# Prove that a flow is volume-preserving if and only if the trace of corresponding ODE matrix is $0$

I am working on the problem stated as below:

Consider the linear equation $$\dot{x}=Ax$$, where $$x\in\mathbb{R}^{n}$$ and $$A$$ is an $$n\times n$$ matrix. Let $$\varphi_{t}$$ be the flow generated. Verify that $$\varphi_{t}(x)=e^{At}x,\ \text{and}\ \det(\varphi_{t})=\det(e^{At})=e^{(Tr A)t},$$ so that $$\varphi_{t}$$ is volume-preserving if and only if $$Tr(A)=0$$.

I have done all the verifications. I will post my verifications after I explain my confusions.

When we define a flow, we always use a vector field, say $$X=(X_{1},\cdots, X_{n})$$ on $$\mathbb{R}^{n}$$, and we say $$\varphi_{t}$$ is a flow generated by this vector field, i.e. $$\varphi_{t}:X\rightarrow X$$ a map and $$\{\varphi_{t}\}$$ satisfies $$\varphi_{0}=id\ \text{and}\ \varphi_{s+t}=\varphi_{t}\circ\varphi_{s},\ \text{for all}\ s,t\in\mathbb{R}.$$

Then, we define the divergence of $$X$$ by $$div(X)=\sum_{i=1}^{n}\dfrac{\partial X_{i}}{\partial x_{i}}=Tr(DX),$$ and accordingly we have the following proposition

Proposition: $$\varphi_{t}$$ is volume-preserving if and only if $$div(X)=0$$ everywhere on $$\mathbb{R}^{n}.$$

However, when I work on problems, they always use a system of ODE, instead of vector fields.

How could I connect the flow generated by linear equation $$\dot{x}=Ax$$ to the flow generated by a vector field?

If I can understand this part, I believe I can solve this problem, since I believe $$Tr(A)$$ must be somehow connected to $$Tr(DX)$$.

Thank you!

Below is my verifications:

To verify, firstly we have $$\dfrac{d}{dt}\varphi_{t}(x)=\dfrac{d}{dt}e^{At}x=Ae^{At}x=A\varphi_{t}(x).$$

Then, $$\det(\varphi_{t})=\det(e^{At})$$ immediately, and thus it remains to show $$\det(e^{At})=e^{(TrA)t}$$.

Firstly, we need to note that for any $$n\times n$$ matrix $$A$$ and for any invertible $$n\times n$$ matrix $$P$$, we can write $$A^{n}=(P^{-1}AP)(P^{-1}AP)\cdots (P^{-1}AP)=P^{-1}A^{n}P,\ \text{for all}\ n,$$ such that \begin{align*} \exp(A)&=I+A+\dfrac{1}{2!}A^{2}+\dfrac{1}{3!}A^{3}+\cdots,\\ &=P^{-1}(I+A+\dfrac{1}{2!}A^{2}+\dfrac{1}{3!}A^{3}+\cdots,)P\\ &=P^{-1}\exp(A)P. \end{align*}

If $$A$$ is diagonalizable, then $$A$$ can be written as $$A=P^{-1}JP,$$ where $$P$$ is invertible and $$J$$ is a diagonal matrix, such that we have \begin{align*} \det(e^{A})&=\det(e^{P^{-1}JP})=\det(P^{-1}e^{J}P)\\ &=\det(P^{-1})\det(e^{J})\det(P)=\det(e^{J})\\ &=\pi_{i=1}^{n}e^{j_{ii}}=e^{\sum_{i=1}^{n}j_{ii}}\\ &=e^{tr(J)}=e^{tr(A)}, \ \text{since trace equals the sum of eigenvalues}. \end{align*}

If $$A$$ is nilpotent, then since every nilpotent matrix is similar to a upper triangular matrix $$D$$ with $$0$$s on the diagonal, we have $$A=P^{-1}DP,$$ and then by the same calculation above we have $$\det(e^{A})=e^{tr(D)}=e^{tr(A)}.$$

Finally, if $$A$$ is a general $$n\times n$$ matrix, then since any matrix $$A$$ can be written as $$A=D+N,$$ where $$D$$ a diagonal matrix and $$N$$ a nilpotent matrix, and since such $$D$$ and such $$N$$ commute (so that we can write $$e^{A}=e^{D}e^{N}$$), we have $$\det(e^{A})=\det(e^{D}e^{N})=\det(e^{D})\det(e^{N})=e^{Tr(D)}e^{Tr(N)}=e^{Tr(D)+Tr(N)}=e^{Tr(A)}.$$

Therefore, for any $$n\times n$$ matrix, we have $$\det(e^{A})=e^{Tr(A)},$$ and thus we have $$\det(e^{At})=e^{Tr(At)}=e^{(TrA)t}.$$

• Note that $\int_{\varphi_t(S)}1=\int_S \lvert\det d\varphi_t\rvert$ for any $S$. The rest are simple computations. Commented Sep 10, 2019 at 13:17
• @John B yes you are right, I was being dumb. Do you want to post your answer so that I can vote you up and approve? You could directly post your answer here if you want. Commented Sep 10, 2019 at 14:25
• Quite fine as it is. :) Commented Sep 10, 2019 at 16:44
• @JohnB no problem. Thank you! Commented Sep 10, 2019 at 17:08

The solution here follows exactly from what John B said.

Proof: Let $$S\subset\mathbb{R}^{n}$$ be a domain, note that $$\varphi_{t}$$ is volume-preserving if and only if $$m\Big(\varphi_{t}(S)\Big)=m(S),$$ which is equivalent to $$\int_{S}1dx=\int_{\varphi_{t}(S)}1dx=\int_{S}\Big|\det\dfrac{\partial\varphi_{t}}{\partial x}\Big|dx:=\int_{S}|\det M|dx,$$ where $$M$$ is defined to be the Jacobian matrix $$\varphi_{t}'$$ of $$\varphi_{t}$$.

On the other hand, $$\dfrac{\partial\varphi_{t}(x)}{\partial x}=\dfrac{\partial}{\partial x}e^{At}x=e^{At}\dfrac{\partial x}{\partial x}=e^{At}.$$

Therefore, $$\det(M)=\det(e^{At})=e^{Tr(A)t}.$$

Thus, $$\varphi_{t}$$ is volume preserving $$\iff |\det(M)|=1\iff Tr(A)=0.$$

Finally, some words about my questions of the connection between ODE and vector field.

They are just equivalent, since you could define a vector field $$F:\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}$$ by $$F(x):=Ax.$$