Find the QR decomposition of the 5-by-5 Pascal matrix. Specify one output argument to return just the upper-triangular factor.

A = pascal(5);
X = qr(A)

X = 5×5

   -2.2361   -6.7082  -15.6525  -31.3050  -56.3489
    0.3090    3.1623   11.0680   26.5631   53.1263
    0.3090   -0.1744    1.8708    7.4833   19.2428
    0.3090   -0.4565    0.3548    0.6325    2.8460
    0.3090   -0.7387   -0.0281   -0.7490   -0.1195

Extract the upper-triangular factor R from X.

R = triu(X)

R = 5×5

   -2.2361   -6.7082  -15.6525  -31.3050  -56.3489
         0    3.1623   11.0680   26.5631   53.1263
         0         0    1.8708    7.4833   19.2428
         0         0         0    0.6325    2.8460
         0         0         0         0   -0.1195

Compare the R in the Q-less QR decomposition to the R factor in a full QR decomposition.

[Q1,R1] = qr(A)

Q1 = 5×5

   -0.4472   -0.6325    0.5345   -0.3162   -0.1195
   -0.4472   -0.3162   -0.2673    0.6325    0.4781
   -0.4472    0.0000   -0.5345    0.0000   -0.7171
   -0.4472    0.3162   -0.2673   -0.6325    0.4781
   -0.4472    0.6325    0.5345    0.3162   -0.1195

R1 = 5×5

   -2.2361   -6.7082  -15.6525  -31.3050  -56.3489
         0    3.1623   11.0680   26.5631   53.1263
         0         0    1.8708    7.4833   19.2428
         0         0         0    0.6325    2.8460
         0         0         0         0   -0.1195

Compute the full QR decomposition of a magic square test matrix by specifying two output arguments.

A = magic(5);
[Q,R] = qr(A)

Q = 5×5

   -0.5234    0.5058    0.6735   -0.1215   -0.0441
   -0.7081   -0.6966   -0.0177    0.0815   -0.0800
   -0.1231    0.1367   -0.3558   -0.6307   -0.6646
   -0.3079    0.1911   -0.4122   -0.4247    0.7200
   -0.3387    0.4514   -0.4996    0.6328   -0.1774

R = 5×5

  -32.4808  -26.6311  -21.3973  -23.7063  -25.8615
         0   19.8943   12.3234    1.9439    4.0856
         0         0  -24.3985  -11.6316   -3.7415
         0         0         0  -20.0982   -9.9739
         0         0         0         0  -16.0005

Verify that $\mathit{A}=\mathrm{QR}$, within machine precision.

norm(A-Q*R)

ans = 1.2569e-14

Specify three output arguments to return a permutation matrix or vector that reduces fill-in in the R factor of the QR decomposition.

Compute the QR decomposition of the west0479 sparse matrix. Specify three outputs to return a permutation matrix that satisfies $\mathrm{AP}=\mathrm{QR}$.

load west0479
A = west0479;
[Q,R,P] = qr(A);

Verify that A*P = Q*R for the permutation matrix P, within machine precision.

norm(A*P-Q*R,'fro')

ans = 3.3386e-10

Now specify the 'vector' option to return p as a permutation vector.

[Q,R,p] = qr(A,'vector');

Verify that A(:,p) = Q*R for the permutation vector p, within machine precision.

norm(A(:,p) - Q*R,'fro')

ans = 3.3386e-10

Verify that the use of a permutation matrix or permutation vector in the decomposition results in an R factor with fewer nonzeros for sparse inputs compared to a nonpermuted decomposition.

[Q1,R1] = qr(A);
spy(R1)

spy(R)

The results show that the permuted decomposition produces an R factor with substantially fewer nonzeros.

Use the economy-size QR decomposition of a coefficient matrix to solve the linear system $\mathrm{Ax}=\mathit{b}$.

Create a 10-by-5 coefficient matrix by using the first five columns of magic(10). For the right-hand side of the linear equation $\mathrm{Ax}=\mathit{b}$, use the row sums of the matrix. With this setup, the solution to the equation $\mathit{x}$ should be a vector of ones.

A = magic(10);
A = A(:,1:5)

A = 10×5

    92    99     1     8    15
    98    80     7    14    16
     4    81    88    20    22
    85    87    19    21     3
    86    93    25     2     9
    17    24    76    83    90
    23     5    82    89    91
    79     6    13    95    97
    10    12    94    96    78
    11    18   100    77    84

b = sum(A,2)

b = 10×1

   215
   215
   215
   215
   215
   290
   290
   290
   290
   290

Compute the economy-size QR decomposition of A. Then solve the linear system $\mathrm{QRx}=\mathit{b}$ with x(p,:) = R\(Q\b). Because Q is orthogonal, this equation is the same as x(p,:) = R\(Q'*b).

[Q,R,p] = qr(A,0)

Q = 10×5

   -0.0050   -0.4775   -0.0504    0.5193    0.0399
   -0.0349   -0.5001   -0.0990   -0.1954   -0.2006
   -0.4384    0.1059   -0.4660    0.4464    0.0628
   -0.0947   -0.4151   -0.2923   -0.2542    0.5274
   -0.1246   -0.4117   -0.2812   -0.1326   -0.4130
   -0.3787    0.0209    0.2702    0.4697    0.0390
   -0.4085   -0.0017    0.2217   -0.2450   -0.2015
   -0.0648   -0.3925    0.6939    0.0669    0.1225
   -0.4683    0.0833    0.0283   -0.3038    0.5265
   -0.4982    0.0867    0.0394   -0.1822   -0.4138

R = 5×5

 -200.7112  -55.5026 -167.6040  -84.7237 -168.7997
         0 -192.1053  -40.3557 -152.4040  -39.2814
         0         0  101.3180  -89.4254   96.0172
         0         0         0   41.0248  -14.9083
         0         0         0         0   24.6386

p = 1×5

     3     1     5     2     4

x(p,:) = R\(Q\b)

x = 5×1

    1.0000
    1.0000
    1.0000
    1.0000
    1.0000

Make a semilog plot of the diagonal of R to confirm that the permuted decomposition produces an R factor with abs(diag(R)) decreasing. Plot the singular values of A in the same plot for comparison. In practice, the diagonal values of R behave in a similar way to the singular values of A. Therefore, you can use the diagonal values of R as a measure for how close to singular the matrix A is.

semilogy(abs(diag(R)),'-o')
hold on
semilogy(svd(A),'r-o')
legend('Diagonal of R','Singular Values of A')

Solve a sparse linear system and use the results to see how much of vector b lies in the column space of S.

Create a random 500-by-20 sparse matrix with 10% density and a vector of ones. Use qr to factorize the matrix into the factors R and C = Q'*b.

S = sprand(500,20,0.1);
b = ones(500,1);
[C,R] = qr(S,b,0);

Use the results to solve $\mathrm{Sx}=\mathit{b}$ with x = R\C.

x = R\C;

Consider the identity${\Vert \mathit{b}\Vert }^2={\Vert \mathrm{Sx}-\mathit{b}\Vert }^2+{\Vert \mathit{C}\Vert }^2$.

Dividing through by the norm of b, you get a new identity that shows how much of b lies in the column space of S:

$\frac{{\Vert \mathrm{Sx}-\mathit{b}\Vert }^2}{{\Vert \mathit{b}\Vert }^2}+\frac{{\Vert \mathit{C}\Vert }^2}{{\Vert \mathit{b}\Vert }^2}=1$.

The first term tells how much of b does not lie in the column space of S, while the second term tells how much of b does lie in the column space of S.

t1 = norm(S*x-b)^2/norm(b)^2

t1 = 0.4000

t2 = norm(C)^2/norm(b)^2

t2 = 0.6000

Use qr to solve the matrix equation $\mathrm{Sx}=\mathit{B}$ with a rectangular sparse coefficient matrix $\mathit{S}$.

Load the west0479 sparse matrix and use the first 200 columns as the rectangular coefficient matrix in a linear system. For the right-hand side of the equation, use the row sums of $\mathit{S}$. With this setup, the solution to $\mathrm{Sx}=\mathit{B}$ is a vector of ones.

load west0479
S = west0479(:,1:200);
B = sum(S,2);

Solve $\mathrm{Sx}=\mathit{B}$ using qr with two inputs and three outputs. The solution to the linear system is x = P*(R\C).

[C,R,P] = qr(S,B);
x = P*(R\C);

Verify that $\mathrm{Sx}-\mathit{B}=0$, within machine precision.

norm(S*x-B)

ans = 8.4349e-11

Note: To calculate the upper-triangular factor R and permutation matrix P, but avoid computing the orthogonal matrix Q (which is often the most computationally expensive part of a call to qr), you can specify B as an empty matrix:

emptyB = zeros(size(S,1),0);
[~,R,P] = qr(S,emptyB);