7

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

copper.hat
  • 172,524
preeti
  • 1,327
  • 1
  • 18
  • 26

1 Answers1

17

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 $$ \nabla U((V,W)) = \begin{bmatrix} W_2 & -V_2 \\ -W_1 & V_1\end{bmatrix}$$

copper.hat
  • 172,524
  • Can you please explain to me, how did you evaluate the derivative of determinant along $(H, K)$? what is the formula? Thank you. – GA316 Apr 02 '20 at 07:31
  • The formula is on the last line of the answer? – copper.hat Apr 02 '20 at 16:56
  • Thanks. but can tell me the intuition behind the formula. why it works and what is it? It looks like some sort of inner product of some vectors to me. – GA316 Apr 03 '20 at 06:24
  • I think my answer has a bug. – copper.hat Apr 03 '20 at 07:08
  • ok. can you please edit it? Thank you. – GA316 Apr 03 '20 at 07:24
  • It is after midnight here. The bottom part of the answer (after note) is correct. The derivative I gave is incorrect. Not sure how it lasted all this time. – copper.hat Apr 03 '20 at 07:38
  • Sorry to disturb you. No problem. Whenever it is possible please edit it. Thank you :). – GA316 Apr 03 '20 at 09:52
  • @GA316: I corrected my answer, not sure what I was thinking before. Hopefully this helps. – copper.hat Apr 03 '20 at 21:31
  • Perfect. Thanks a lot :) – GA316 Apr 04 '20 at 00:27
  • @GA316: Thanks for catching the error. – copper.hat Apr 04 '20 at 00:54
  • You are welcome :) – GA316 Apr 04 '20 at 06:32
  • What you've called the "Frobenius inner product" is different from Wikipedia's definition. They claim $\langle AB\rangle=\text{trace}(A^B)$, while yours appears to be $\langle AB\rangle=\text{trace}[\text{adjug}(A^)B]$, where $\text{adjug}$ denotes the adjugate. – WillG Sep 23 '22 at 16:24
  • Can you comment at all on the difference between these inner products and why the latter appears here, rather than the one on Wikipedia? – WillG Sep 23 '22 at 16:26
  • 1
    @WillG I am not sure what you mean. I am using the standard inner product corresponding to the Frobenius norm, $\langle A,B \rangle = \operatorname{tr} (A^B)$. The derivative of the determinant map is $D \det (A) H = \operatorname{tr} ( \operatorname{adj} (A)H) = \langle \operatorname{adj} (A)^, H \rangle$, all of which is consistent with the above and usual usage. – copper.hat Sep 23 '22 at 17:44
  • 1
    @copper.hat Ok I see now. I misunderstood your second to last line. – WillG Sep 23 '22 at 18:05