I am given two vectors $B,F$ in $\mathbb{R}^3$ with $B=B_1e_1+B_2e_2+B_3e_3$ and $F=F_1e_1+F_3e_3$, where $(e_1,e_2,e_3)$ is the standard basis. Now I am given the differential equation $\dot{B}=F\times B$, where $\times$ is the cross product of two vectors.
I am to show that this describes a gyration of the vector $B$ about the vector $F$ with the angular frequency $\omega=\sqrt{F^2_1+F^2_3}$. First I am going to give you the solution for this differential equation and then I am going to pose a question about it because I do not understand the solution.
First I calculated the cross product: $$F\times B=\begin{pmatrix}F_3B_2\\ F_3B_1 - F_1B_3\\ F_1B_2\end{pmatrix}.$$
This can also be written as the product of a matrix with the vector $B$: $$\begin{pmatrix}F_3B_2\\ F_3B_1 - F_1B_3\\ F_1B_2\end{pmatrix}=\begin{pmatrix}0 & F_3 & 0 \\ F_3 & 0 & -F_1 \\ 0 & F_1 & 0\end{pmatrix}\begin{pmatrix}B_1\\ B_2\\ B_3\end{pmatrix}.$$
Thus, we have $$\begin{pmatrix}\dot{B}_1\\ \dot{B}_2\\ \dot{B}_3\end{pmatrix}=\begin{pmatrix}0 & F_3 & 0 \\ F_3 & 0 & -F_1 \\ 0 & F_1 & 0\end{pmatrix}\begin{pmatrix}B_1\\ B_2\\ B_3\end{pmatrix}.$$
I went on to diagonalise the matrix by calculating the eigenvalues $\{-F,0,F\}$, where $F:=\sqrt{F^2_3-F^2_1}$.
By calculating the eigenvectors we get the transformation matrix $$S=\begin{pmatrix}-F_3/F_1 & F_1/F_3 & F_3/F_1 \\ F/F_1 & 0 & F/F_1 \\ 1 & 1 & 1\end{pmatrix}.$$
Now, in the eigenbasis we have the differential equation $$\begin{pmatrix}\dot{u}_1\\ \dot{u}_2\\ \dot{u}_3\end{pmatrix}=\begin{pmatrix}-F^2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & F^2\end{pmatrix}\begin{pmatrix}u_1\\ u_2\\ u_3\end{pmatrix},$$ which has the solution $$\begin{pmatrix}c_1e^{-F^2t}\\ c_2\\ c_3e^{F^2t}\end{pmatrix}$$ with arbitrary constants $c_{1,2,3}\in\mathbb{R}$.
The constant component of the vector in the eigenbasis seems to be the component parallel to the field vector. However, the other components do not describe a gyrating moition about that axis. Rather, they describe an exponential decay/growth of the component depending on whether $F_1<F_3$,$F_1>F_3$ or $F_1=F_3$. I would expect a term like $cos,sin,\exp{i\omega t}$ to appear somewhere. Or maybe I have made a mistake somewhere? I was not sure of whether I can simply write the cross product as a product of a matrix with a vector. Maybe this is not applicable in this case to decouple the differential equations.
I am not asking for an answer! Any hints, however, are welcome.
This will be a short answer as you have nearly solved it all yourself. The only thing that went wrong is in the calculation of the crossproduct at the beginning. The first component should be $-B_2F_3$.
When you have this the eigenvalues becomes: 0, $\pm \sqrt{-(F_1^2 + F_3^2)} = \pm i \sqrt{F_1^2 + F_2^2}$, and now doing the reasoning as earlier you have your "gyrating frequency".