If $a$ and $b$ are column vectors of equal dimension, is there an analytic formula for the SVD of $ab^T+ba^T$?
From a few trials I ran with Mathematica, it appears that there are only two non-zero eigenvalues, $a\cdot b\pm|a||b|$, but I don't know how to prove that or how to find the associated eigenvectors.
Short answer:
You have a $n-2$ dimensional nullspace orthogonal to $Range([a,b])$ and two eigenvectors: $\left\| a\right\|b + \left\| b\right\|a$ and $\left\| a\right\|b - \left\| b\right\|a$ with eigenvalues $a\cdot b + \left\| a\right\|\left\| b\right\| $ and $a\cdot b - \left\| a\right\|\left\| b\right\| $ respectively.
Slightly longer answer: Looking at the matrix $M=ab^T + ba^T$, you realize that the $Range(M) = Range([a,b])$, hence you have a $n-2$ dimensional nullspace (if $a$ not colinear with $b$). So one straight forward way to find the eigenvectors and eigenvalues, is simply to solve now the equation \begin{align} M(\alpha a + \beta b) = \lambda (\alpha a + \beta b) \end{align}
for $\alpha$, $\beta$ and $\lambda$. After some transformations of this equation, you get \begin{align} \frac{\alpha^2}{\beta^2} = \frac{\left\|b\right\|^2}{\left\|a\right\|^2} \Leftrightarrow \frac{\alpha}{\beta} =\pm\frac{\left\|b\right\|}{\left\|a\right\|} \end{align} and \begin{align} \lambda = a\cdot b \pm \left\| a\right\|\left\| b\right\|. \end{align} With the choices of $\alpha_{1,2} = \pm\left\| b\right\|$ and $\beta_{1,2} = \left\| a\right\|$ you get the mentioned eigenvectors. Let me know if you need more detail on any step.