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would like someone to look over this and assure me I'm not making a silly mistake....

Given a $3\times9$ matrix $V$:

$$ \small\begin{bmatrix} 1.0814 & -0.1251 & -0.1726 & -1.4443 & -0.2240 & -0.1665 & 0.5450 & 0.7832 & -1.2032 \\ -0.1885 & 1.1356 & -0.1545 & -0.0974 & -1.4325 & -0.1313 & 0.9151 & -0.9659 & 0.5941 \\ -0.0367 & -0.0531 & 1.2606 & -0.0458 & 1.2000 & -1.2934 & -0.0510 & -0.0277 & -0.0615 \\ \end{bmatrix} $$

I want to form a $9\times9$ matrix whereby its row $n$ is of the following form: $$ \begin{bmatrix} V^2_{1,n} & V^2_{2,n} & V^2_{3,n} & V_{1,n}V_{2,n} & V_{1,n}V_{3,n} & V_{2,n}V_{3,n} & V_{1,n} & V_{2,n} & V_{3,n} \end{bmatrix}. $$

So, for the first two rows I have: $$\small\begin{bmatrix} 1.0814^2 & -0.1885^2 & -0.0367^2 & -0.204 & -0.04 & 0.006918 & 1.0814 & -0.1885 & -0.0367 \\ -0.1251^2 & 1.1356^2 & -0.0531^2 & -0.142 & 0.006643 & -0.06 & -0.1251 & 1.1356 & -0.0531 \\ \vdots\end{bmatrix} $$

Continuing in this manner gives me the $9\times9$ matrix. Then, taking the $2$-norm condition number of the $9\times9$ matrix gives me $ 6365.85 $, however, the paper I am trying to replicate says that the condition number should be $266446.5$.

Am I constructing the $9\times9$ matrix correctly?

Thank you.

user1551
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Mike
  • 461
  • This seems extracted from the "The little, bad book about nightmarish numerical questions for pure mathematicians", from Editorial Evil Engineers Ltd....I'm not sure about your definition: for ex., it seems to be $,V_{1,1}^2,$ is the entry $,1-1,$ in your new matrix, but what is that square there?? – DonAntonio May 01 '13 at 12:06
  • I think I got it: your matrix is $,V=(v_{ij}),$ , and you're defining a new matrix by taking some entries\products of entries of the first matrix, right? Gee, this looks messy... – DonAntonio May 01 '13 at 12:11
  • The way I read it is that $ V_{1,1} $ relates to the element at row 1, column 1 of $ V $. So, that would be $ 1.0814 $. Now, the $ V^2_{1,1} $ gives 1.0814^2. That is my understanding, although I may be abusing notation! – Mike May 01 '13 at 12:16
  • I get, for $ L_{1} $ the condition number $ 10570 $, for $ L_{2} $ I get $ 6365.85 $, for the Euclidean Norm $ 10068.851 $ and for the Infinity Norm I get $ 23565.91 $. – Mike May 01 '13 at 12:20

1 Answers1

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I got different answers using MATLAB:

V=[1.0814 -0.1251 -0.1726 -1.4443 -0.2240 ...
   -0.1665 0.5450 0.7832 -1.2032; ...
   -0.1885 1.1356 -0.1545 -0.0974 -1.4325 ...
   -0.1313 0.9151 -0.9659 0.5941; ...
   -0.0367 -0.0531 1.2606 -0.0458 1.2000 ...
   -1.2934 -0.0510 -0.0277 -0.0615];
W=zeros(9,9);
for k=1:9
  v=V(:,k);  % v is the k-th column of V
  W(k,:)=[v'.^2, v(1)*v(2), v(1)*v(3), v(2)*v(3), v'];
end
% cond. nos w.r.t. the 2-, 1-, infinity- norms and Frobenius norm
[cond(W), cond(W,1), cond(W,inf), cond(W,'fro')]

and the output is

ans =
  1.0e+003 *
    1.0941    1.1659    5.3552    1.8293
user1551
  • 139,064
  • Hi, Thankyou for trying. I have been using Mathcad and will investigate some more this evening. I will report back any differences. – Mike May 01 '13 at 14:36