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.

Suppose we have a matrix $M$ whose elements are rational functions of $x$ with the same denominator.

A simple example

$M=\left[\begin{matrix} \frac{x}{a^2+x^2} & \frac{-a}{a^2+x^2}\\ \frac{a}{a^2+x^2} & \frac{x}{a^2+x^2}\end{matrix}\right]$


$M=\left[\begin{matrix} a^2+x^2 & 0\\ 0 & a^2+x^2\end{matrix}\right]^{-1}\left[\begin{matrix} x & -a\\ a & x\end{matrix}\right]$

but also

$M=\left[\begin{matrix} x & a\\ -a & x\end{matrix}\right]^{-1}\left[\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right]$

The first expression is obvious, but the second is lower-degree and preferable. Is there a systematic way to find the lowest-degree expression of the form $M=(\text{numerator matrix})^{-1}(\text{denominator matrix})$, given $M$ ?

(i.e. the numerator matrix and denominator matrix must contain polynomials in $x$ whose degree is as low as possible)


Thinking further, I notice that $\text{det}\left( \left[\begin{matrix} x & a\\ -a & x\end{matrix}\right]\right)=x^2+a^2$ and $\text{det}\left(\left[\begin{matrix} a^2+x^2 & 0\\ 0 & a^2+x^2\end{matrix}\right]\right)=(x^2+a^2)^2$ , i.e. the determinants of the "numerator" and "denominator" of the first expression have common zeros, which is not the case for the second expression. It could well be that this lack of common zeros uniquely determines the "simplest" expression.

share|improve this question
what exactly do you mean by 'lower degree'? Only linear in $x$? This is not quite correct, calculating the inverse your second expression involves calculation of the determinant, which is quadratic in $x$. It only seems to look linear. –  user20266 Dec 20 '11 at 17:56
This occurs in an equation, and I can multiply both sides with the "numerator matrix" to get rid of the inverse. The result is an equality between matrices which contain polynomials, and I seek the lowest-degree form of that expression. –  Wouter Dec 20 '11 at 18:15

Your Answer


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

Browse other questions tagged or ask your own question.