TITLE: Derivative of Determinant Map
QUESTION [5 upvotes]: For $ V= ( V_1, V_2) $  and $ W= ( W_1, W_2) $, given a determinant map $ \det : \mathbb{R}^2\times \mathbb{R}^2\rightarrow \mathbb{R}$ defined as $ \det (V,W)=  V_1W_2-V_2W_1$. Then have to find the derivative of the determinant map at $( V, W)\in R^2$ evaluated at $(H,K)\in \mathbb{R}^2$ . 
Please help me with this terminology. 
It seems to me if $ U= V_1W_2-V_2W_1$, 
then derivative of $U  =  V_1W_2-V_2W_1$. (by using the Jacobian technique)
Then, in that case 
derivative of U at $(H, K) = \det (H, K)$.

REPLY [15 votes]: Write $U((V_1,V_2,W_1,W_2)) = V_1 W_2-V_2 W_1$. Then just compute the partial derivatives: $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_1} = W_2$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_2} = -W_1$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_1} = - V_2$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_2} = V_1$. Then the derivative at $(V,W)$ in the direction $(H,K)$ is
given by
$$DU((V,W))((H,K)) = W_2 H_1-W_1 H_2-V_2 K_1+V_1 K_2 = \det(H,W)+\det(V,K)$$
This can also be written in terms of the Frobenius inner product as
$$DU((V,W))((H,K)) =\langle \begin{bmatrix} W_2 & -V_2 \\ -W_1 & V_1\end{bmatrix}, \begin{bmatrix} H_1 & K_1 \\ H_2 & K_2 \end{bmatrix} \rangle$$
and so we can write the gradient
$$ \nabla U((V,W)) = \begin{bmatrix} W_2 & -V_2 \\ -W_1 & V_1\end{bmatrix}$$