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?
On
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 $w^T (u v^T + v u^T) w = - 2 (w^T u)(w^T v) < 0$.
On
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.
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 $v = \begin{pmatrix} -1 \\ 0 \end{pmatrix}.$