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Derivative of Determinant Map

consider $v=(v_1,v_2)\in \mathbb{R}^2$ ,$w=(w_1,w_2)\in\mathbb{R}^2$ consider the determinant map det:$\mathbb{R}^2\times \mathbb{R}^2$ define by $\det(v,w)=v_1w_2-w_1v_2$. 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

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

$2$. $\det(H,K)$

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

$4$. $\det(V,W) + \det(K,W)$.

Well, I have no idea how to solve this one.

  • $\begingroup$ and so $V = v = (v_1, v_2)$ and $W = w = (w_1, w_2)$ ? $\endgroup$ – Thomas Jun 11 '12 at 17:16
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    $\begingroup$ Your map "det" is seen as a function from $\mathbb{R}^2 \times \mathbb{R}^2$ to $\mathbb{R}$. Hence its derivative is simply the gradient. Try to differentiate with respect to $v$ and to $w$ separately. $\endgroup$ – Siminore Jun 11 '12 at 17:19
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    $\begingroup$ math.stackexchange.com/questions/146433/… $\endgroup$ – copper.hat Jun 11 '12 at 17:23
  • $\begingroup$ yahh thomas.... $\endgroup$ – Marso Jun 11 '12 at 18:55

Compute the partials using the fact that $\det$ is multilinear:

Thus the derivative at $(V,W)$ in the direction $(H,0)$ will be easily computed from the expression $\det(V+H,W) = \det(V,W)+\det(H,W)$. In particular, the derivative at $(V,W)$ in the direction $(H,0)$ is $\det(H,W)$.

Similarly, the derivative at $(V,W)$ in the direction $(0,K)$ will be given by $\det(V,K)$.

Since the derivative is linear, we have that the derivative at $(V,W)$ in the direction $(H,K)$ is just the sum of the derivatives in the direction $(H,0)$ and $(0,K)$.

Hence the result is $\det(H,W)+\det(V,K)$.


If $A=(a_{ij})$ is an $n\times n$ matrix we have a formula $$ \delta_{ij} \det A = \sum_k a^{*}_ {ik}a_{jk} $$ where $A^* = (a_{ij}^*)$ is the cofactor matrix of $A$ and $\delta_{ij}$ the Kronecker $\delta$.

By standard results from linear algebra $a^*_{ij}$ is obtained by taking the determinant of the matrix $A$ with the $i$-th column and $j$-th row removed and the $(i,j)$ entry set to $1$. This implies that $a^*_{ij}$ does not depend on $a_{ik}$ or $a_{kj}$ for any $k$. So we can differentiate the formula for $\det A$ wrt $a_{il}$ and sum over $i$ to obtain $$\sum_i\delta_{ij}\frac{\partial}{\partial a_{il}}\det A = \frac{\partial}{\partial a_{jl}}\det A= \sum_{k,i}a^*_{ik}\delta_{jk,il} =a^*_{jl} $$ Now the first formula also tells us that $a^*_{jl}=\det(A) (A^{-T})_{jl}$ with $A^{-T}$ denoting the transpose of the inverse. Inserting this in the last one implies $$ \frac{\partial}{\partial a_{jl}}\det A = \det(A) (A^{-1})_{lj} $$ for invertible $n\times n$ matrices $A$


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