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

Question

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 $$\det(e^{At})=e^{Tr(At)}=e^{(TrA)t}.$$

Answer 1

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 $F(x):=Ax.$