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Given complex eigenvalues (occurring in conjugate pairs) how to get a single instance of a real matrix which has these eigenvalues. I know the matrix is not unique as eigenvectors are not fixed but in my case any real matrix will suffice.

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    $\begingroup$ If $\lambda_i\ne\lambda_j$, this is simple. For each eigenvalue $\lambda=a+bi$ you can take matrix $\begin{pmatrix}a & -b\\b & a\end{pmatrix}$; whole matrix is block-diagonal. $\endgroup$ – Michael Galuza Jul 1 '15 at 11:50
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If eigenvalue is $a \pm bi$, the matrix is $$ \begin{bmatrix} a & -b \\ b & a \end{bmatrix}. $$

For more eigenvalue pairs, place $2 \times 2$ blocks like this one down the diagonal of your $2k \times 2k$ matrix.

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If the given eigenvalues are all distinct, then you can take the companion matrix of the polynomial $(x-\lambda_1)\cdots(x-\lambda_n)$, which has real coefficients because the roots occur in conjugate pairs.

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