Given complex eigenvalues, with its complex-conjugates also as eigenvalues, how can I construct a matrix with real entries?

Given a list of complex eigenvalues, where the eigenvalues' complex-conjugates are also in the list, how can I construct a matrix with real entries that has this list as its spectrum?

• It suffices to solve this problem for a single complex eigenvalue and its conjugate. Do you know the eigenvalues of a rotation matrix? – Qiaochu Yuan Aug 27 '16 at 21:52
• Note that the block matrix $$\pmatrix{A&0\\0&B}$$ has the eigenvalues from $A$ combined with the eigenvalues from $B$. – Omnomnomnom Aug 27 '16 at 21:57
• @user58865: almost; you need to multiply by $r$ after putting your pairs of complex eigenvalues into polar form. But yes, that's the argument I had in mind. – Qiaochu Yuan Aug 27 '16 at 22:41

• generate the characteristic polynomial $$p(x):=\Pi_{k=1}^n(x-\lambda_k)(x-\bar \lambda_k)=\Pi_{k=1}^n(x^2-2 (\Re{\lambda_k})x+|\lambda_k|^2)$$ which has real coefficients.
• Compute its characteristic polynomial $p$ with the leading coefficient $1$ and other coefficients $a_i$.
$$\left[\begin{matrix} & 1 & & & \\ & & 1 && \\ & & & \ddots& \\ & & & & 1 \\ a_{n - 1} & & \cdots& a_1 & a_0 \end{matrix}\right]$$