LSQR method
x = lsqr(A,b)
lsqr(A,b,tol)
lsqr(A,b,tol,maxit)
lsqr(A,b,tol,maxit,M)
lsqr(A,b,tol,maxit,M1,M2)
lsqr(A,b,tol,maxit,M1,M2,x0)
[x,flag] = lsqr(A,b,tol,maxit,M1,M2,x0)
[x,flag,relres] = lsqr(A,b,tol,maxit,M1,M2,x0)
[x,flag,relres,iter] = lsqr(A,b,tol,maxit,M1,M2,x0)
[x,flag,relres,iter,resvec] = lsqr(A,b,tol,maxit,M1,M2,x0)
[x,flag,relres,iter,resvec,lsvec]
= lsqr(A,b,tol,maxit,M1,M2,x0)
x = lsqr(A,b)
attempts
to solve the system of linear equations A*x=b
for x
if A
is
consistent, otherwise it attempts to solve the least squares solution x
that
minimizes norm(bA*x)
. The m
byn
coefficient
matrix A
need not be square but it should be large
and sparse. The column vector b
must have length m
.
You can specify A
as a function handle, afun
,
such that afun(x,'notransp')
returns A*x
and afun(x,'transp')
returns A'*x
.
Parameterizing Functions explains how to provide additional
parameters to the function afun
, as well as the
preconditioner function mfun
described below,
if necessary.
If lsqr
converges, a message to that effect
is displayed. If lsqr
fails to converge after
the maximum number of iterations or halts for any reason, a warning
message is printed displaying the relative residual norm(bA*x)/norm(b)
and
the iteration number at which the method stopped or failed.
lsqr(A,b,tol)
specifies
the tolerance of the method. If tol
is []
,
then lsqr
uses the default, 1e6
.
lsqr(A,b,tol,maxit)
specifies
the maximum number of iterations.
lsqr(A,b,tol,maxit,M)
and lsqr(A,b,tol,maxit,M1,M2)
use n
byn
preconditioner M
or M
= M1*M2
and effectively solve the system A*inv(M)*y
= b
for y
, where y = M*x
.
If M
is []
then lsqr
applies
no preconditioner. M
can be a function mfun
such
that mfun(x,'notransp')
returns M\x
and mfun(x,'transp')
returns M'\x
.
lsqr(A,b,tol,maxit,M1,M2,x0)
specifies
the n
by1
initial guess. If x0
is []
,
then lsqr
uses the default, an all zero vector.
[x,flag] = lsqr(A,b,tol,maxit,M1,M2,x0)
also
returns a convergence flag.
Flag  Convergence 

0 

1 

2  Preconditioner 
3 

4  One of the scalar quantities calculated during 
Whenever flag
is not 0
,
the solution x
returned is that with minimal norm
residual computed over all the iterations. No messages are displayed
if you specify the flag
output.
[x,flag,relres] = lsqr(A,b,tol,maxit,M1,M2,x0)
also
returns an estimate of the relative residual norm(bA*x)/norm(b)
.
If flag
is 0
, relres
<= tol
.
[x,flag,relres,iter] = lsqr(A,b,tol,maxit,M1,M2,x0)
also
returns the iteration number at which x
was computed,
where 0 <= iter <= maxit
.
[x,flag,relres,iter,resvec] = lsqr(A,b,tol,maxit,M1,M2,x0)
also
returns a vector of the residual norm estimates at each iteration,
including norm(bA*x0)
.
[x,flag,relres,iter,resvec,lsvec]
= lsqr(A,b,tol,maxit,M1,M2,x0)
also returns a vector of
estimates of the scaled normal equations residual at each iteration: norm((A*inv(M))'*(BA*X))/norm(A*inv(M),'fro')
.
Note that the estimate of norm(A*inv(M),'fro')
changes,
and hopefully improves, at each iteration.
n = 100; on = ones(n,1); A = spdiags([2*on 4*on on],1:1,n,n); b = sum(A,2); tol = 1e8; maxit = 15; M1 = spdiags([on/(2) on],1:0,n,n); M2 = spdiags([4*on on],0:1,n,n); x = lsqr(A,b,tol,maxit,M1,M2);
displays the following message:
lsqr converged at iteration 11 to a solution with relative residual 3.5e009
This example replaces the matrix A
in Example
1 with a handle to a matrixvector product function afun
.
The example is contained in a function run_lsqr
that
Calls lsqr
with the function
handle @afun
as its first argument.
Contains afun
as a nested function,
so that all variables in run_lsqr
are available
to afun
.
The following shows the code for run_lsqr
:
function x1 = run_lsqr n = 100; on = ones(n,1); A = spdiags([2*on 4*on on],1:1,n,n); b = sum(A,2); tol = 1e8; maxit = 15; M1 = spdiags([on/(2) on],1:0,n,n); M2 = spdiags([4*on on],0:1,n,n); x1 = lsqr(@afun,b,tol,maxit,M1,M2); function y = afun(x,transp_flag) if strcmp(transp_flag,'transp') % y = A'*x y = 4 * x; y(1:n1) = y(1:n1)  2 * x(2:n); y(2:n) = y(2:n)  x(1:n1); elseif strcmp(transp_flag,'notransp') % y = A*x y = 4 * x; y(2:n) = y(2:n)  2 * x(1:n1); y(1:n1) = y(1:n1)  x(2:n); end end end
When you enter
x1=run_lsqr;
MATLAB^{®} software displays the message
lsqr converged at iteration 11 to a solution with relative residual 3.5e009
[1] Barrett, R., M. Berry, T. F. Chan, et al., Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, SIAM, Philadelphia, 1994.
[2] Paige, C. C. and M. A. Saunders, "LSQR: An Algorithm for Sparse Linear Equations And Sparse Least Squares," ACM Trans. Math. Soft., Vol.8, 1982, pp. 4371.