Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
7  
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 en.wikipedia.org/wiki/Matrix_calculus –  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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.