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In Brubeck, Nakatsukasa, and Trefethen - Vandermonde with Arnoldi (example 3) they solve the following linear system: $$\operatorname{Re}\left(\begin{array}{ccc}1 & \cdots & z_{1}^{n} \\ 1 & \cdots & z_{2}^{n} \\ \vdots & \ddots & \vdots \\ 1 & \cdots & z_{m}^{n}\end{array}\right)\left(\begin{array}{c}a_{0} \\ \vdots \\ a_{n}\end{array}\right)-\operatorname{Im}\left(\begin{array}{ccc}z_{1} & \cdots & z_{1}^{n} \\ z_{2} & \cdots & z_{2}^{n} \\ \vdots & \ddots & \vdots \\ z_{m} & \cdots & z_{m}^{n}\end{array}\right)\left(\begin{array}{c}b_{1} \\ \vdots \\ b_{n}\end{array}\right)\approx\left(\begin{array}{c}f_{1} \\ f_{2} \\ \vdots \\ f_{m}\end{array}\right).$$

$A=\left(\begin{array}{ccc}1 & \cdots & z_{1}^{n} \\ 1 & \cdots & z_{2}^{n} \\ \vdots & \ddots & \vdots \\ 1 & \cdots & z_{m}^{n}\end{array}\right)$.

For solving it, they use the following 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 equivalent to creating a vector c=[a,b] where $\operatorname{Re}(A)a\approx f$ and $\operatorname{Im}(A(:,2:n+1))b \approx f$ and the second one means $c=a-[0,bi]$. I was wondering how it can be solved in this way, in fact I reproduced the code of the paper in Mathematica and the results are not the same. Is there any typo in this procedure?

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No, the first line is not equivalent to what you claim in the text, I believe: that's one linear system containing a subtraction and not two separate linear systems.

In detail, the matrix of this linear system is obtained by concatenating horizontally M = real(A) and N = imag(A(:,2:n+1)). If you split the unknown into $$ c = \begin{bmatrix} a\\-b \end{bmatrix} $$ then this linear system is $$ f = \begin{bmatrix}M & N\end{bmatrix} \begin{bmatrix} a\\-b \end{bmatrix} = Ma - Nb, $$ which corresponds to your first formula. But, ultimately, you are solving one linear system with matrix $\begin{bmatrix} M & N\end{bmatrix}$. The instruction [real(A) imag(A(:,2:n+1))] stacks matrices one next to the another; it is not a "vectorized" syntax to solve two linear systems with different matrices and the same RHS.

I suspect that's the source of the issue (and, also, possibly, not solving the linear system in the least-squares sense.)

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  • $\begingroup$ Thank you for your answer. I have included $\approx$ to indicate the least-squares equalities. Are you saying that '[real(A) imag(A(:,2:n+1))]\f' is a substraction?. I have typed it on MATLAB and, effectively, it is the solution of two linear systems $\endgroup$
    – Gaussian
    Jan 10 at 14:44
  • $\begingroup$ No, that line computes the solution (in the least-squares sense) of one linear system. I have expanded, I hope this solves the issue. This is just basic linear algebra, in the end. $\endgroup$ Jan 10 at 14:55

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