Matrix Differential Equation with a Skew-Symmetric Matrix

Question

From a bank of masters exams:

Say the position of a particle moving in $\mathbb{R}^n$ is given by a smooth vector-valued function $\vec{x}(t)$. Suppose that $\vec{x}(t)$ satisfies a differential equation, $$ \frac{d\vec{x}}{dt} = A(t)\vec{x},$$ where $A(t)$ is a real anti-symmetric matrix depending smoothly on $t$. Show that this particle moves on a sphere, that is, $||\vec{x}(t)||$ is constant.

By the spectral theorem, $A$ is normal and therefore has a complete basis of eigenvectors in $\mathbb{C}^n$. I am familiar with the "standard" method of solving for matrix exponentials, i.e. finding the eigenvalues and eigenvectors of $A$, and then using linear combinations of $e^{\lambda t}\vec{x}$ as the solutions, but there is not a complete basis of eigenvectors in $\mathbb{R}$. Taking the matrix exponential $e^A$ doesn't seem to do anything.

Answer 1

Taking from user8268 and Shiyu:

Compute the time derivative of $||\vec{x}||^2 = \vec{x} \cdot \vec{x}$, which becomes

$ \begin{align} \frac{d}{dt} ||\vec{x}||^2 &= \frac{d}{dt} (\vec{x} \cdot \vec{x}) \\ &= \frac{d\vec{x}}{dt} \cdot \vec{x} + \vec{x} \cdot \frac{d\vec{x}}{dt} \\ &= 2 \left( \vec{x} \cdot \frac{d\vec{x}}{dt} \right) \\ &= 2 \left( \vec{x} \cdot A \vec{x} \right) \\ &= 2 \left( \vec{x}^T A \vec{x} \right) = 2(0) = 0 \end{align} $

The last line is true because $\vec{x}^TA\vec{x} = 0$ for all $\vec{x}$ if $A$ is skew-symmetric. Therefore $||\vec{x}||^2$ is constant, implying that $||\vec{x}|| \geq 0$ is constant.

Answer 2

Skew-symmetric matrices have pure imaginary eigenvalues (see http://en.wikipedia.org/wiki/Skew-symmetric_matrix). This means that the matrix will rotate a vector by $\pi/2$ (for odd dimensions there is also a 0 eigenvalue). This implies that the direction of change is always perpendicular to position. Sounds like a sphere to me.