# matrices – Solution of complicated linear system Answer

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matrices – Solution of complicated linear system

In Brubeck, Nakatsukasa, and Trefethen – Vandermonde with Arnoldi (instance 3) they decipher the next linear system:
$$operatorname{Re}left(commence{array}{ccc}1 & cdots & z_{1}^{n} 1 & cdots & z_{2}^{n} vdots & ddots & vdots 1 & cdots & z_{m}^{n}aim{array}privilege)left(commence{array}{c}a_{0} vdots a_{n}aim{array}privilege)-operatorname{Im}left(commence{array}{ccc}z_{1} & cdots & z_{1}^{n} z_{2} & cdots & z_{2}^{n} vdots & ddots & vdots z_{m} & cdots & z_{m}^{n}aim{array}privilege)left(commence{array}{c}b_{1} vdots b_{n}aim{array}privilege)approxleft(commence{array}{c}f_{1} f_{2} vdots f_{m}aim{array}privilege).$$

$$A=left(commence{array}{ccc}1 & cdots & z_{1}^{n} 1 & cdots & z_{2}^{n} vdots & ddots & vdots 1 & cdots & z_{m}^{n}aim{array}privilege)$$.

For fixing it, they employ the next MATLAB code:

``````c = [real(A) imag(A(:,2:n+1))]f;
c = c(1:n+1) - 1i*[0; c(n+2:2*n+1)];
``````

The first line is equal to making a vector `c=[a,b]` the place $$operatorname{Re}(A)aapprox f$$ and $$operatorname{Im}(A(:,2:n+1))b approx f$$ and the second means $$c=a-[0,bi]$$. I used to be questioning the way it can breathe solved on this route, in truth I reproduced the code of the paper in Mathematica and the outcomes are usually not the identical. Is there any typo on this process?

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