构造满足X'=AX且所有解均为2π周期的3×3矩阵A的技术问询

AI 小北

2026-4-15

Good afternoon, I've been asked on a homework problem to do the following:
Give an example of a $3 \times 3$ matrix $A$ for which all nonequilibrium solutions of $X^\prime = AX$ are periodic with period $2\pi$.
From studying these types of differential equations, I've found that $3 \times 3$ matrices $A$ with pairs of complex eigenvalues of the form $bi$ for some constant $b$ yield solutions which are periodic in two of the three directions without dealing with the pesky $e^{\lambda_j t}$, but I can't seem to avoid the $e^{\lambda_j t}$ in the real eigenvalue section no matter my construction. I am not certain if this problem allow...

Hey Ashton, great question! Let's work through this to get rid of that annoying exponential term you're stuck on.

First, a quick key reminder about real matrices: complex eigenvalues always come in conjugate pairs. So a 3×3 real matrix can only have one real eigenvalue plus a pair of complex conjugate eigenvalues. For all non-equilibrium solutions to be periodic (no growing/decaying exponentials), two rules must hold:

All eigenvalues must have a real part of 0 (so no non-zero real eigenvalues—those would give you the $e^{\lambda t}$ terms that break periodicity)
The matrix must be diagonalizable over the complex numbers (no Jordan blocks), because Jordan blocks with pure imaginary eigenvalues lead to solutions with $t \times \cos(bt)$ or $t \times \sin(bt)$ terms, which aren't periodic.

Since we're limited to real 3×3 matrices, the only valid real eigenvalue here is 0 (it's the only real number with a real part of 0). Pair that with the conjugate pure imaginary eigenvalues $\pm i$ (this gives us a period of $2\pi$, since period = $2\pi / b$ where $b=1$).

A straightforward example of such a matrix is a block-diagonal combination of a 2×2 rotation matrix (for the $\pm i$ eigenvalues) and a 1×1 zero block (for the 0 eigenvalue):

A = [
    [0, -1, 0],
    [1,  0, 0],
    [0,  0, 0]
]

Let's confirm why this works:

The eigenvalues of $A$ are $i$, $-i$, and $0$.
Non-equilibrium solutions (where the first two components aren't both zero) take the form:
$$X(t) = \begin{pmatrix} c_1 \cos t - c_2 \sin t \ c_1 \sin t + c_2 \cos t \ c_3 \end{pmatrix}$$
Here, $c_1$ and $c_2$ aren't both zero (so it's non-equilibrium), and $c_3$ is any constant. The first two components repeat every $2\pi$, and the third is constant—so the entire solution is periodic with period $2\pi$.
Equilibrium solutions are the constant vectors where $c_1 = c_2 = 0$, which are excluded from the "nonequilibrium" requirement.

You might wonder if we can use a non-zero pure imaginary eigenvalue for the third component—but we can't with a real 3×3 matrix. Non-zero pure imaginary eigenvalues must come in conjugate pairs, which would require an even number of them, and 3 is odd. So the 0 eigenvalue is your only valid real option here.

That should fix your exponential term issue—just make sure your real eigenvalue is 0, not any non-zero real number!

备注：内容来源于stack exchange，提问作者Ashton B