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I need to construct real-valued matrices with specific complex eigenvalues. I have seen the companion matrix, which sort of does my job, but there are some other desirable properties as well, so I'm looking for a more general construction. If anyone could shed some light, that'd be awesome.

The eigenvalues will be distinct. They will lie on the unit circle. Preferably, I could choose the eigenvectors as well (I know some restrictions will apply, but it's fine). It would be amazing, if the matrix had as few zeros as possible.

So, given I am asking for a lot, I will appreciate the most general construction.

I have seen this answer (https://math.stackexchange.com/q/1345699), but I wonder if there is a method that provides greater freedom over the matrix.

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    $\begingroup$ Every matrix with those eigenvalues is obtained by factoring the polynomial $\prod(x-\lambda_i)$, with $\lambda_i$ the eigenvalues together with multiplicity, over the reals (not necessarily in smallest factors, just some choice of factors), building a block diagonal matrix with the companion matrices of each of the factors, and the pre- and post multiplying the resulting matrix with $P^{-1}$ and $P$, for an arbitrary real matrix $P$. $\endgroup$ – logarithm May 8 at 20:50
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Build $M$ with $2 \times 2$ block matrices along the diagonal as in the linked question. Then for any invertible matrix $A$ the matrix $AMA^{-1}$ will have the eigenvalues you want. You can design $A$ to get the other properties you care about.

The original $M$ is likely to be the one with the most zeroes.

A random square matrix $A$ will be invertible (the probability that the determinant is $0$ is $0$) and I'm pretty sure $AMA^{-1}$ will have no $0$ entries with probability $1$.

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  • $\begingroup$ thanks for the answer. is there way I can choose A to have the desired eigenvectors? $\endgroup$ – Ussiane Lepsiq May 9 at 18:41
  • $\begingroup$ You can arrange the eigenvectors for the transformed $M$ (not those of $A$). Let $A$ be the transformation that maps the canonical basis of $\mathbb{R}^n$ to the basis (of eigenvectors) of your choice. But I suspect a problem. The complex eigenvalues occur in complex conjugate pairs. The corresponding eigenvectors span a two dimensional real subspace on which $M$ acts essentially as a rotation. There won't be eigenvectors when you think in terms of action on a vector space over $\mathbb{R}$. How you manage this depends on information about your application that's not in the question. $\endgroup$ – Ethan Bolker May 9 at 19:22

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