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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 $(H, K) = \det (H, K)$.

1 Answers 1

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Write your function $U(V_1,V_2,W_1,W_2) = V_1 W_2-V_2 W_1$. Then just compute the partial derivatives: $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_1} = W_2$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_2} = -W_1$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_1} = - V_2$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_2} = V_1$. So, the derivative at $(H,K)$ is $\begin{bmatrix} K_2 & -H_2 \\ -K_1 & H_1\end{bmatrix}.$

Note: My answer needs some elaboration.

The derivation is correct, but the expression of the answer needs some elaboration. The derivative is a linear map $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$. To evaluate the derivative in a particular direction ($(H,K)$ in this instance) the above expression must be multiplied componentwise by the corresponding components of the direction.

The derivative of $\det$ evaluated at $(V,W)$ in the direction $(H,K)$ is given by $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_1} H_1+\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_2} H_2 +\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_1} K_1 +\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_2} K_2$, which simplifies to $W_2 H_1-W_1 H_2-V_2 K_1+V_1 K_2 = \det(H,W)+\det(V,K)$.