Rewrite Matrix such that it has only real eigenvalues

Question

I am looking for a way to rewrite a real matrix $A\in\mathbb{R}^{s\times s}$, which has at least one complex eigenvalue, to a matrix $B$ with the properties:

1. $B$ is real and has twice the dimension of $A$, i.e. $B\in\mathbb{R}^{2s\times 2s}$
2. The entries in $B$ derive from the entries in $A$ (e.g. are just the real/imaginary parts of the entries of $A$).
3. $B$ has only real eigenvalues
4. The spectral radii of the two matrices $\rho(A)$, $\rho(B)$ are compareable. For example in a way $\rho(A)^n=\rho(B)$ for some $n\in\mathbb{N}$ or $C_1 \rho(A) \leq \rho(B) \leq C_2 \rho(B)$, $0<C_1\leq C_2<\infty$.

Answer 1

This question is not fully specified, but the first thing that comes to mind is $B=\begin{bmatrix} 0 & A^* \\ A & 0 \end{bmatrix}$ (where $*$ is the conjugate transpose, which is just the transpose since you said $A$ is real). Properties:

1. Obvious.
2. Maybe obvious? Since $A$ is real, each $b_{ij}$ is some $a_{k\ell}$, and the relation between the indices is not complicated.
3. Standard (maybe not obvious)
4. The spectral radius of $A$ is less than or equal to that of $B$. Specifically, the eigenvalues of $B$ are the singular values of $A$, and the variational interpretation of the largest singular value as $\sigma_1=\max_{\| x \|=1} \| Ax \|$ gives the inequality. The reverse is not true, even with a constant factor, because the spectral radius of $B$ is a norm for $A$ (the operator Euclidean norm), whereas the spectral radius of $A$ is not a norm (because, e.g., $\rho \left ( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right ) = 0$).