Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let a square $n\times n$ real matrix ${\bf A}$ and two vectors ${\bf x}$ and ${\bf b}$ of length $n$, such that $${\bf A}{\bf x}={\bf b}.$$ Solving for ${\bf x}$ through standard Gaussian Elimination yields an aggregate complexity of almost $O(n^3)$. However, there are cases where solving (or $\epsilon$-approximately solving) for ${\bf x}$ costs $O(n\log^\rho n)$, such as systems where ${\bf A}$ is a symmetric and diagonally dominant matrix (e.g., a Laplacian) [1].

Which other families of linear systems (i.e., matrices) admit linear (or nontrivial poly(n)) time solutions? If we consider finite fields instead of real matrices, are there any families of matrices there that admit nearly linear time solutions?

[1] http://www.cs.yale.edu/homes/spielman/Research/linsolve.html

share|improve this question

2 Answers 2

Trigonometric matrix algebras (circulant matrices, Hartley, DCT/DST) are another example.

share|improve this answer

Which other families of linear systems (i.e., matrices) admit linear (or nontrivial poly(n)) time solutions?

Besides the symmetric diagonally dominant matrices you mention, I think that any positive definite matrix $A$ that can be Cholesky decomposed quickly gives you $x$ quickly.

Added: You can find $x$ quickly if $A$ has "low displacement rank" and therefore has "displacement structure."

http://web.eecs.utk.edu/~dongarra/etemplates/node389.html

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.8792

http://www.sciencedirect.com/science/article/pii/S0024379500002615

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.