MATLAB爱好者论坛-LabFans.com - Matrix Condition and the QR Decomposition

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[B]Contents[/B]
[LIST][*][URL="#1"]Query[/URL][*][URL="#2"]Short answer[/URL][*][URL="#3"]QR Decomposition[/URL][*][URL="#4"]tricond[/URL][*][URL="#5"]qr function[/URL][*][URL="#12"]Gallery[/URL][*][URL="#20"]Tests[/URL][*][URL="#21"]Plot[/URL][/LIST][B]Query[/B]

Professor Magdy Hanna of the Department of Engineering Mathematics and Physics at Fayoum University in Fayoum, Egypt, asked me:

If we have the QR decomposition of a matrix, with or without pivoting, is there an easy way to get the condition number of the matrix without starting from scratch and using the SVD?[B]Short answer[/B]

The short answer to Prof. Hanna's question is:

Yes. The ratio of the largest and smallest diagonal elements of qr(A) is often a good estimate of cond(A).Compare this estimate to the exact condition of a magic square. Produce a score measuring how well the estimate does.

A = magic(11); R = qr(A); d = abs(diag(R)); est = max(d)/min(d); kappa = cond(A); disp(" est kappa score") disp([est kappa est/kappa]) est kappa score 2.6964 11.1021 0.2429[B]QR Decomposition[/B]

A more complete answer involves column pivoting.

The QR decomposition [B]without[/B] column pivoting is an orthogonal Q and an upper triangular R so that

A = Q*RThe QR composition [B]with[/B] column pivoting is an orthogonal Q, an upper triangular R and a permutation P that reorders the columns of A so that

A*P = Q*RIn either case, Q and P are orthogonal, so

cond(R) = cond(A)Any estimate of the condition of R provides an estimate of the condition of A.

[B]tricond[/B]

An estimate the condition of a triangular matrix is the ratio of the largest and smallest elements on the diagonal.

type tricond function cond = tricond(T) % Estimate condition of lower or upper triangular matrix. if istril(T) || istriu(T) D = diag(abs(T)); cond = max(D)/min(D); else cond = NaN; end end[B]qr function[/B]

Here is the 4-by-4 [URL="https://blogs.mathworks.com/cleve/2018/02/19/fun-with-the-pascal-triangle/"]Pascal matrix[/URL]

A = pascal(4)A = 1 1 1 1 1 2 3 4 1 3 6 10 1 4 10 20This matrix is reasonably well conditioned.

kappa = cond(A)kappa = 691.9374With one output, the qr function provides the "Q-less" QR decomposition with no pivoting.

(With two outputs, the qr function produces the same R and also reveals the orthogonal Q.)

R = qr(A)R = -2.0000 -5.0000 -10.0000 -17.5000 0 -2.2361 -6.7082 -14.0872 0 0 1.0000 3.5000 0 0 0 -0.2236For this matrix tricond without pivoting produces a poor estimate of the exact condition.

trico = tricond(R); kappa = cond(A); disp(" tricond kappa score") disp([trico kappa trico/kappa]) tricond kappa score 10.0000 691.9374 0.0145With three outputs, qr does column pivoting and returns a different R.

[Q,R,P] = qr(A); RR = -22.7376 -5.2336 -1.5393 -12.0065 0 -1.6153 -1.2034 -1.3387 0 0 0.4270 -0.2170 0 0 0 -0.0638Compare tricond and the exact condition.

trico = tricond(R); disp(" tricond kappa score") disp([trico kappa trico/kappa]) tricond kappa score 356.6259 691.9374 0.5154Column pivoting produces a much better tricond estimate for this matrix.

[B]Gallery[/B]

Let's do more tests using these matrices from the [URL="https://blogs.mathworks.com/cleve/2019/06/24/bohemian-matrices-in-the-matlab-gallery"]MATLAB Gallery[/URL].

funs = ["moler", "parter", "binomial", "chebspec", "cauchy",... "chebspec", "chebvand","circul", "frank", "ris"];One of these Gallery matrices is the "moler" matrix, even though I didn't invent it. A precursor to the "moler" matrix is a badly conditioned upper triangular matrix with ones on the diagonal, minus ones above the diagonal and zeros below.

n = 5; U = eye(n) - triu(ones(n),1)U = 1 -1 -1 -1 -1 0 1 -1 -1 -1 0 0 1 -1 -1 0 0 0 1 -1 0 0 0 0 1The "moler" matrix is

M = U'*UM = 1 -1 -1 -1 -1 -1 2 0 0 0 -1 0 3 1 1 -1 0 1 4 2 -1 0 1 2 5The "moler" matrix squares the condition of U. The condition of M grows like 4^n.

kappa = cond(M)kappa = 865.9801The tricond score without column pivoting is very poor.

R = qr(M); trico = tricond(R); disp(" tricond kappa score") disp([trico kappa trico/kappa]) tricond kappa score 20.7364 865.9801 0.0239With pivoting the score is much better.

[~,R,~] = qr(M); trico = tricond(R); disp(" tricond kappa score") disp([trico kappa trico/kappa]) tricond kappa score 555.0198 865.9801 0.6409These scores are in the first line of the following output and in the upper left-hand corners of the 3-D bar graphs.

[B]Tests[/B]

[Dnopiv,Dpiv,xt,yt] = tester(funs); fun n nopiv piv moler 5 0.024 0.641 10 0.000 0.426 15 0.000 0.337 20 0.000 0.287 25 0.000 0.372 parter 5 0.610 0.804 10 0.564 0.776 15 0.536 0.757 20 0.517 0.744 25 0.502 0.735 binomial 5 0.795 0.926 10 0.387 0.817 15 0.150 0.768 20 0.056 0.696 25 0.021 0.674 chebspec 5 0.120 0.110 10 0.236 0.095 15 0.183 0.358 20 0.132 0.165 25 0.022 0.318 cauchy 5 0.217 0.376 10 0.034 0.242 15 0.016 0.247 20 0.058 0.116 25 0.102 0.222 chebspec 5 0.120 0.110 10 0.236 0.095 15 0.183 0.358 20 0.132 0.165 25 0.022 0.318 chebvand 5 0.055 0.577 10 0.002 0.326 15 0.000 0.338 20 0.000 0.492 25 0.000 0.082 circul 5 0.370 0.370 10 0.252 0.252 15 0.209 0.209 20 0.180 0.180 25 0.162 0.162 frank 5 0.390 0.390 10 0.257 0.296 15 0.191 0.248 20 Inf 0.013 25 0.039 0.100 ris 5 0.610 0.804 10 0.564 0.776 15 0.536 0.757 20 0.517 0.744 25 0.502 0.735 [B]Plot[/B]

[c,ax1,ax2] = ploter(funs,Dnopiv,Dpiv,xt,yt);[IMG]https://blogs.mathworks.com/cleve/files/QR_blog_01.png[/IMG]
[URL="https://www.mathworks.com/products/matlab/"]Published with MATLAB® R2026a[/URL]

[url=https://blogs.mathworks.com/cleve/2026/04/13/matrix-condition-and-the-qr-decomposition/?s_tid=feedtopost]More...[/url]