Determining the derivation of a determinant

Question

Let $\Phi\colon E\to M$ with $E\subset \mathbb{R}\times M$ and $M\subset\mathbb{R}^n$ open. Consider the function given by $x\mapsto \Phi(t,x)$ for fixed $t\in\mathbb{R}$. (1) Determine $$ \frac{\partial}{\partial t}\text{det}D_x\Phi(t,x), $$ wherat $D_x\Phi(t,x)$ is the Jacobi matrix of the function given by $x\mapsto\Phi(t,x)$ for fixed $t\in\mathbb{R}$. (2) Calculate $\frac{\partial}{\partial t}\text{det}D_x\Phi(0,x)$ if $D_x\Phi(0,x)=I_n$ and $\frac{\partial}{\partial t}\Phi(t,x)=f(x)$ for all $x\in M$, whereat $f\colon M\to\mathbb{R}^n$ is any function.

Hello!

(1) First of all it is $$ D_x\Phi(t,x)=\begin{pmatrix}\frac{\partial\Phi_1(t,x)}{\partial x_1} & \ldots & \frac{\partial\Phi_1(t,x)}{\partial x_n}\\\vdots & \ddots & \vdots\\\frac{\partial\Phi_n(t,x)}{\partial x_1} & \ldots & \frac{\partial\Phi_n(t,x)}{\partial x_n} \end{pmatrix} $$ so I have to determine $$ \frac{\partial}{\partial t}\text{det}\begin{pmatrix}\frac{\partial\Phi_1(t,x)}{\partial x_1} & \ldots & \frac{\partial\Phi_1(t,x)}{\partial x_n}\\\vdots & \ddots & \vdots\\\frac{\partial\Phi_n(t,x)}{\partial x_1} & \ldots & \frac{\partial\Phi_n(t,x)}{\partial x_n} \end{pmatrix} $$ This is the same as determining $$ \frac{\partial}{\partial t}\text{det}(\text{grad}\Phi_1(t,x),\ldots,\text{grad}\Phi_n(t,x)). $$ To my knowledge now it is $$ \frac{\partial}{\partial t}\text{det}(\text{grad}\Phi_1(t,x),\ldots,\text{grad}\Phi_n(t,x))=\sum_{j=1}^{n}\text{det}(\text{grad}\Phi_1(t,x),\ldots,\frac{\partial}{\partial t}\text{grad}\Phi_j(t,x),\ldots,\text{grad}\Phi_n(t,x)). $$ So, I think that with $\frac{\partial}{\partial t}\text{grad}\Phi_j(t,x)$ it is meant $$ \left(\frac{\partial}{\partial t}\frac{\partial\Phi_j(t,x)}{\partial x_1},\ldots,\frac{\partial}{\partial t}\frac{\partial\Phi_j(t,x)}{\partial x_n}\right)^T $$ and because of Schwartz I think it is $$ \left(\frac{\partial}{\partial t}\frac{\partial\Phi_j(t,x)}{\partial x_1},\ldots,\frac{\partial}{\partial t}\frac{\partial\Phi_j(t,x)}{\partial x_n}\right)^T=\left(\frac{\partial}{\partial x_1}\frac{\partial\Phi_j(t,x)}{\partial t},\ldots,\frac{\partial}{\partial x_n}\frac{\partial\Phi_j(t,x)}{\partial t}\right)^T\\ =\text{grad}\frac{\partial}{\partial t}\Phi_j(t,x). $$ So what is searched for to my opinion is $$ \sum_{j=1}^{n}\text{det}(\text{grad}\Phi_1(t,x),\ldots,\text{grad}\frac{\partial}{\partial t}\Phi_j(t,x),\ldots,\text{grad}\Phi_n(t,x)) $$

(2)

It is $\text{grad}\frac{\partial}{\partial t}\Phi_j(t,x)=\text{grad}f_j(x)$. The j-th summand is given by $$ \text{det}(e_1,e_2,\ldots,e_{j-1},\text{grad}f_j(x),e_{j+1},\ldots,e_n) $$ with $e_i=(0,\ldots,0,1,0,\ldots,0)^T$ with the 1 on i-th position.

This determinant can be calculated by making a triangle matrix out of it. So this determinant is $\frac{\partial}{\partial x_j}f_j(x)$.

So it is $$ \frac{\partial}{\partial t}D_x\Phi(t,x)=\sum_{j=1}^{n}\frac{\partial}{\partial x_j}f_j(x)=\text{div}f. $$

So that are my results for (1) and (2).

I would really like to know if I am right. Would be great! Ciao!

With greetings and kind regards!

Yours math12

Answer 1

Your calculations are right, the time derivative is applied as in the product formula, since det is multilinear in its rows resp. columns.

The next step is to assume that the Jacobian is regular at the point of derivation. Then you can write $$ \frac{∂}{∂t} \operatorname{grad}\, \Phi_j(t,x)=\sum_{k=1}^n a_{jk} \operatorname{grad} \Phi_k(t,x) $$ so that $$ \det(\operatorname{grad}\,Φ_1(t,x),…,\frac∂{∂t}\operatorname{grad}\,Φ_j(t,x),…,\operatorname{grad}\,Φ_n(t,x))=a_{jj}\det(D_xΦ) $$

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In general it holds for matrix valued functions that $$ \tfrac{d}{dt}\det(A(t))=tr\left(A(t)^{-1}\tfrac{d}{dt}A(t)\right)\det(A(t))=tr\left(A(t)^{\#}\tfrac{d}{dt}A(t)\right) $$ with $A^\#$ the adjugate/adjoint matrix of $A$.

Answer 2

The determinant of an $n \times n$ matrix $A(t)$ can be defined as $$\det A = \sum_{S_n} \epsilon_{i_1,\cdots, i_n} a_{1,i_1} a_{2,i_2} \cdots a_{n,i_n}$$ where the sum is over all permutations $(i_1 \; i_2 \; \cdots \; i_n)$ of $\{1, 2, \dots , n\}$ and the term $\epsilon_{i_1,\cdots, i_n}$ is the Levi-Civita symbol that takes on the values $0, \pm 1$ in some complicated fashion. $S_n$ represents all permutations of the set we are interested in. Calm down and let this sink in for a bit. Don't try to really understand it, just remember the form. That's all that matters for our purposes.

Next recall the product rule $$(b_1 \cdots b_n)' = \sum_{k=1}^n b_1 \cdots b_k' \cdots b_n.$$

Then use this on the expression for the determinant: $$\frac{d}{dt}\det A = \sum_{S_n} \sum_{k=1}^n \epsilon_{i_1,\cdots, i_n} a_{1,i_1} \cdots a_{k,i_k}' \cdots a_{n,i_n} $$

Swap the sums: $$\frac{d}{dt}\det A = \sum_{k=1}^n \sum_{S_n} \epsilon_{i_1,\cdots, i_n} a_{1,i_1} \cdots a_{k,i_k}' \cdots a_{n,i_n} $$

But what is the inner sum? It is the determinant of the matrix $A$ but with $a_{k,i}$ replaced with $a_{k,i}'$ for fixed $k$ and $i=1,2,\dots, n$. That is, we take the derivative of every entry in the $k$th row. If we let $A_k$ be the matrix with the $k$th row changed in this manner, then

$$\det A = \sum_{k=1}^n \det A_k.$$