Time dependent ODE involving cross product Let $\vec{A}$ be any time dependent vector quantity, and $\vec{\alpha}$ any constant vector. I was told that a solution to the differential equation
$$
\dot{\vec{A}} = \vec{\alpha}\times\vec{A}
$$
is that the vector $A$ rotates around the vector $\vec{\alpha}$ with velocity $\left|\vec{\alpha}\right|$. I tried to solve the equation, and then check if $\vec{A}\cdot\vec{\alpha}\equiv const$, but it yielded no results. Is this true?
 A: The equation is quite relevant in physics. Think of the precession of a
magnetic dipole in a magnetic field, NMR etc.
\begin{eqnarray*}
\frac{d\mathbf{A}}{dt} &=&\mathbf{\alpha }\times \mathbf{A} \\
\frac{d\mathbf{\alpha \cdot A}}{dt} &=&\mathbf{\alpha \cdot \alpha }\times
\mathbf{A}=0\Rightarrow \mathbf{\alpha \cdot A}\;\mathrm{const} \\
\frac{d}{dt}\mathbf{A}^{2} &=&\mathbf{A\cdot \alpha }\times \mathbf{A}
=0\Rightarrow \mathbf{A}^{2}\;\mathrm{const}
\end{eqnarray*}
Let $\mathbf{\alpha }$ be a unit vector (scale $t$ with $\alpha =|\mathbf{
\alpha }|$, $t\rightarrow \alpha t$). Then ($\mathsf{U}$ is the unit $
3\times 3$ matrix)
\begin{eqnarray*}
\mathbf{A} &=&\mathbf{A}^{//}+\mathbf{A}^{\perp } \\
\mathbf{A}^{//} &=&(\mathbf{\alpha \cdot A})\mathbf{\alpha } \\
\mathbf{A}^{\perp } &=&(\mathsf{U}-\mathbf{\alpha \alpha })\cdot \mathbf{A}=%
\mathbf{\alpha }\times (\mathbf{\alpha }\times \mathbf{A})
\end{eqnarray*}
is the component 0f $\mathbf{A}$ perpendicular to $\mathbf{\alpha }$ and $
\mathbf{\alpha }\times \mathbf{A}=\mathbf{\alpha }\times \mathbf{A}^{\perp }$
.
\begin{eqnarray*}
\frac{d}{dt}\mathbf{A}^{\perp } &=&(\mathsf{U}-\mathbf{\alpha \alpha })\cdot
\frac{d\mathbf{A}}{dt}=(\mathsf{U}-\mathbf{\alpha \alpha })\cdot \mathbf{
\alpha }\times \mathbf{A=\alpha }\times \mathbf{A}=\mathbf{\alpha }\times
\mathbf{A}^{\perp } \\
\frac{d^{2}}{dt^{2}}\mathbf{A}^{\perp } &=&\frac{d}{dt}\mathbf{\alpha }
\times \mathbf{A}^{\perp }=\mathbf{\alpha }\times (\mathbf{\alpha }\times
\mathbf{A}^{\perp })=\alpha ^{2}\mathbf{A}^{\perp }=\mathbf{A}^{\perp }
\end{eqnarray*}
A: One way to see that $\vec{A}(t)$ rotates around $\vec{\alpha}$ is that its tangent (or derivative) is always perpendicular to $\vec{\alpha}$ and $\vec{A}$. In order to give more definitive proof it might be easier to write the differential equation is matrix form
$$
\begin{bmatrix}
\dot{A}_x \\ \dot{A}_y \\ \dot{A}_z
\end{bmatrix} = 
\begin{bmatrix}
0 & -\alpha_z & \alpha_y \\
\alpha_z & 0 & -\alpha_x \\
-\alpha_y & \alpha_x & 0
\end{bmatrix}
\begin{bmatrix}
A_x \\ A_y \\ A_z
\end{bmatrix},
$$
where the subscripts $x$, $y$ and $z$ refer to the vector components of $\vec{A}$ and $\vec{\alpha}$. The eigenvalues of the matrix in that equation are $\lambda_1=0$, $\lambda_2=i\|\vec{\alpha}\|$ and $\lambda_3=-i\|\vec{\alpha}\|$. The corresponding eigenvectors are
$$
\vec{v}_1 = \vec{\alpha},
$$
$$
\vec{v}_2 = 
\begin{bmatrix}
\alpha_x \alpha_y - i \alpha_z \|\vec{\alpha}\| \\
\alpha_y^2 - \|\vec{\alpha}\| \\
\alpha_y \alpha_z + i \alpha_x \|\vec{\alpha}\|
\end{bmatrix},
$$
$$
\vec{v}_2 = 
\begin{bmatrix}
\alpha_x \alpha_y + i \alpha_z \|\vec{\alpha}\| \\
\alpha_y^2 - \|\vec{\alpha}\|^2 \\
\alpha_y \alpha_z - i \alpha_x \|\vec{\alpha}\|
\end{bmatrix}.
$$
The first eigenvalue/eigenvector pair shows that any initial condition which is a multiple of $\vec{\alpha}$ will yield a constant solution. The other two pairs are each others complex conjugate and the eigenvalues a completely imaginary, which means a periodic solution. The total general solution can therefore be written as
$$
\vec{A}(t) = C_1 \vec{\alpha} + 
C_2 \left(\cos\left(\|\vec{\alpha}\| t\right) \begin{bmatrix}\alpha_x \alpha_y \\ \alpha_y^2 - \|\vec{\alpha}\|^2 \\ \alpha_y \alpha_z\end{bmatrix} + \sin\left(\|\vec{\alpha}\| t\right) \begin{bmatrix}\alpha_z \|\vec{\alpha}\| \\ 0 \\ -\alpha_x \|\vec{\alpha}\|\end{bmatrix}\right) + 
C_3 \left(\cos\left(\|\vec{\alpha}\| t\right) \begin{bmatrix}-\alpha_z \|\vec{\alpha}\| \\ 0 \\ \alpha_x \|\vec{\alpha}\|\end{bmatrix} + \sin\left(\|\vec{\alpha}\| t\right) \begin{bmatrix}\alpha_x \alpha_y \\ \alpha_y^2 - \|\vec{\alpha}\|^2 \\ \alpha_y \alpha_z\end{bmatrix}\right).
$$
The angular frequency of the periodic movement is thus equal to $\|\vec{\alpha}\|$, however this does not mean that it is equal to the speed of $\vec{A}(t)$ ($\|\dot{\vec{A}}(t)\|$), because this scales linearly with the shortest distance from $\vec{A}$ to the line defined by $\{\mu\vec{\alpha}\; |\; \mu \in \mathbb{R}\}$. One other thing that can be noted is that when taking the dot product between any periodic solution and $\vec{\alpha}$ always yield zero.
