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I would like to form a matrix from its eigenvalues located within the unit circle, not necessarily real. For example, I would like to place them very close to the unit circle, though inside, and then form a real matrix. The entries in the matrix are really not important as long as the eigenvalues match the ones I want to place and the entries are real.

How to do this?

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  • $\begingroup$ In general, that can't be done. $\endgroup$ – José Carlos Santos Sep 26 '18 at 13:31
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    $\begingroup$ In a real matrix, complex eigenvalues always appear in pairs. $\endgroup$ – 5xum Sep 26 '18 at 13:44
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This problem is not solvable for any arbitrary collection of complex numbers, for the following reason: if the matrix $A$ is real, any complex eigenvalues occur in conjugate pairs. So for example it is impossible to construct a real matrix with eigenvalues $1, 2, 2 + 3i$, but it is possible to construct one with eigenvalues $1, 2, 2 + 3i, 2 - 3i$.

The construction is not too difficult. For a real eigenvalue, place it on the diagonal. For the complex conjugate pair $a \pm bi$, place the real $2 \times 2$ matrix

$$ \begin{pmatrix} a & b \\ -b & a \end{pmatrix} $$

on the diagonal. (Check that this matrix has $a \pm ib$ as its eigenvalues).

For example, starting with the set of desired eigenvalues $1, 2, 2 + 3i, 2-3i$, we construct the matrix $$ A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & -3 & 2 \end{pmatrix} $$ which is real, and has precisely the set $1, 2, 2 \pm 3i$ as its eigenvalues.

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It's not possible in general, for the following reason : the eigenvalues are the root of the characteristic polynomial of your matrix. Given eigenvalues $\lambda_1, ..., \lambda_n \in \mathcal{C}$, the characteristic polynomial of your matrix should be $$P(X)=-\prod_{i=1}^n (X- \lambda_i)$$ but for arbitrary $\lambda_i \in \mathbb{C}$, this polynomial has no reason to have real coefficients (as it would have for a real matrix).

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