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For an arbitrary square matrix $A$, I'm looking for the set of vectors $x$ that $A$ maps orthogonal to $x$ (in other words, $x^T Ax = 0$). How can I go about solving this problem? What would I expect the solution set to look like (a hyperplane, maybe)?


Editing in some more info:

$x = 0$ is a solution, but it's not one I care about much. In general there aren't necessarily more solutions ($A = I$, for example), but I sort of lied to you: the form of $A$ is such that I expect at least one other.

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up vote 1 down vote accepted

$$x^\mathrm{T}Ax=0 \implies x^\mathrm{T}A^\mathrm{T}x=0 \implies x^\mathrm{T}\left(A+A^\mathrm{T}\right)x=0.$$ Because $A+A^\mathrm{T}$ is always diagonalizable, expressing $x$ as a linear combination of the eigenvectors of $A+A^\mathrm{T}$ shows that $x$ must be in the nullspace of $A+A^\mathrm{T}$. It is straightforward to check that all $x\in\mathrm{null}\left(A+A^\mathrm{T}\right)$ satisfy the desired condition.

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Brilliant! Thank you. – GMB Sep 22 '12 at 16:19

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