# THE FACTORIZE OBJECT for solving linear systems

Copyright 2009, Timothy A. Davis, University of Florida. June 4, 2009. davis@cise.ufl.edu http://www.cise.ufl.edu/~davis

This is a demonstration of the FACTORIZE object for solving linear systems and least-squares problems, and for computations with the matrix inverse and pseudo-inverse.

## Contents

- Rule Number One: never multiply by the inverse, inv(A)
- Rule Number Two: never break Rule Number One
- How to use BACKSLASH solve A*x=b
- BACKSLASH versus INV ... let the battle begin
- LU and LINSOLVE are fast and accurate but complicated to use
- INV is easy to use, but slow and inaccurate
- So the winner is ... nobody
- The FACTORIZE object to the rescue
- Least-squares problems
- Underdetermined systems
- Computing selected entries in the inverse or pseudo-inverse
- Computing the entire inverse or pseudo-inverse
- Update/downdate of a dense Cholesky factorization
- Caveat Executor
- Summary

## Rule Number One: never multiply by the inverse, inv(A)

Use backslash or a matrix factorization instead (LU, CHOL, or QR).

## Rule Number Two: never break Rule Number One

However, the problem with Rule Number One is that it can be hard to figure out which matrix factorization to use and how to use it. Using LU, CHOL, or QR is complicated, particularly if you want the best performance. BACKSLASH (MLDIVIDE) is great, but it can't be reused when solving multiple systems (x=A\b and y=A\c). Its syntax doesn't match the use of the inverse in mathematical expressions, either.

The goal of the FACTORIZE object is to solve this problem ...

"Don't let that INV go past your eyes; to solve that system, FACTORIZE!"

## How to use BACKSLASH solve A*x=b

First, let's create a square matrix A and a right-hand-side b for a linear system A*x=b. There are many ways to solve this system. The best way is to use x=A\b. The residual r is a vector of what's left over in each equation, and its norm tells you how accurately the system was solved.

    format compact ;
    A = rand (3)
    b = rand (3,1)
    x = A\b
    r = b-A*x ;
    norm (r)

    A =
        0.8241    0.0121    0.9128
        0.7273    0.4032    0.5886
        0.5890    0.9468    0.7197
    b =
        0.3478
        0.5430
        0.6993
    x =
        0.6294
        0.4942
       -0.1937
    ans =
         0

## BACKSLASH versus INV ... let the battle begin

The backslash operation x=A\b is mathematically the same as x=inv(A)*b. However, backslash is faster and more accurate since it uses a matrix factorization instead of multiplying by the inverse. Even though your linear algebra textbook might write x=A^(-1)*b as the solution to the system A*x=b, your textbook author never means for you to compute the inverse.

These next statements give the same answer, so what's the big deal?

    S = inv(A) ;
    x = S*b
    x = A\b

    x =
        0.6294
        0.4942
       -0.1937
    x =
        0.6294
        0.4942
       -0.1937

The big deal is that you should care about speed and you should care even more about accuracy. BACKSLASH relies on matrix factorization (LU, CHOL, QR, or other specialized methods). It's faster and more reliable than multiplying by the inverse, particularly for large matrices and sparse matrices. Here's an illustration of how pathetic inv(A)*b can be.

    A = gallery ('frank',16) ; xtrue = ones (16,1) ; b = A*xtrue ;

    x = inv(A)*b ; norm (b-A*x)
    x = A\b      ; norm (b-A*x)

    ans =
        0.0145
    ans =
       1.7764e-15

The performance difference between BACKSLASH and INV for even small sparse matrices is striking.

    load west0479 ;
    A = west0479 ;
    n = size (A,1)
    b = rand (n,1) ;
    tic ; x = A\b ; toc
    norm (b-A*x)
    tic ; x = inv(A)*b ; toc
    norm (b-A*x)

    n =
       479
    Elapsed time is 0.002805 seconds.
    ans =
       9.8678e-11
    Elapsed time is 0.116907 seconds.
    ans =
       1.5276e-09

What if you want to solve multiple systems? Use a matrix factorization. But which one? And how do you use it? Here are some alternatives using LU for the sparse west0479 matrix, but some are faster than others.

    tic ; [L,U]     = lu(A) ; x1 = U \ (L \ b)         ; t1=toc ; nz1=nnz(L+U);
    tic ; [L,U,P]   = lu(A) ; x2 = U \ (L \ P*b)       ; t2=toc ; nz2=nnz(L+U);
    tic ; [L,U,P,Q] = lu(A) ; x3 = Q * (U \ (L \ P*b)) ; t3=toc ; nz3=nnz(L+U);

    fprintf ('1: nnz(L+U): %5d time: %8.4f resid: %e\n', nz1,t1, norm(b-A*x1));
    fprintf ('2: nnz(L+U): %5d time: %8.4f resid: %e\n', nz2,t2, norm(b-A*x2));
    fprintf ('3: nnz(L+U): %5d time: %8.4f resid: %e\n', nz3,t3, norm(b-A*x3));

    1: nnz(L+U): 16151 time:   0.0070 resid: 1.426954e-10
    2: nnz(L+U): 15826 time:   0.0147 resid: 9.176936e-11
    3: nnz(L+U):  3704 time:   0.0052 resid: 9.058758e-11

## LU and LINSOLVE are fast and accurate but complicated to use

A quick look at ``help lu'' will scroll off your screen. For full matrices, [L,U,p] = lu (A,'vector') is fastest. Then for the forward/backsolves, use LINSOLVE instead of BACKSLASH for even faster performance. But for sparse matrices, use the optional 'Q' output of LU so you get a good fill-reducing ordering. But you can't use 'Q' if the matrix is full. But LINSOLVE doesn't work on sparse matrices.

But ... Ack! That's getting complicated ...

Here's the best way to solve A*x=b and A*y=c when A is full and unsymmetric:

    n = 1000 ;
    A = rand (n) ;
    b = rand (n,1) ;
    c = rand (n,1) ;
    tic ; [L,U,p] = lu (A, 'vector') ; LUtime = toc

    tic ; x = U \ (L \ b (p,:)) ;
          y = U \ (L \ c (p,:)) ; toc

    tic ; opL = struct ('LT', true) ;
          opU = struct ('UT', true) ;
          x = linsolve (U, linsolve (L, b(p,:), opL), opU) ;
          y = linsolve (U, linsolve (L, c(p,:), opL), opU) ; toc

    LUtime =
        0.3043
    Elapsed time is 0.014511 seconds.
    Elapsed time is 0.006586 seconds.

## INV is easy to use, but slow and inaccurate

Oh bother! Using LU and LINSOLVE is too complicated. You just want to solve your system. Let's just compute inv(A) and use it twice. Easy to write, but slower and less accurate ...

    S = inv (A) ;
    x = S*b ; norm (b-A*x)
    y = S*c ; norm (c-A*y)

    ans =
       3.5716e-11
    ans =
       7.7556e-12

Sometimes using the inverse seems inevitable. For example, your textbook might show the Schur complement formula as S = A-B*inv(D)*C. This can be done without inv(D) in one of two ways: SLASH or BACKSLASH (MRDIVIDE or MLDIVIDE to be precise).

inv(A)*B and A\B are mathematically equivalent, as are B*inv(A) and B/A, so these three methods give the same results (ignoring computational errors, which are worse for inv(D)). Only the first equation looks like the equation in your textbook, however.

    A = rand (200) ; B = rand (200) ; C = rand (200) ; D = rand (200) ;

    tic ; S1 = A - B*inv(D)*C ; toc ;
    tic ; S2 = A - B*(D\C) ;    toc ;
    tic ; S3 = A - (B/D)*C ;    toc ;

    Elapsed time is 0.020362 seconds.
    Elapsed time is 0.015951 seconds.
    Elapsed time is 0.017257 seconds.

## So the winner is ... nobody

BACKSLASH: mostly simple to use (except remember that Schur complement formula?). Fast and accurate ... but slow if you want to solve two linear systems with the same matrix A.

LU, QR, CHOL: fast and accurate. Awful syntax to use. Drag out your linear algebra textbook if you want to use these in MATLAB. Whenever I use them I have to derive them from scratch, even though I wrote most of the sparse factorizations used in MATLAB!

INV: slow and inaccurate. Wins big on ease-of-use, though, since it's a direct plug-in for all your nice mathematical formulas.

No method is best on all three criterion: speed, accuracy, and ease of use.

Is there a solution? Yes ... keeping reading ...

## The FACTORIZE object to the rescue

The FACTORIZE method is just as easy to use as INV, but just as fast and accurate as BACKSLASH, LU, QR, CHOL, and LINSOLVE.

F = factorize(A) computes the factorization of A and returns it as an object that you can reuse to solve a linear system with x=F\b. It picks LU, QR, or Cholesky for you, just like BACKSLASH.

S = inverse(A) is simpler yet. It does NOT compute inv(A), but factorizes A. When multiplying S*b, it doesn't mulitply by the inverse, but uses the correct forward/backsolve equations to solve the linear system.

    n = 1000 ;
    A = rand (n) ;
    b = rand (n,1) ;
    c = rand (n,1) ;

    tic ;                       x = A\b ; y = A\c ; toc
    tic ; S = inv(A) ;          x = S*b ; y = S*c ; toc
    tic ; F = factorize(A) ;    x = F\b ; y = F\c ; toc
    tic ; S = inverse(A) ;      x = S*b ; y = S*c ; toc

    Elapsed time is 0.638810 seconds.
    Elapsed time is 0.813682 seconds.
    Elapsed time is 0.317389 seconds.
    Elapsed time is 0.319094 seconds.

## Least-squares problems

Here are some different methods for solving a least-squares problem when your system is over-determined. The last two methods are the same.

    A = rand (1000,200) ;
    b = rand (1000,1) ;

    tic ; x = A\b            ; toc, norm (A'*A*x-A'*b)
    tic ; x = pinv(A)*b      ; toc, norm (A'*A*x-A'*b)
    tic ; x = inverse(A)*b   ; toc, norm (A'*A*x-A'*b)
    tic ; x = factorize(A)\b ; toc, norm (A'*A*x-A'*b)

    Elapsed time is 0.056897 seconds.
    ans =
       2.7016e-12
    Elapsed time is 0.338749 seconds.
    ans =
       2.3034e-12
    Elapsed time is 0.079205 seconds.
    ans =
       3.0968e-12
    Elapsed time is 0.079404 seconds.
    ans =
       3.0968e-12

FACTORIZE is better than BACKSLASH because you can reuse the factorization for different right-hand-sides. For full-rank matrices, it's better than PINV because it's faster (and PINV fails for sparse matrices).

    A = rand (1000,200) ;
    b = rand (1000,1) ;
    c = rand (1000,1) ;

    tic ;                  ; x = A\b ; y = A\c ; toc
    tic ; S = pinv(A)      ; x = S*b ; y = S*c ; toc
    tic ; S = inverse(A)   ; x = S*b ; y = S*c ; toc
    tic ; F = factorize(A) ; x = F\b ; y = F\c ; toc

    Elapsed time is 0.116545 seconds.
    Elapsed time is 0.341485 seconds.
    Elapsed time is 0.081160 seconds.
    Elapsed time is 0.082303 seconds.

## Underdetermined systems

The under-determined system A*x=b where A has more columns than rows has many solutions. x=A\b finds a basic solution (some of the entries in x are zero). pinv(A)*b finds a minimum 2-norm solution, but it's slow. QR factorization will do the same if A has full rank. That's what the factorize(A) and inverse(A) methods do.

    A = rand (200,1000) ;
    b = rand (200,1) ;

    tic ; x = A\b            ; toc, norm (x)
    tic ; x = pinv(A)*b      ; toc, norm (x)
    tic ; x = inverse(A)*b   ; toc, norm (x)
    tic ; x = factorize(A)\b ; toc, norm (x)

    Elapsed time is 0.078900 seconds.
    ans =
        3.2259
    Elapsed time is 0.356762 seconds.
    ans =
        0.4610
    Elapsed time is 0.082058 seconds.
    ans =
        0.4610
    Elapsed time is 0.082241 seconds.
    ans =
        0.4610

## Computing selected entries in the inverse or pseudo-inverse

If you want just a few entries from the inverse, it's still better to formulate the problem as a system of linear equations and use a matrix factorization instead of computing inv(A). The FACTORIZE object does this for you, by overloading the subsref operator.

    A = rand (1000) ;

    tic ; S = inv (A)     ; S (2:3,4), toc
    tic ; S = inverse (A) ; S (2:3,4), toc

    ans =
        0.4632
       -0.0876
    Elapsed time is 0.812422 seconds.
    ans =
        0.4632
       -0.0876
    Elapsed time is 0.324822 seconds.

## Computing the entire inverse or pseudo-inverse

Rarely, and I mean RARELY, you really do need the inverse. More frequently what you want is the pseudo-inverse. You can force a factorization to become a plain matrix by converting it to double. Note that inverse(A) only handles full-rank matrices (either dense or sparse), whereas pinv(A) works for all dense matrices (not sparse).

The explicit need for inv(A) (or S=A\eye(n), which is the same thing) is RARE. If you ever find yourself multiplying by the inverse, then you know one thing for sure. You know with certainty that you don't know what you're doing.

    A = rand (500) ;
    tic ; S1 = inv (A) ;            ; toc
    tic ; S2 = double (inverse (A)) ; toc
    norm (S1-S2)

    A = rand (500,400) ;
    tic ; S1 = pinv (A)             ; toc
    tic ; S2 = double (inverse (A)) ; toc
    norm (S1-S2)

    Elapsed time is 0.108800 seconds.
    Elapsed time is 0.147705 seconds.
    ans =
       3.1817e-11
    Elapsed time is 1.680637 seconds.
    Elapsed time is 0.193965 seconds.
    ans =
       3.8189e-14

## Update/downdate of a dense Cholesky factorization

Wilkinson considered the update/downdate of a matrix factorization to be a key problem in computational linear algebra. The idea is that you first factorize a matrix. Next, make a low-rank change to A, and patch up (or down...) the factorization so that it becomes the factorization of the new matrix. In MATLAB, this only works for dense symmetric positive definite matrices, via cholupdate. This is much faster than computing the new factorization from scratch.

    n = 1000 ;
    A = rand (n) ;
    A = A*A' + n*eye (n) ;
    w = rand (n,1) ; t = rand (n,1) ; b = rand (n,1) ;
    F = factorize (A) ;

    tic ; F = F + w ; x = F\b ; toc
    tic ; y = (A+w*w')\b ;      toc
    norm (x-y)

    tic ; F = F - t ; x = F\b ; toc
    tic ; y = (A+w*w'-t*t')\b ; toc
    norm (x-y)

    Elapsed time is 0.032744 seconds.
    Elapsed time is 0.198138 seconds.
    ans =
       3.3726e-17
    Elapsed time is 0.035119 seconds.
    Elapsed time is 0.207019 seconds.
    ans =
       3.4408e-17

## Caveat Executor

One caveat: If you have a large number of very small systems to solve, the object-oriented overhead of creating and using an object can dominate the run time, at least in MATLAB R2009a. For this case, if you want the best performance, stick with BACKSLASH, or LU and LINSOLVE (just extract the appropriate formulas from the M-files in the FACTORIZE package).

Hopefully the object-oriented overhead will drop in future versions of MATLAB, and you can ignore this caveat.

    A = rand (10) ; b = rand (10,1) ; F = factorize (A) ;

    tic ; for k = 1:10000, x = F\b ; end ; toc

    tic ; for k = 1:10000, x = A\b ; end ; toc

    [L,U,p] = lu (A, 'vector') ;
    opL = struct ('LT', true) ;
    opU = struct ('UT', true) ;
    tic ;
    for k = 1:10000
        x = linsolve (U, linsolve (L, b(p,:), opL), opU) ;
    end
    toc

    Elapsed time is 2.324888 seconds.
    Elapsed time is 0.336457 seconds.
    Elapsed time is 0.219211 seconds.

## Summary

So ... don't use INV, and don't worry about how to use LU, CHOL, or QR factorization. Just install the FACTORIZE package, and you're on your way. Assuming you are now in the Factorize/ directory, cut-and-paste these commands into your command window:

     addpath (pwd)
     savepath

And remember ...

"Don't let that INV go past your eyes; to solve that system, FACTORIZE!"