Let $\mathbf{a}\in\mathbb{R}^{4}$ and the entries of a $3$ by $4$ real matrix $A$ be the variables of the polynomial equations, $\det\left(A\right)=0$, $S_{1},S_{2}\subseteq\left\{ 1,2,3,4\right\}$ with $\leftS_{1}\right=\leftS_{2}\right=3$ , and $A\mathbf{a}=0$. Is the projectivization of the affine variety whose variables are the entries of $A$ over $\mathbb{R}$ empty for any $\mathbf{a}\in\mathbb{R}^{4}$?
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$\begingroup$ What is $a^*$? What is $X$? If it's a generic matrix, how do you projectivize it? $\endgroup$– Anton FonarevCommented Dec 18, 2013 at 0:36

$\begingroup$ What I'm reading is "the projectivization of $X$ over…" Also, what is $a^*$? Please, modify your question so that it becomes clear. $\endgroup$– Anton FonarevCommented Dec 18, 2013 at 0:44

$\begingroup$ I'm sorry, but this question is very poorly written and might still be unclear not only to me. It seems like your are asking if the exists a nonzero real skewsymmetric matrix $A$ of rank 2 that satisfies $a^*Aa$ for a given $a\in\mathbb{C}^4$. This is definitely not a research level question. Please, try solving the problem yourself or asking at math.stackexchange. $\endgroup$– Anton FonarevCommented Dec 18, 2013 at 1:08

$\begingroup$ @nataliea, the question is really unclear. You are unlikely to receive helpful answers if you do not explain your notation. $\endgroup$– Boris BukhCommented Dec 18, 2013 at 3:26

$\begingroup$ The intersection won't be empty as long as the dimension of the projective variety corresponding to skewsymmetric matrices of rank $\leq 2$ has dimension at least $1$. (You can see this using the rule $\operatorname{codim}(U \cap V) \leq \operatorname{codim}(U)+\operatorname{codim}(V)$.) On the other hand, this does not immediately imply that the intersection has $\mathbb{R}$points, which is what you wanted. (That's why I removed the comment.) $\endgroup$– R.P.Commented Dec 18, 2013 at 7:02

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No. Write $\mathbf{a}=x+iy $, with $x,y\in\mathbb{R}^4$. Since $X$ is skewsymmetric, the condition $\mathbf{a}^*X\mathbf{a}=0$ is equivalent to $y^*Xx=0$. Assuming $x\neq 0$, all skewsymmetric matrices $X\neq 0$ with $Xx=0$ are contained in your variety: since the rank of a skewsymmetric matrix is even, they are of rank 2, hence they satisfy the conditions on the minors. If $x=0$ take $Xy=0$ instead.

1$\begingroup$ Great answer. One little remark: if $x=0$ (or if $y=0$), then any rank $2$ skewsymmetric matrix will do. $\endgroup$– R.P.Commented Dec 18, 2013 at 6:21

$\begingroup$ This is a classical linear algebra result. See e.g. Lang, Algebra, ch. XV,Theorem 8.1. $\endgroup$– abxCommented Dec 18, 2013 at 9:20

$\begingroup$ Skew 4x4 matrices lead to the projective space $P^5$, the subvariety given by vanishing of determinant is therefore of dimension $4$, and the single equation $a^{\ast}Xa=0$ (which looks like two real equations, but is only one because of the skewness) gets you to a variety of dimension $3$ (except in the case when $x$ and $y$ are linearly dependent, in which case the equation is always satisfied). As a matter of language, algebraic geometers would not call this variety 'empty" even it had no real points (which it does). $\endgroup$ Commented Dec 18, 2013 at 16:56

$\begingroup$ The real points are those $X$ such that $Xx$ is orthogonal to $y$. $\endgroup$ Commented Dec 18, 2013 at 16:58

$\begingroup$ No, you cannot use this kind of argument over $\Bbb{R}$. Two nonempty hypersurfaces may very well have an empty intersection  think of 2 spheres in $\Bbb{P}^3(\Bbb{R})$. $\endgroup$– abxCommented Dec 19, 2013 at 4:57