Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Ax = b. I need a way to analyze a square matrix A to see if its solution vector x will always be positive when b is positive.

This question arises from solving the radiosity equation:

alt text

I'm interested to know when A is incorrect, which would be when x has negative values even though b is positive.

share|cite|improve this question
up vote 5 down vote accepted

If all entries of $A^{-1}$ are positive numbers, then $A$ has the property you desire.

(Edit: This condition is both sufficient and necessary. If one entry $A^{-1}_{ij}$ is negative, then the choice $b=e_j$ will make the $i$-th component of $x=A^{-1}b$ equal to $A^{-1}_{ij}$, but this is negative.)

The following argument shows that your particular example matrix does have this property: your matrix $A$ has the form

$A = 1 - R F$

where $R=\mathrm{diag}(\rho_1,\rho_2,\dots,\rho_n)$ is the diagonal matrix of reflectivities and $F$ the matrix of form factors. The physical interpretation of $R$ and $F$ makes sure that the norm of $RF$ will be smaller than 1, so that $A^{-1}$ can be expressed as a geometric series

$A^{-1} = (1- RF)^{-1} = 1 + RF + (RF)^2 + (RF)^3 + \dots$

Now, since all entries of $RF$ are positive numbers, this formula implies that the entries of $A^{-1}$ are positive numbers as well.

share|cite|improve this answer

Your Answer


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.