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Does $u v^T + v u^T$ have exactly one positive and one negative eigenvalue when $u \not \propto v$?

$u$ and $v$ are column vectors in $\mathbb{R}^n$.

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if u=-v it doesn't work. if u,v are orthogonal it definitely works. if they aren't orthogonal, i haven't solved for the eigenvectors, but i suspect there is some simple formula for them that should finish it. – Jack Schmidt Oct 25 '11 at 17:49
@Jack: sorry i screwed up the question a bit, I'll change it. – opt Oct 25 '11 at 17:50
Without loss of generality, you can take $u$ to be $(1, 0, ..., 0)^T$. – wnoise Oct 25 '11 at 17:57
Another WLOG is n=2 since the subspace spanned by u and v is invariant. so v=[a,b] but I haven't had a chance to finish. – Jack Schmidt Oct 25 '11 at 18:29
When n=2, char poly is negative at x=0 and positive at the ends, so one pos and one neg root. Seems long and ugly compared to the orthogonal case where u+v and u-v are the eigenvectors and this is basically "symmetric and anti-symmetric parts". – Jack Schmidt Oct 25 '11 at 18:33
up vote 6 down vote accepted

I presume that $u$, $v$ are nonzero. By absorbing the length of $u$ into $v$, we may assume WLOG that $\|u\|=1$. Let $v=\alpha u + \beta w$ for some unit vector $w\perp u$. Note that $\beta\not=0$ because $u \not \propto v$. Also, $uv^\top+vu^\top = 2\alpha uu^\top+\beta uw^\top+\beta wu^\top$. Hence $$ uv^\top+vu^\top \sim \begin{pmatrix} 2\alpha&\beta\\ \beta&0 \end{pmatrix}\oplus0_{n-2} $$ and the result follows immediately.

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Looks good. This is what I was trying to outline above, but you made it look simple and smooth. I especially like the simple fix for non-orthogonal v. In case anyone misses the last tilde: Choose a basis {u,w} and extend it with things perpendicular to both u and w, and then uvt+vut has the simple form given by user1551. – Jack Schmidt Oct 25 '11 at 19:09

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