3
$\begingroup$

I'm wondering, if it's possible to bring every $ uv^{\top} + vu^{\top} $, where $u,v \in \mathbb{R}^{n} $ in the form of $ xx^{\top} + yy^{\top} $, with $x,y \in \mathbb{R}^{n} $, since $ rank(uv^{\top} + vu^{\top}) = 2.$

Is there a explicit formula for such a decomposition? I tried to calculate it directly on paper, but need to calculate a nonlinear equation system. Is there a trick or an idea, what I could do to find such a formula?

$\endgroup$
1
  • 1
    $\begingroup$ $\operatorname{rank}(uv^T+vu^T) \leq 2$. The rank is 1 when $v = u$. $\endgroup$ – Arin Chaudhuri Oct 5 '16 at 16:54
5
$\begingroup$

No.

$xx^T + yy^T$ is positive semi definite but $uv^T + vu^T$ need not be. Consider $u = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $v = \begin{pmatrix} -1 \\ 0 \end{pmatrix}.$

$\endgroup$
5
$\begingroup$

In general $u v^T + v u^T$ might not be positive semidefinite, so you wouldn't have such a decomposition. However you can write $u v^T + v u^T = x x^T - y y^T$, where $x = (u+v)/\sqrt{2}$ and $y = (u-v)/\sqrt{2}$.

In fact, the only case where $u v^T + v u^T$ is positive semidefinite is when $u$ and $v$ are linearly dependent. Otherwise you could take $w$ such that $w^T (u + v) = 0$ while $w^T u \ne 0$, and $w^T (u v^T + v u^T) w = - 2 (w^T u)(w^T v) < 0$.

$\endgroup$
0
$\begingroup$

Notable fact to supplement the answers: we can rewrite these matrices as $$ \pmatrix{u&v}\pmatrix{0&1\\1&0} \pmatrix{u&v}^T, \qquad \pmatrix{x&y} \pmatrix{1&0\\0&1} \pmatrix{x&y}^T $$ we may then reach Robert Israel's conclusion using Sylvester's law of inertia.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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