Differential of determinant in the Itô calculus I'm trying to answer a question about a particular Itô equation $$\mathrm{d}X_t = AX_t\mathrm{d}t+X_tB(\mathrm{d}W_t)$$ where $X_t$ and $A$ are $n\times n$ square matrices and $B : M_n\to M_n$ ($M_n$ being the space of $n\times n$ matrices) is linear and obeys $\mathrm{tr}(B(\mathrm{d}W_t)) = 0$. In the simple example I'm working through, $n = 2$ and $$B\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & -a_{11} \end{pmatrix}$$ Namely, I'm trying to write down an equation for $\mathrm{d}(\det(X_t))$ so that I can show some conservation laws for a particular set of constraints on $A$ (i.e. that for some choice of $A$ the SDE evolves such that $\det(X_t)\equiv \det(X_0)\neq 0$). I know that in the non-stochastic case, $$\mathrm{d}(\det(X)) = \mathrm{tr}(\mathrm{adj}(X)\,\mathrm{d}X)$$ due to Jacobi's formula. I also have Itô's lemma from here. Unfortunately, I'm pretty sure the quadratic covariations in the example I'm working through are nonzero. Does this mean that I need to find a way of succinctly expressing the Hessian matrix of the determinant, or is there a better way to do this? Thanks!
Edit: I see that I could use the Stratanovich form of the SDE, which is more amenable to the chain rule and would allow me to use Jacobi's formula in a straightforward way, but I think I would still need to calculate the second derivative of the determinant somehow in order to do the conversion.
 A: Define the scalar variables
$$\eqalign{
\alpha &= \det(X) \cr
\beta &= {\rm tr}(X) &\implies d\beta = I:dX \cr
\lambda &= {\rm tr}(X^2) &\implies d\lambda = 2X^T:dX \cr
}$$ where a colon is the Frobenius product notation for the trace
$$A:B = {\rm tr}(A^TB)$$ 
Write characteristic equation of $X$, take its trace, and solve for $\alpha$
$$\eqalign{
\alpha I &= \beta X-X^2 \cr
2\,\alpha &= \beta^2-\lambda \cr
\alpha &= \tfrac{1}{2}(\beta^2-\lambda) \cr
}$$
The differential of the last equation is
$$\eqalign{
 d\alpha &= (\beta\,I-X^T):dX \cr
}$$
Now define a constant "off-diagonal" matrix $F$ which has zeros on the diagonal and ones elsewhere. Also define the constant matrix $E_k$ whose $(k,k)$ element is one and all other elements zero. Finally, the matrix $Y=(\beta X-X^2)^T$ will prove convenient. 
Using these constant matrices and the elementwise/Hadamard product $(\odot)$, we can express the $B$ function as
$$\eqalign{B(dW) &= F\odot dW + (E_1-E_2) (E_1:dW)}$$
The differential of $X$ can be written as
$$\eqalign{
dX &= AX\,dt + XB(dW) \cr
 &= AX\,dt + X\big(F\odot dW + (E_1-E_2) (E_1:dW)\big) \cr
}$$
Substitute this into the previous differential and simplify
$$\eqalign{
d\alpha
 &= (\beta\,I-X^T):\Big(AX\,dt + X\big(F\odot dW + (E_1-E_2) (E_1:dW)\big)\Big) \cr
 &= Y:\Big(A\,dt + \big(F\odot dW + (E_1-E_2) (E_1:dW)\big)\Big) \cr
 &= Y:A\,dt + Y:(F\odot dW) + Y:(E_1-E_2)(E_1:dW) \cr
 &= Y:A\,dt + (F\odot Y):dW + \big(Y:(E_1-E_2)\big)E_1:dW \cr
d\,\det(X)
 &= (Y:A)\,dt + \Big(F\odot Y + (Y:E_1)E_1-(Y:E_2)E_1\Big):dW \cr
 &= {\rm tr}(A^TY)\,dt + \Big(F\odot Y + {\rm tr}(E_1Y)E_1-{\rm tr}(E_2Y)E_1\Big):dW \cr
}$$
Not sure if this expression will be of any help solving your problem, but it was an interesting exercise.  
