I'm trying to write a fokker-planck equation as an SDE. I know my diffusion matrix, $D(\mathbf{x})$, where $D = \frac{1}{2}\sigma \sigma^T$.
How can I find this sigma, to then use in the SDE?
Edit: I don't think the fokker-planck bit is relevant to the question, it's just there for context (unless someone comes along and suggests a better way of going about it, which would be much appreciated!)
Edit 2: I've seen another answer that gives the answer if the matrix is positive definite. However, the entries in my matrix can vary, and I think it's unlikely all the eigenvalues will remain positive
If the matrix $B$ is not positive semi-definite, then there is no such matrix $A$.
For suppose that $$ Bv = c v $$ for some negative number $c$, and unit vector $v$, and suppose that such an $A$ exists. Then \begin{align} \langle Bv, v \rangle &= c < 0. \end{align} But we can also say
\begin{align} \langle Bv, v \rangle &= \langle AA^tv, v \rangle \\ &=(AA^tv)^t v \\ &=(v^t A^t A) v \\ &=(v^t A^t) (A v) \\ &= \langle Av, Av \rangle \\ &= \| Av \|^2 \ge 0\\ \end{align} which is a contradiction.
So: for the matrix $B$ to have a "square root" like $A$ requires that $B$ be positive semidefinite. By the way, since $AA^t$ is symmetric, it also requires that $B$ be symmetric.
Those other proofs you saw: they weren't just being stupid by requiring symmetric positive semi-definiteness. They were actually proving the strongest theorem possible.
For actually finding the matrix $A$, @celtschk's suggestion is spot-on. The reason it didn't work in your test case is that your input matrix, $B$, wasn't symmetric. When it IS symmatric, you get $$ B = U D V^t $$ but it'll turn out that $V$ is actually equal to $U$, and $U^t U = I$, so you get \begin{align} B &= U D U^t \end{align} at which point you can define $E = D^\frac12$, adn note that $E$ is diagonal so that $E^t = E$, and thus \begin{align} B &= U D U^t\\ &= U E^2 U^t\\ &= U E^t E U^t\\ &= U E^t I E U^t\\ &= U E^t (U^t U) E U^t\\ &= (U E^t U^t) (U E U^t)\\ &= S^t S \end{align} where $S = UEU^t$.
You can shove the factor of $\frac12$ in there wherever you'd like it.