Let $$\Phi_t=\begin{cases}\mathbb R^n\to\mathbb R^n,\\ x\mapsto\begin{pmatrix}e^{t\theta_1}\\&\ddots\\&&e^{t\theta_n}\end{pmatrix}x\end{cases}$$ where $\theta_j\in i\mathbb R$ for all $j=1,\dots,n$.
Show that there is a sequence $t_m\to\infty$ such that $\Phi_{t_m}\to\text{id}_{\mathbb{C}^n}$
At first I have tried to construct such a sequence $t_m$ where even $\Phi_{t_m}=\text{id}$ but I have come to think that this is not possible and I've ran out of ideas. This is an exercise from a problem sheet for differential equations so maybe one needs to use certain theorems. All I know in this regard is that $\Phi_t$ is the flow of the matrix $A=\text{diag}(\theta_1,\dots,\theta_n)$ and the equation $\dot x=Ax$ has the general solution$$x(t)=\lambda_1e^{t\theta_1}\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}+\dots+\lambda_n e^{t\theta_n}\begin{pmatrix}0\\\vdots\\0\\1\end{pmatrix}$$ where $\lambda_i\in\mathbb C$. Maybe one can use an asymptotic behaviour or something else. Any ideas?
I will write $\theta_k=2\pi\, i\, a_k$, thus introducing by this formula the "beter variables", the "system" $$ (a_1,a_2,\dots,a_n)\in [0,1)\ . $$ We can (and do) adapt the above sequence for our purposes to have the terminal value $a_n=1$. "(re)scaling".
We can suppose that the above values are $\Bbb Q$-linearly independent.
Else we eliminate such $\Bbb Q$-linear dependencies, passing to refined new subsystem, so that each variable of the old system can be $\Bbb Z$-linearly expressed in terms of the variables in the new system.
We eliminate now the terminal $1$ from the list, and after adjusting $n$ correspondingly (sorry) get an other system (also with other possible entries) $$ (a_1,a_2,\dots,a_n)\in [0,1)\ . $$ If the new system is empty, we are done. Just take $(t_N)$ to run in $\Bbb N$ to infinity. (This is the case iff all $a_1,a_2, \dots$ are rational after rescaling.)
Else we have to work, this is the hard case.
Now we will use Weyl's Criterion on Equidistributed_sequence, Generalized version .
Then let us show that the sequence $(Na)_{N\in\Bbb N}$, $$ Na:=(Na_1, Na_2,\dots, Na_n)\ ,\qquad N\in\Bbb N $$ is uniformly distributed modulo one (in each component).
Due to Weyl's Criterion, it is enough (and we need) to show that for all $$ l=(l_1,l_2,\dots,l_n)\in\Bbb Z^n$$ we have $$ 0=\lim_{N\to\infty} \frac 1N\sum_{0\le j<N}e^{2\pi\, i\, (Na_1, Na_2,\dots, Na_n)\cdot(l_1,l_2,\dots,l_n)} \ . $$ This is so because the exponential sum can be easily bounded, $$ \begin{aligned} &\sum_{0\le j<N}\exp\Big(\ 2\pi\, i\, j\underbrace{(a_1l_1+a_2l_2+\dots+a_nl_n)}_{\text{Notation: }T}\ \Big) \\ &\qquad=\sum_{0\le j<N}e^{2\pi i\, jT} \\ &\qquad=\frac {\displaystyle 1-e^{2\pi i\, NT}} {\displaystyle 1-e^{2\pi i\, T}} \ , \end{aligned} $$ and the denominator is fixed (after fixing $l$) and $\ne 0$, and the numerator is in modulus $\le 2$. Then the factor $\frac 1N$ under the limit in Weyl's criterion is bringing the whole expression to zero for $N\to\infty$.
So the sequence $Na$ is uniformly distributed in the corresponding torus. In particular, for each given $\epsilon>0$, we find at least one $N_\epsilon$ so that $N_\epsilon \cdot a$ has all components taken modulo one in the interval $[0,\epsilon)$. This can be converted to a sequence $(N_{1/K})_{K\ge 1}$, so that all components of $N_{1/K}a$ have rational parts that are closer and closer to zero. Use then this sequence.
This is a dry exposition, even after some prosaic reformulations. So i try to give some numerical data that show why the problem is hard, where it is so, and how far we have to go with the choice of $N_\epsilon$ for a given $\epsilon$.
Let us consider first the case $\theta_1=2\pi i\,\sqrt 2$, $\theta_2=2\pi i$.
Then $a_1=\sqrt 2$, $a_2=1$, from this system we eliminate the $1$, and try to get multiples of $a=a_1$ that come closer and closer to integers. This is relatively simple, for instance we consider the convergents of the continued fraction associated to $a=\sqrt 2$, some sage
The fractional part converges (alternatively) better and better to zero. The sequence $(t_K)_K$ would be in this case the sequence of denominators of the convergents of $\sqrt 2$, this is $2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741\dots$.
What happens now if we start with $\theta_1,\theta_2,\theta_3$, so that the corresponding $a$-system is $(a_1=\sqrt 2,\ a_2=\sqrt 3,\ a_3=1)$ ?
We remove the terminal one, deal with the refined system $a=(a_1=\sqrt 2,\ a_2=\sqrt 3)$.
Then the first value of $N$ so that the fractional parts of $Na$ are both in the interval $[0,\ 0.01]$ is...
In this case we have no suitable algorithm to simultaneously get integer values approximated better and better. For this reason we need the ergodic argument given...