Plot condition number of Hilbert matrix against matrix size n.

**adashiu** · March 16th 2009, 01:02 AM

Plot condition number of Hilbert matrix against matrix size n. Propose set of linear equations with Hilbert matrix of coefficients and with solution

.Try to solve the equation for different n and plot the norm of the residual against n.

Honestly I have no idea how to cooperate with it. I've just found out that i have to use hilb(n).

**CaptainBlack** · March 16th 2009, 02:56 AM

Originally Posted by adashiu

Plot condition number of Hilbert matrix against matrix size n. Propose set of linear equations with Hilbert matrix of coefficients and with solution

.Try to solve the equation for different n and plot the norm of the residual against n.

Honestly I have no idea how to cooperate with it. I've just found out that i have to use hilb(n).

As:

$ H_n = \begin{bmatrix} 1 & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n+1} \\ : & : & \ddots & : \\ \frac{1}{n} & \frac{1}{n+1} & ... & \frac{1}{2n-1} \\ \end{bmatrix} $

You want the solution of:

$ H_nx=c $

to be:

$ x=\begin{bmatrix}1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} $

in which case we want:

$ c=\begin{bmatrix}\sum_{r=1}^n \frac{1}{r} \\ \\ \sum_{r=2}^{n+1} \frac{1}{r} \\ \\ \vdots \\ \\ \sum_{r=n}^{2n-1} \frac{1}{r}\end{bmatrix} $

CB

**adashiu** · March 16th 2009, 07:32 AM

Ummm.... It looks for me like a magic

And how to write an m. file concerning solution of this problem? Would you be so kind and tell me?

**CaptainBlack** · March 17th 2009, 10:10 AM

Originally Posted by CaptainBlack

As:

$ H_n = \begin{bmatrix} 1 & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n+1} \\ : & : & \ddots & : \\ \frac{1}{n} & \frac{1}{n+1} & ... & \frac{1}{2n-1} \\ \end{bmatrix} $

You want the solution of:

$ H_nx=c $

to be:

$ x=\begin{bmatrix}1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} $

in which case we want:

$ c=\begin{bmatrix}\sum_{r=1}^n \frac{1}{r} \\ \\ \sum_{r=2}^{n+1} \frac{1}{r} \\ \\ \vdots \\ \\ \sum_{r=n}^{2n-1} \frac{1}{r}\end{bmatrix} $

CB

Untested:

Code:

Hn=hilb(n);
x=ones(n,1);
c=(sum(Hn))';  % or Hn*x
x1=Hn\c;
err=sqrt((x1-x)'*(x1-x));
cn=cond(Hn);

CB

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