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For $V = (V_1,V_2 )\in\mathbb{R}^2$ and $W = (W_1,W_2 )\in\mathbb{R}^2$ , Consider the determinant map $$\det :\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$$ defined by $$\det(V,W) = V_1W_2 -V_2W_1$$

Then the derivative of the determinant map at $(V,W)\in\mathbb{R}^2 \times \mathbb{R}^2$ evaluated on $(H, K )\in\mathbb{R}^2 \times\mathbb{R}^2$ is:

  1. $\det (H,W) + \det(V, K )$
  2. $\det (H, K )$
  3. $\det (H,V ) + \det(W, K )$
  4. $\det (V,W) + \det(K,W)$

Can anyone help me to solve this problem? Thank you.

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Just write down $\det(V+ H, W+K) - \det(V,W)$ and toss away the higher order term. –  martini Oct 5 '12 at 12:43
    
Hint: Multilinear maps satisfy the product rule. –  Alexander Thumm Oct 5 '12 at 12:44
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This is your fourth question that appears to be copied from somewhere. Please cite your sources, tell us whether a question is homework, and explain what you've tried so far. –  Noah Snyder Oct 5 '12 at 14:49
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In particular, please see how you can improve your problem based on advice given here. –  Willie Wong Oct 5 '12 at 15:09
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1 Answer 1

up vote 0 down vote accepted

If you have a continuous bi-linear map(in general multilinear) $ f: E_1 \times E_2 \rightarrow F $ , where $ E_1, E_2,F $ are Banach spaces, then given

$ v_1 \in E_1 $ and $ v_2 \in E_2 $ ,

$ Df(v_1, v_2)(h_1,h_2) = f(v_1, h_2) +f(h_1, v_2) $ .

This follows from the definition of Frechet derivative and the expression therein. Martini has already written it down explicitly.

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