Don’t invert that matrix

There is hardly ever a good reason to invert a matrix.

What do you do if you need to solve Ax = b where A is an n x n matrix? Isn’t the solution A-1 b? Yes, theoretically. But that doesn’t mean you need to actually find A-1. Solving the equation Ax = b is faster than finding A-1. Books might write the problem as x = A-1 b, but that doesn’t mean they expect you to calculate it that way.

What if you have to solve Ax = b for a lot of different b‘s? Surely then it’s worthwhile to find A-1. No. The first time you solve Ax = b, you factor A and save that factorization. Then when you solve for the next b, the answer comes much faster. (Factorization takes O(n3) operations. But once the matrix is factored, solving Ax = b takes only O(n2) operations. Suppose n = 1,000. This says that once you’ve solved Ax = b for one b, the equation can be solved again for a new b 1,000 times faster than the first one. Buy one get one free.)

What if, against advice, you’ve computed A-1. Now you might as well use it, right? No, you’re still better off solving Ax = b than multiplying by A-1, even if the computation of A-1 came for free. Solving the system is more numerically accurate than the performing the matrix multiplication.

It is common in applications to solve Ax = b even though there’s not enough memory to store A-1. For example, suppose n = 1,000,000 for the matrix A but A has a special sparse structure — say it’s banded — so that all but a few million entries of A are zero.  Then A can easily be stored in memory and Ax = b can be solved very quickly. But in general A-1 would be dense. That is, nearly all of the 1,000,000,000,000 entries of the matrix would be non-zero.  Storing A requires megabytes of memory, but storing A-1 would require terabytes of memory.

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Related post: Applied linear algebra

106 thoughts on “Don’t invert that matrix

  1. I have systems of equations that fit nicely in an Ax=b matrix format and usually solve them by inverting A. They’re relatively small (also sparse but not banded and not necessarily positive definite), so this works well. However, I have to make incremental changes in the values of A (say, change two of the three non-zero values in one row) and find the corresponding x and have to do this many, many times. Can factoring help me preserve the work from previous iterations and reduce my ridiculous run times?

  2. If you change A then you have a new problem and so you can’t reuse the old factorization.

    On the other hand, if your change to A is small, say A’ = A + E where E has small norm, then a solution x to Ax = b may be approximately a solution to A’x = b. If you’re solving the latter by an iterative method, then you could give it a head start by using the old x as your starting point.

  3. Great Post!

    Here is a follow-up question: what if I have a matrix A and vector b and I need the quantity b’A^{-1}b. Of course I can get A^{-1} explicitly and compute the product. This is very unstable, even in small examples. What would be the “solving the system” formulation of this problem?

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