TITLE: Matrix Differential Equation with a Skew-Symmetric Matrix
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: 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.