minres

Solve system of linear equations — minimum residual method

Description

example

x = minres(A,b) attempts to solve the system of linear equations A*x = b for x using the Minimum Residual Method. When the attempt is successful, minres displays a message to confirm convergence. If minres fails to converge after the maximum number of iterations or halts for any reason, it displays a diagnostic message that includes the relative residual norm(b-A*x)/norm(b) and the iteration number at which the method stopped.

example

x = minres(A,b,tol) specifies a tolerance for the method. The default tolerance is 1e-6.

example

x = minres(A,b,tol,maxit) specifies the maximum number of iterations to use. minres displays a diagnostic message if it fails to converge within maxit iterations.

example

x = minres(A,b,tol,maxit,M) specifies a symmetric positive definite preconditioner matrix M and computes x by effectively solving the system ${H}^{-1}A\text{\hspace{0.17em}}{H}^{-T}y={H}^{-1}b$ for y, where $y={H}^{T}x$ and $H={M}^{1/2}={\left({M}_{1}{M}_{2}\right)}^{1/2}$. The algorithm does not form H explicitly. Using a preconditioner matrix can improve the numerical properties of the problem and the efficiency of the calculation.

example

x = minres(A,b,tol,maxit,M1,M2) specifies factors of the preconditioner matrix M such that M = M1*M2.

example

x = minres(A,b,tol,maxit,M1,M2,x0) specifies an initial guess for the solution vector x. The default is a vector of zeros.

example

[x,flag] = minres(___) returns a flag that specifies whether the algorithm successfully converged. When flag = 0, convergence was successful. You can use this output syntax with any of the previous input argument combinations. When you specify the flag output, minres does not display any diagnostic messages.

example

[x,flag,relres] = minres(___) also returns the residual error in the computed solution. If flag is 0, then relres <= tol.

example

[x,flag,relres,iter] = minres(___) also returns the iteration number iter at which x was computed.

example

[x,flag,relres,iter,resvec] = minres(___) also returns a vector of the residual norm at each iteration, including the first residual norm(b-A*x0).

example

[x,flag,relres,iter,resvec,resveccg] = minres(___) also returns a vector of the conjugate gradients residual norms at each iteration.

Examples

collapse all

Solve a square linear system using minres with default settings, and then adjust the tolerance and number of iterations used in the solution process.

Create a sparse symmetric tridiagonal matrix A as the coefficient matrix. Use the row sums of A as the vector b for the right-hand side of $\mathrm{Ax}=\mathit{b}$ so that the solution $\mathit{x}$ is expected to be a vector of ones.

n = 400;
on = ones(n,1);
A = spdiags([-2*on 4*on -2*on],-1:1,n,n);
b = sum(A,2);

Solve $\mathrm{Ax}=\mathit{b}$ using minres. The output display includes the value of the relative residual error $\frac{‖\mathit{b}-\mathrm{Ax}‖}{‖\mathit{b}‖}$.

x = minres(A,b);
minres stopped at iteration 20 without converging to the desired tolerance 1e-06
because the maximum number of iterations was reached.
The iterate returned (number 20) has relative residual 0.017.

By default minres uses 20 iterations and a tolerance of 1e-6, and the algorithm is unable to converge in those 20 iterations for this matrix. Since the residual is on the order of 1e-2, it is a good indicator that more iterations are needed. You also can use a larger tolerance to make it easier for the algorithm to converge.

Solve the system again using a tolerance of 1e-4 and 250 iterations.

x = minres(A,b,1e-4,250);
minres converged at iteration 200 to a solution with relative residual 7e-13.

Examine the effect of using a preconditioner matrix with minres to solve a linear system.

Create a symmetric positive definite, banded coefficient matrix.

A = delsq(numgrid('S',102));

Define b so that the true solution to $\mathrm{Ax}=\mathit{b}$ is a vector of all ones.

b = sum(A,2);

Set the tolerance and maximum number of iterations.

tol = 1e-12;
maxit = 100;

Use minres to find a solution at the requested tolerance and number of iterations. Specify six outputs to return information about the solution process:

• x is the computed solution to A*x = b.

• fl0 is a flag indicating whether the algorithm converged.

• rr0 is the relative residual of the computed answer x.

• it0 is the iteration number when x was computed.

• rv0 is a vector of the residual history for $‖\mathit{b}-\mathrm{Ax}‖$.

• rvcg0 is a vector of the conjugate gradient residual history for $‖{\mathit{A}}^{\mathit{T}}\mathit{A}\text{\hspace{0.17em}}\mathit{x}-{\mathit{A}}^{\mathit{T}}\mathit{b}‖$.

[x,fl0,rr0,it0,rv0,rvcg0] = minres(A,b,tol,maxit);
fl0
fl0 = 1
rr0
rr0 = 0.0013
it0
it0 = 100

fl0 is 1 because minres does not converge to the requested tolerance 1e-12 within the requested 100 iterations.

To aid with convergence, you can specify a preconditioner matrix. Since A is symmetric, use ichol to generate the preconditioner $\mathit{M}=\mathit{L}\text{\hspace{0.17em}}{\mathit{L}}^{\mathit{T}}$. Specify the 'ict' option to use incomplete Cholesky factorization with threshold dropping, and specify a diagonal shift value of 1e-6 to avoid nonpositive pivots. Solve the preconditioned system by specifying L and L' as inputs to minres.

setup = struct('type','ict','diagcomp',1e-6,'droptol',1e-14);
L = ichol(A,setup);
[x1,fl1,rr1,it1,rv1,rvcg1] = minres(A,b,tol,maxit,L,L');
fl1
fl1 = 0
rr1
rr1 = 2.4539e-15
it1
it1 = 4

The use of an ilu preconditioner produces a relative residual less than the prescribed tolerance of 1e-12 at the fourth iteration. The output rv1(1) is norm(b), and the output rv1(end) is norm(b-A*x1).

You can follow the progress of minres by plotting the relative residuals at each iteration. Plot the residual history of each solution with a line for the specified tolerance.

semilogy(0:length(rv0)-1,rv0/norm(b),'-o')
hold on
semilogy(0:length(rv1)-1,rv1/norm(b),'-o')
yline(tol,'r--');
legend('No preconditioner','ICHOL preconditioner','Tolerance','Location','East')
xlabel('Iteration number')
ylabel('Relative residual') Examine the effect of supplying minres with an initial guess of the solution.

Create a tridiagonal sparse matrix. Use the sum of each row as the vector for the right-hand side of $\mathrm{Ax}=\mathit{b}$ so that the expected solution for $\mathit{x}$ is a vector of ones.

n = 900;
e = ones(n,1);
A = spdiags([e 2*e e],-1:1,n,n);
b = sum(A,2);

Use minres to solve $\mathrm{Ax}=\mathit{b}$ twice: one time with the default initial guess, and one time with a good initial guess of the solution. Use 200 iterations and the default tolerance for both solutions. Specify the initial guess in the second solution as a vector with all elements equal to 0.99.

maxit = 200;
x1 = minres(A,b,[],maxit);
minres converged at iteration 27 to a solution with relative residual 9.5e-07.
x0 = 0.99*e;
x2 = minres(A,b,[],maxit,[],[],x0);
minres converged at iteration 7 to a solution with relative residual 6.7e-07.

In this case supplying an initial guess enables minres to converge more quickly.

Returning Intermediate Results

You also can use the initial guess to get intermediate results by calling minres in a for-loop. Each call to the solver performs a few iterations and stores the calculated solution. Then you use that solution as the initial vector for the next batch of iterations.

For example, this code performs 100 iterations four times and stores the solution vector after each pass in the for-loop:

x0 = zeros(size(A,2),1);
tol = 1e-8;
maxit = 100;
for k = 1:4
[x,flag,relres] = minres(A,b,tol,maxit,[],[],x0);
X(:,k) = x;
R(k) = relres;
x0 = x;
end

X(:,k) is the solution vector computed at iteration k of the for-loop, and R(k) is the relative residual of that solution.

Solve a linear system by providing minres with a function handle that computes A*x in place of the coefficient matrix A.

One of the Wilkinson test matrices generated by gallery is a 21-by-21 tridiagonal matrix. Preview the matrix.

A = gallery('wilk',21)
A = 21×21

10     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
1     9     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
0     1     8     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
0     0     1     7     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
0     0     0     1     6     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
0     0     0     0     1     5     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0
0     0     0     0     0     1     4     1     0     0     0     0     0     0     0     0     0     0     0     0     0
0     0     0     0     0     0     1     3     1     0     0     0     0     0     0     0     0     0     0     0     0
0     0     0     0     0     0     0     1     2     1     0     0     0     0     0     0     0     0     0     0     0
0     0     0     0     0     0     0     0     1     1     1     0     0     0     0     0     0     0     0     0     0
⋮

The Wilkinson matrix has a special structure, so you can represent the operation A*x with a function handle. When A multiplies a vector, most of the elements in the resulting vector are zeros. The nonzero elements in the result correspond with the nonzero tridiagonal elements of A. Moreover, only the main diagonal has nonzeros that are not equal to 1.

The expression $\mathrm{Ax}$ becomes:

$\mathrm{Ax}=\left[\begin{array}{cccccccccc}10& 1& 0& \cdots & & \cdots & & \cdots & 0& 0\\ 1& 9& 1& 0& & & & & & 0\\ 0& 1& 8& 1& 0& & & & & ⋮\\ ⋮& 0& 1& 7& 1& 0& & & & \\ & & 0& 1& 6& 1& 0& & & ⋮\\ ⋮& & & 0& 1& 5& 1& 0& & \\ & & & & 0& 1& 4& 1& 0& ⋮\\ ⋮& & & & & 0& 1& 3& \ddots & 0\\ 0& & & & & & 0& \ddots & \ddots & 1\\ 0& 0& \cdots & & \cdots & & \cdots & 0& 1& 10\end{array}\right]\left[\begin{array}{c}{\mathit{x}}_{1}\\ {\mathit{x}}_{2}\\ {\mathit{x}}_{3}\\ {\mathit{x}}_{4}\\ {\mathit{x}}_{5}\\ ⋮\\ \\ ⋮\\ \\ {\mathit{x}}_{21}\end{array}\right]=\left[\begin{array}{c}10{\mathit{x}}_{1}+{\mathit{x}}_{2}\\ {\mathit{x}}_{1}+9{\mathit{x}}_{2}+{\mathit{x}}_{3}\\ {\mathit{x}}_{2}+8{\mathit{x}}_{3}+{\mathit{x}}_{4}\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20}+10{\mathit{x}}_{21}\end{array}\right]$.

The resulting vector can be written as the sum of three vectors:

$\mathrm{Ax}=\left[\begin{array}{c}0+10{\mathit{x}}_{1}+{\mathit{x}}_{2}\\ {\mathit{x}}_{1}+9{\mathit{x}}_{2}+{\mathit{x}}_{3}\\ {\mathit{x}}_{2}+8{\mathit{x}}_{3}+{\mathit{x}}_{4}\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20}+10{\mathit{x}}_{21}+0\end{array}\right]$=$\left[\begin{array}{c}0\\ {\mathit{x}}_{1}\\ ⋮\\ {\mathit{x}}_{20}\end{array}\right]+\left[\begin{array}{c}10{\mathit{x}}_{1}\\ 9{\mathit{x}}_{2}\\ ⋮\\ 10{\mathit{x}}_{21}\end{array}\right]+\left[\begin{array}{c}{\mathit{x}}_{2}\\ ⋮\\ {\mathit{x}}_{21}\\ 0\end{array}\right]$.

In MATLAB®, write a function that creates these vectors and adds them together, thus giving the value of A*x:

function y = afun(x)
y = [0; x(1:20)] + ...
[(10:-1:0)'; (1:10)'].*x + ...
[x(2:21); 0];
end

(This function is saved as a local function at the end of the example.)

Now, solve the linear system $\mathrm{Ax}=\mathit{b}$ by providing minres with the function handle that calculates A*x. Use a tolerance of 1e-12 and 50 iterations.

b = ones(21,1);
tol = 1e-12;
maxit = 50;
x1 = minres(@afun,b,tol,maxit)
minres converged at iteration 11 to a solution with relative residual 4.1e-16.
x1 = 21×1

0.0910
0.0899
0.0999
0.1109
0.1241
0.1443
0.1544
0.2383
0.1309
0.5000
⋮

Check that afun(x1) produces a vector of ones.

afun(x1)
ans = 21×1

1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
⋮

Local Functions

function y = afun(x)
y = [0; x(1:20)] + ...
[(10:-1:0)'; (1:10)'].*x + ...
[x(2:21); 0];
end

Input Arguments

collapse all

Coefficient matrix, specified as a symmetric matrix or function handle. This matrix is the coefficient matrix in the linear system A*x = b. Generally, A is a large sparse matrix or a function handle that returns the product of a large sparse matrix and column vector. You can use issymmetric to confirm that A is symmetric.

Specifying A as a Function Handle

You can optionally specify the coefficient matrix as a function handle instead of a matrix. The function handle returns matrix-vector products instead of forming the entire coefficient matrix, making the calculation more efficient.

To use a function handle, use the function signature function y = afun(x). Parameterizing Functions explains how to provide additional parameters to the function afun, if necessary. The function call afun(x) must return the value of A*x.

Data Types: double | function_handle
Complex Number Support: Yes

Right-hand side of linear equation, specified as a column vector. The length of b must be equal to size(A,1).

Data Types: double
Complex Number Support: Yes

Method tolerance, specified as a positive scalar. Use this input to trade-off accuracy and runtime in the calculation. minres must meet the tolerance within the number of allowed iterations to be successful. A smaller value of tol means the answer must be more precise for the calculation to be successful.

Data Types: double

Maximum number of iterations, specified as a positive scalar integer. Increase the value of maxit to allow more iterations for minres to meet the tolerance tol. Generally, a smaller value of tol means more iterations are required to successfully complete the calculation.

Preconditioner matrices, specified as separate arguments of matrices or function handles. You can specify a preconditioner matrix M or its matrix factors M = M1*M2 to improve the numerical aspects of the linear system and make it easier for minres to converge quickly. You can use the incomplete matrix factorization functions ilu and ichol to generate preconditioner matrices. You also can use equilibrate prior to factorization to improve the condition number of the coefficient matrix. For more information on preconditioners, see Iterative Methods for Linear Systems.

minres treats unspecified preconditioners as identity matrices.

Specifying M as a Function Handle

You can optionally specify any of M, M1, or M2 as function handles instead of matrices. The function handle performs matrix-vector operations instead of forming the entire preconditioner matrix, making the calculation more efficient.

To use a function handle, use the function signature function y = mfun(x). Parameterizing Functions explains how to provide additional parameters to the function mfun, if necessary. The function call mfun(x) must return the value of M\x or M2\(M1\x).

Data Types: double | function_handle
Complex Number Support: Yes

Initial guess, specified as a column vector with length equal to size(A,2). If you can provide minres with a more reasonable initial guess x0 than the default vector of zeros, then it can save computation time and help the algorithm converge faster.

Data Types: double
Complex Number Support: Yes

Output Arguments

collapse all

Linear system solution, returned as a column vector. This output gives the approximate solution to the linear system A*x = b. If the calculation is successful (flag = 0), then relres is less than or equal to tol.

Whenever the calculation is not successful (flag ~= 0), the solution x returned by minres is the one with minimal residual norm computed over all the iterations.

Convergence flag, returned as one of the scalar values in this table. The convergence flag indicates whether the calculation was successful and differentiates between several different forms of failure.

Flag Value

Convergence

0

Success — minres converged to the desired tolerance tol within maxit iterations.

1

Failure — minres iterated maxit iterations but did not converge.

2

Failure — The preconditioner matrix M or M = M1*M2 is ill conditioned.

3

Failure — minres stagnated after two consecutive iterations were the same.

4

Failure — One of the scalar quantities calculated by the minres algorithm became too small or too large to continue computing.

5

Failure — Preconditioner matrix M is not symmetric positive definite.

Relative residual error, returned as a scalar. The relative residual error is an indication of how accurate the returned answer x is. minres tracks the relative residual and conjugate gradients residual at each iteration in the solution process, and the algorithm converges when either residual meets the specified tolerance tol. The relres output contains the value of the residual that converged, either the relative residual or the conjugate gradients residual:

• The relative residual error is equal to norm(b-A*x)/norm(b) and is generally the residual that meets the tolerance tol when minres converges. The resvec output tracks the history of this residual over all iterations.

• The conjugate gradients residual error is equal to norm(A'*A*x - A'*b). This residual causes minres to converge less frequently than the relative residual. The resveccg output tracks the history of this residual over all iterations.

Data Types: double

Iteration number, returned as a scalar. This output indicates the iteration number at which the computed answer for x was calculated.

Data Types: double

Residual error, returned as a vector. The residual error norm(b-A*x) reveals how close the algorithm is to converging for a given value of x. The number of elements in resvec is equal to the number of iterations. You can examine the contents of resvec to help decide whether to change the values of tol or maxit.

Data Types: double

Conjugate gradients residual norms, returned as a vector. The number of elements in resveccg is equal to the number of iterations.

Data Types: double

collapse all

Minimum Residual Method

The MINRES and SYMMLQ methods are variants of the Lanczos method that underpins the conjugate gradients method PCG. Like PCG, the coefficient matrix still needs to be symmetric, but MINRES and SYMMLQ allow it to be indefinite (not all eigenvalues need to be positive). This is achieved by avoiding the implicit LU factorization normally present in the Lanczos method, which is prone to breakdowns when zero pivots are encountered with indefinite matrices.

MINRES minimizes the residual in the 2-norm, while SYMMLQ solves a projected system using an LQ factorization and keeps the residual orthogonal to all previous ones. The GMRES method was developed to generalize MINRES to nonsymmetric problems .

Tips

• Convergence of most iterative methods depends on the condition number of the coefficient matrix, cond(A). You can use equilibrate to improve the condition number of A, and on its own this makes it easier for most iterative solvers to converge. However, using equilibrate also leads to better quality preconditioner matrices when you subsequently factor the equilibrated matrix B = R*P*A*C.

• You can use matrix reordering functions such as dissect and symrcm to permute the rows and columns of the coefficient matrix and minimize the number of nonzeros when the coefficient matrix is factored to generate a preconditioner. This can reduce the memory and time required to subsequently solve the preconditioned linear system.

 Barrett, R., M. Berry, T. F. Chan, et al., Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, SIAM, Philadelphia, 1994.

 Paige, C. C. and M. A. Saunders, “Solution of Sparse Indefinite Systems of Linear Equations.” SIAM J. Numer. Anal., Vol.12, 1975, pp. 617-629.