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.

I have a large matrix, around 10x10. Each individual element in the matrix itself is also a very large number, roughly of the order 10^30.

I know that matrices can be used to solve linear equations. So if there is an equation ax + by = c then can I denote the 10x10 matrix, a, using x&y which are just 2 real numbers and b&c which are matrices. Can this be used to compress the larger matrix?

If not, then is there any other way that I can use so that I can use some form of short expression which can be evaluated into the larger matrix?

share|improve this question
    
For an arbitrary matrix, you can't compress it. Do you know any other information about the structure of the matrix? Could you perhaps post a sample matrix? (If you do, use Stack Exchange's code formatting: put 4 spaces at the start of each line of the matrix.) –  Snowball Oct 21 '12 at 15:29
    
There are many ways to generate large matrices from the small seed (like $n \times n$ identity matrix), but it is generally very hard to do it the opposite direction (i.e. from matrix generate a seed that could be later used to regenerate it) unless the matrix in question has very precisely defined structure (and even then it probably won't be easy). Good luck! –  dtldarek Oct 21 '12 at 15:31
1  
By the way, the theorem under section 9.2 of the comp.compression FAQs justifies my first sentence. Don't be discouraged, though: if your matrix does have some further structure, you should be able to compress it. –  Snowball Oct 21 '12 at 15:33
    
apart from the dimensions of the matrix there is no other pattern in the structure. Perhaps I will find some luck converting the matrix into something else... –  ritratt Oct 21 '12 at 17:33
    
@ritratt: If there really is no pattern at all, then you won't be able to compress it. See that theorem I linked. –  Snowball Oct 21 '12 at 22:14

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.