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Could you tell me what is equal to $K_{\infty}(A)$ (condition number of matrix $A$, using the infinity norm) if $A$ is is the Hilbert matrix with dimension $n=100$ ??

RedRose
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evinda
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I used Mathematica with 200 digit arithmetic, and got roughly 1.27e+151. I suspect that is close to the correct answer, because I got the same answer using 300 digit arithmetic, and I didn't get any error messages when I computed the inverse of the Hilbert matrix.

Then I used the formula given in The inverse of the matrix $\{1/(i+j-1)\}$ and got 1426700494452210808740532234308706330681789850377576943636058227514675\ 2815065190703595823722727878319842564448332758716086650462089419171226\ 497083443807596469250/1125849977 which is about 1.26722e+151.

Stephen Montgomery-Smith
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  • I found my result ( 4.6629e+020 ),by writing a function in matlab.Why is my result so different from yours?? :/ – evinda Nov 24 '13 at 09:54
  • When I run my code,I get this warning message: Matrix is close to singular or badly scaled.Results may be inaccurate.RCOND:2.144574e-021. – evinda Nov 24 '13 at 10:47
  • Because the matrix is very ill-conditioned, there is a lot of floating point error in computing the inverse of the matrix. And to compute the condition number, you need to know the inverse. Matlab uses IEEE double precision, which is only about 17 digits. That is nowhere near enough digits to compute this. – Stephen Montgomery-Smith Nov 24 '13 at 15:25
  • I have also an other question..If we have the tridiagonal matrix,that has the number 4 at the main diagonal,the number 1 at the first diagonal below the main diagonal,and also the number 1 at the first diagonal above the main diagonal,no matter wich dimension I give,I get that the condition number,using the infinity norm,is 3..Is this right???If yes,why does this happen??Why isn't there any change of the condition number?? – evinda Nov 24 '13 at 16:45
  • Ask this as another question. I gotta go now. – Stephen Montgomery-Smith Nov 24 '13 at 16:51
  • Hey, if you like the answer I gave you, click the "accept" button. It gives me points. Also, I think I answered your other question as well. – Stephen Montgomery-Smith Nov 24 '13 at 19:28