Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

Thanks.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
Brilliant! Thank you. –  GMB Sep 22 '12 at 16:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.