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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.

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just compute the time derivative of $\vec{x}\cdot\vec{x}$ – user8268 Apr 1 '11 at 22:01
user8268 is right. in order to prove $\|x\|$ is constant, just compute its derivative with respect to time. $\|x\|^2=x^Tx$, $dx^Tx=2x^Tdx=2x^TAxdt$. Since $A$ is skew-symmetric, $x^TAx=0$ – Shiyu Apr 2 '11 at 11:04
@Shiyu or user8268: could you explain the differentiation step $d(x^Tx) = 2x^Tdx$? I don't quite understand how differentiation interacts with the transpose. – Michael Chen Apr 2 '11 at 14:38
please refer to – Shiyu Apr 3 '11 at 10:19
@Shiyu: If you'd like to post your comment as an answer, I will accept it. – Michael Chen Apr 5 '11 at 18:03

2 Answers 2

up vote 4 down vote accepted

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.

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Skew-symmetric matrices have pure imaginary eigenvalues (see 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.

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Do you know the solution of that ODE – Rejo_Slash Nov 15 '14 at 11:48
Please provide an example – Tpofofn Nov 15 '14 at 13:35
I am facing same issue. It occurred to me while solving following problem in the link. Please click HERE for stack exchange link. Equation (2) in the link is the example. If I can solve that ODE,it will be very helpful. That ODE in eq 2 is in same format as in this question. Thanks – Rejo_Slash Nov 16 '14 at 6:03

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