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This is basically an inverse Eigenvector problem:

Given a vector $\vec x\in\mathbb R^n$ ($\mathbb C^n$), what is the highest rank $m$ a matrix $A\in\mathbb R^{m\times n}$ ($\mathbb C^{m\times n}$) can have such that $\vec x$ is a zero-Eigenvector (i.e. $A\vec x = \vec 0$)?


It is clear that each of the $m$ rows of $A$ must be a vector $\vec y^T$ orthogonal to $\vec x$, and as noted elsewhere, given a skew-symmetric matrix $S$ the vector $\vec x^T S$ is orthogonal to $\vec x$ since $$\vec x^T S\vec x = -\vec x^T S^T\vec x = -(\vec x^T S\vec x)^T = -\vec x^T S\vec x = 0.$$

Assuming the converse to be true, since there are $n(n-1)/2$ parameters for a skew-symmetric matrix, for $n\geqslant 4$ one would actually obtain a too large $n\times n(n-1)/2$ matrix $A$ to start with, the rank of which obviously cannot exceed $n$ (it is interesting to note that for $n=3$, $A$ itself can also be chosen skew-symmetric, $A=\begin{pmatrix}0&z&-y\\-z&0&x\\y&-x&0\end{pmatrix}$, meaning its rank must be 2 then (or zero, but that implied $\vec x=\vec 0$); whether this skew-symmetry of $A$ is also the case for $n>3$ I don't know). However I don't know how to determine the actual rank from this or whether there is a better way to determine it...

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up vote 3 down vote accepted

By a change of basis, we can always assume that $x=(1,0,0,\ldots,0)^\top$. So, $Ax=0$ if and only if the first column of $A$ is zero. Hence the maximum possible rank of $A$ is $\min(m,n-1)$.

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Wow, that was much simpler than I thought... Thanks! That also means my level set question is answered by "For non-critical points, the level set is always $n-1$ dimensional" –  Tobias Kienzler Oct 21 '13 at 12:35
    
@TobiasKienzler Don't jump to the conclusion so quickly. In this question (or strictly speaking, in the question inside the above grey box), we have no restrictions on $A$ except that $Ax=0$. In this case, the highest possible rank of $A$ is $\min(m,n-1)$. However, when extra conditions are imposed on $A$ (e.g. if $A$ is required to be a cross-product matrix, as you illustrated in the question's discussion), this upper bound $\min(m,n-1)$ may not be attainable. –  user1551 Oct 21 '13 at 12:56
    
Hm, in that other question the restriction is that $A$ is the Jacobian of a $\mathbb R^m\to\mathbb R^n$ function where $m\leqslant n$ is chosen as small as possible without reducing the rank beneath the maximum. Doesn't that imply $n-1$ (unless the respective vector were zero of course)? –  Tobias Kienzler Oct 21 '13 at 13:06
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@TobiasKienzler I don't know. The question and all the answers there are so long that I lack the concentration to read them right now ;-D . Anyway, have fun. –  user1551 Oct 21 '13 at 13:34
    
:D It's a separate discussion anyway, thanks for this one though! –  Tobias Kienzler Oct 21 '13 at 13:36

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