Derivative of determinant of a matrix

Question

I would like to know how to calculate:

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

Readers of this thread may be interested in the "Matrix Cookbook" by Petersen & Pedersen, which has explicit formulas for tons of things like the derivatives of the determinant, eigenvalues, and inverse of a matrix. A few proofs, but mostly just references. — Jess Riedel, Jan 04 '23 at 02:08

Agustí Roig · Answer 1 · 2013-01-13T17:53:47.797

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

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

obtaining:

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

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

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

Bombyx mori · Accepted Answer · 2013-01-14T07:47:24.987

18

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.

edited Jan 14 '13 at 07:47

answered Jan 13 '13 at 08:03

Bombyx mori

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we can write it as a sum?, and how? please – kiroro Jan 13 '13 at 08:09
Try to write down the 2 by 2 case yourself and see how it works. – Bombyx mori Jan 13 '13 at 08:17
how to use the formula? please – kiroro Jan 13 '13 at 09:35
Find a DG book on principal bundles. – Bombyx mori Jan 13 '13 at 09:37
what is the relation ? – kiroro Jan 13 '13 at 09:44
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 ? – kiroro Jan 13 '13 at 08:45
I have updated. – Bombyx mori Jan 14 '13 at 08:03

score 17 · Answer 3 · edited Jan 13 '13 at 17:54

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Like product rule:

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

edited Jan 13 '13 at 17:54

Agustí Roig

17,959

answered Jan 13 '13 at 08:15

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Yes, I guess that's what OP wants. – Bombyx mori Jan 13 '13 at 08:16
@kiroro helped you? – Jan 13 '13 at 08:17
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yes this is the formula that I want, but I'm not yet arrived – kiroro Jan 13 '13 at 08:24
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@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!!! – Jan 13 '13 at 08:45
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ok thank you good night – kiroro Jan 13 '13 at 08:47

score 6 · Answer 4 · answered Nov 19 '15 at 14:29

6

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:

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.

answered Nov 19 '15 at 14:29

desmond13

625

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Is there an explicit derivation of second-order derivative of determinant when A is singular? – vulture May 22 '18 at 04:09
Note that when $A$ is invertible, $\mathrm{adj}(A) = A^{-1}\det A$, so the formula is just $(d/dt)\det A(t) = (\det A)\mathrm{Tr}[ A^{-1} (dA/dt)]$. – Jess Riedel Jan 04 '23 at 02:13

Derivative of determinant of a matrix

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