## Saturday, March 10, 2012

### Solving overdetermined systems with the QR decomposition

A system of linear equations is considered overdetermined if there are more equations than unknowns. In practice, we have a system Ax=b where A is a m by n matrix and b is a m dimensional vector b but m is greater than n. In this case, the vector b cannot be expressed as a linear combination of the columns of A. Hence, we can't find x so that satisfies the problem Ax=b (except in specific cases) but it is possible to determine x so that Ax is as close to b as possible. So we wish to find x which minimizes the following error

Considering the QR decomposition of A we have that Ax=b becomes

multiplying by Q^T we obtain

and since Q^T is orthogonal (this means that Q^T*Q=I) we have

Now, this is a well defined system, R is an upper triangular matrix and Q^T*b is a vector. More precisely b is the orthogonal projection of b onto the range of A. And,

The function linalg.lstsq() provided by numpy returns the least-squares solution to a linear system equation and is able to solve overdetermined systems. Let's compare the solutions of linalg.lstsq() with the ones computed using the QR decomposition:
```from numpy import *

# generating a random overdetermined system
A = random.rand(5,3)
b = random.rand(5,1)

x_lstsq = linalg.lstsq(A,b)[0] # computing the numpy solution

Q,R = linalg.qr(A) # qr decomposition of A
Qb = dot(Q.T,b) # computing Q^T*b (project b onto the range of A)
x_qr = linalg.solve(R,Qb) # solving R*x = Q^T*b

# comparing the solutions
print 'qr solution'
print x_qr
print 'lstqs solution'
print x_lstsq
```
This is the output of the script above:
```qr solution
[[ 0.08704059]
[-0.10106932]
[ 0.56961487]]
lstqs solution
[[ 0.08704059]
[-0.10106932]
[ 0.56961487]]
```
As we can see, the solutions are the same.

1. Thanks! This is helping me get some image processing homework done.

1. I'm always happy to help you David :)

2. Succinct. Thank you.

3. If speed is important to you, you can use the triangular nature of the `R` matrix and solve it much faster using solve_triangular (in scipy.linalg):

import scipy
x_qr2 = scipy.linalg.solve_triangular(R, Qb, check_finite=False)

This is 5.5x (or more) faster than numpy.linalg.solve, for 160 by 160 matrixes at least.

4. Is there a way to get nonnegative solution for underdertermined system for Ax=b

5. Is there a way to get nonnegative solution for underdertermined system for Ax=b

1. Of course, but it is a different kind of optimization problem.