Complexity of Matrix InversionDate: 04/25/2001 at 23:44:28 From: Terri Barron Subject: Linear Algebra Dr Math- What is the computational complexity of inverting an nxn matrix? (In general, not special cases such as a triangular matrix.) By my estimation, using Guassian Elimination, the complexity is O(n^2). Is this close? Are there better algorithms (robust, not case-specific)? Thanks, Terri Date: 04/26/2001 at 11:54:04 From: Doctor Douglas Subject: Re: Linear Algebra Hi Terri, and thanks for writing to Ask Dr. Math. Gaussian Elimination leads to O(n^3) complexity. The usual way to count operations is to count one for each "division" (by a pivot) and one for each "multiply-subtract" when you eliminate an entry. Here's one way of arriving at the O(n^3) result: At the beginning, when the first row has length n, it takes n operations to zero out any entry in the first column (one division, and n-1 multiply-subtracts to find the new entries along the row containing that entry. To get the first column of zeroes therefore takes n(n-1) operations. In the next column, we need (n-1)(n-2) operations to get the second column zeroed out. In the third column, we need (n-2)(n-3) operations. The sum of all of these operations is: n n n n(n+1)(2n+1) n(n+1) SUM i(i-1) = SUM i^2 - SUM i = ------------ - ------ i=1 i=1 i=1 6 2 which goes as O(n^3). To finish the operation count for Gaussian Elimination, you'll need to tally up the operations for the process of back-substitution (you can check that this doesn't affect the leading order of n^3). You might think that the O(n^3) complexity is optimal, but in fact there exists a method (Strassen's method) that requires only O(n^log_2(7)) = O(n^2.807...) operations for a completely general matrix. Of course, there is a constant C in front of the n^2.807. This constant is not small (between 4 and 5), and the programming of Strassen's algorithm is so awkward, that often Gaussian Elimination is still the preferred method. Even Strassen's method is not optimal. I believe that the current record stands at O(n^2.376), thanks to Don Coppersmith and Shmuel Winograd. Here is a Web page that discusses these methods: Fast Parallel Matrix Multiplication - Strategies for Practical Hybrid Algorithms - Erik Ehrling http://www.f.kth.se/~f95-eeh/exjobb/background.html These methods exploit the close relation between matrix inversion and matrix multiplication (which is also an O(n^3) task at first glance). I hope this helps! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/