Let $x_t$ be a diffusion process driven by a two-dimension brownian motion $w_t$.
$$dx_t = \mu(t, x_t) dt + \sigma(x_t, t)dw_t$$
where, $\mu$ is 2*1 and $\sigma$ is 2 * 2.
The book I am reading, is saying that the generator of function f is the following manner... $$ Lf(x) = \mu(t, x)Df(x) + \frac12 Tr \{\sigma(t, x) \sigma(t, x)^T D^2f(x)\} $$
I can't wrap my mind around the last term with trace.
Let, $$ \sigma = \left(\begin{array}{cc} \sigma_{1,1} & \sigma_{1,2}\\ \sigma_{2,1} & \sigma_{2,2} \end{array}\right) $$
then I can't get the result consistent with the 1-dimentional setting by doing ... $$tr \left(\begin{array}{cc} \sigma_{1,1} & \sigma_{1,2}\\ \sigma_{2,1} & \sigma_{2,2} \end{array}\right) * \left(\begin{array}{cc} \sigma_{1,1} & \sigma_{1,2}\\ \sigma_{2,1} & \sigma_{2,2} \end{array}\right)^T * \left(\begin{array}{cc} f_{x1,x1} & f_{x1,x2}\\ f_{x2,x1} & f_{x2,x2} \end{array}\right) $$
Can anyone help me take a look at the previous formula, and give me an idea how it is reduced to the familiar formula without trace? (i.e. ($\sigma_{11} f_{xx} + \sigma_{22} f_{yy} + \sigma_{12}f_{xy} + \sigma_{21}f_{yx}$))