Good morning everyone, I would like to know how to calculate:

$\frac{d}{dt}\det \big(A_1(t), A_2(t), \ldots, A_n (t) \big)$

Help me please. Thank you

  • $\begingroup$ What is your mean $A_i(t)$? $i$-th column of $A$? $\endgroup$ – SKMohammadi Jan 2 '16 at 14:32

The formula is $$d(\det(m))=\det(m)Tr(m^{-1}dm)$$ where $dm$ is the matrix with $dm_{ij}$ in the entires. The derivation is based on Cramer's rule, that $m^{-1}=\frac{Adj(m)}{\det(m)}$. It is useful in old-fashioned differential geometry involving principal bundles.

I noticed Terence Tao posted a nice blog entry on it. So I probably do not need to explain more at here.

  • $\begingroup$ we can write it as a sum?, and how? please $\endgroup$ – kiroro Jan 13 '13 at 8:09
  • $\begingroup$ Try to write down the 2 by 2 case yourself and see how it works. $\endgroup$ – Bombyx mori Jan 13 '13 at 8:17
  • $\begingroup$ for n=2, i must have $ \frac{d}{dt}\det \big(A_1(t),A_2(t)\big)=$ $\det \big(A'_1(t),A_2(t)\big)+ \det \big(A_1(t),A'_2(t))\big) $ but how ? $\endgroup$ – kiroro Jan 13 '13 at 8:45
  • $\begingroup$ how to use the formula? please $\endgroup$ – kiroro Jan 13 '13 at 9:35
  • $\begingroup$ Find a DG book on principal bundles. $\endgroup$ – Bombyx mori Jan 13 '13 at 9:37

Think I can provide a proof for Matias' formula.

So, let

$$ A(t) = \mathrm{det}\left( A_1(t), \dots , A_n(t) \right) \ . $$

By definition,

$$ \frac{dA(t)}{dt} = \mathrm{lim}_{h\rightarrow 0} \frac{A(t+h) - A(t)}{h} = \mathrm{lim}_{h\rightarrow 0} \frac{\det (A_1(t+h), \dots, A_n(t+h)) - \det(A_1(t), \dots , A_n(t))}{h} $$

Now, we subtract and add

$$ \det(A_1(t), A_2(t+h), \dots , A_n(t+h)) $$


$$ \frac{dA(t)}{dt} = \mathrm{lim}_{h\rightarrow 0} \frac{\det (A_1(t+h), A_2(t+h),\dots, A_n(t+h)) - \det(A_1(t), A_2(t+h), \dots , A_n(t+h))}{h} + \mathrm{lim}_{h\rightarrow 0}\frac{ \det(A_1(t), A_2(t+h), \dots , A_n(t+h))-\det(A_1(t), \dots , A_n(t))}{h} $$

Now we focus on the first addend, which is

$$ \det \left( \mathrm{lim}_{h\rightarrow 0} \frac{A_1(t+h) - A_1(t)}{h}, \mathrm{lim}_{h\rightarrow 0} A_2(t+h), \dots,\mathrm{lim}_{h\rightarrow 0} A_n(t+h) \right) $$

That is,

$$ \det (A_1'(t), A_2(t), \dots , A_n(t)) \ . $$

Now, let's go for the second addend to which we substract and add

$$ \det(A_1(t), A_2(t), A_3(t+h), \dots , A_n(t+h)) \ . $$

From which we will obtain the term

$$ \det (A_1(t), A'_2(t), A_3(t), \dots , A_n(t)) \ . $$

Keep on doing analogous operations till you get

$$ \det (A_1(t), A_2(t), \dots , A_{n-1}(t), A_n'(t)) \ . $$


Like product rule:

$$\dfrac{d}{dt}\det(A_1(t),A_2(t),...,A_n(t))=\det(A_1^{'}(t),A_2(t),A_n(t))+\det(A_1(t),A_2^{'}(t),...,A_n(t))+...+\det(A_1(t),A_2(t),...,A_n^{'}(t)) $$

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    $\begingroup$ Yes, I guess that's what OP wants. $\endgroup$ – Bombyx mori Jan 13 '13 at 8:16
  • $\begingroup$ @kiroro helped you? $\endgroup$ – user52188 Jan 13 '13 at 8:17
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    $\begingroup$ yes this is the formula that I want, but I'm not yet arrived $\endgroup$ – kiroro Jan 13 '13 at 8:24
  • $\begingroup$ @kiroro I'll be happy to explain later!! In my country it's too late, ok? I'm going to sleep. You need it fast? But it seems that you are having good results. nice!!! $\endgroup$ – user52188 Jan 13 '13 at 8:45
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    $\begingroup$ ok thank you good night $\endgroup$ – kiroro Jan 13 '13 at 8:47

In the previous answers it was not explicitly said that there is also the Jacobi's formula to compute the derivative of the determinant of a matrix.

You can find it here well explained: JACOBI'S FORMULA.

And it basically states that:

enter image description here

Where the adj(A) is the adjoint matrix of A. How to compute the adjugate matrix is explained here: ADJUGATE MATRIX.

I hope it will help someone.

  • $\begingroup$ Is there an explicit derivation of second-order derivative of determinant when A is singular? $\endgroup$ – vulture May 22 '18 at 4:09

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