first column will have its generic element be invertible.

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