Experimenting with Hilbert matrix

The following experiment shows that the Hilbert matrix A of size 10 is very sensitive to changes in b. The condition number of the matrix is 1.6x10^13. The solution to Ax=b changes a lot when b changes from [1 1 1 1 1 1 1 1 1 1]' by a small amount.

ť format long
ť for i=1:10
for j=1:10
A(i,j)=1/(i+j-1);
end
end

% This defines the Hilbert matrix of size 10.

ť cond(A)

ans =

    1.602533632202711e+013

% This shows that the matrix has a large condition number 16025336322027.11, more than 16 trillion.

ť b=ones(10,1);
ť A\b

ans =

  1.0e+006 *

  -0.00000999563844
   0.00098962226204
  -0.02375193941951
   0.24016661264454
  -1.26090953729814
   3.78281563674004
  -6.72513649280694
   6.99974868636912
  -3.93741561223132
   0.92360301627563

% Recall that ones(10,1) means the 10x1 column vector consisting of 1's. The command A\b tells Matlab to solve Ax=b. The solution is x1=-9.99563844, x2=989.62226204, , etc.

ť b=[1 1 1 1.1 1 1.2 0.9 0.92 .8 1.1 ]' 

b =

   1.00000000000000
   1.00000000000000
   1.00000000000000
   1.10000000000000
   1.00000000000000
   1.20000000000000
   0.90000000000000
   0.92000000000000
   0.80000000000000
   1.10000000000000

ť A\b

ans =

  1.0e+011 *

  -0.00002549120031
   0.00219657069024
  -0.04670921198724
   0.42418723371259
  -2.02187301254720
   5.55543605610408
  -9.11167390811971
   8.80322945370647
  -4.62080309845867
   1.01605313866939

% We changed b slightly from [1 1 1 1 1 1 ...]' and solve Ax=b. The result is x1=-2549120.031, x2=219657069.024, .... which is very much different from the solution above. I tested below another b slightly different from [1 1 1 1 1 ...]. The result is again very much different.

ť b=[1.001 0.999 1.002 0.981 1.002 0.978 .980 1.002 1 0.991]';
ť A\b

ans =

  1.0e+010 *

  -0.00000460189437
   0.00043353979203
  -0.00978756996277
   0.09278406450263
  -0.45675557344773
   1.28693129265434
  -2.15356084533401
   2.11508672274315
  -1.12546640162162
   0.25034436600482