I am working with a matrix $A \in \mathbb{R}^{4 \times 4}$ with structure:
$A = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}$
I would like to rescale the first two rows to make sure that the module of the maximum eigenvalue of the resulting matrix $B \in \mathbb{R}^{4 \times 4}$ is lower or equal to a given constant $c$. Any idea?
Let $A=\begin{pmatrix}a&b&c&d\\e&f&g&h\\1&0&0&0\\0&1&0&0\end{pmatrix}$, $p$ be the characteristic polynomial of $A$ and $C_p$ be the companion matrix of $p$.
According to the Gerschgorin theorem, the complex eigenvalues of $A$ are in the following union of euclidean balls in $\mathbb{C}$.
i) With the matrix $C_p$.
$B(0,|-ch+gd|)\cup B(0,|-ha+bg-cf+ed|+1)$
$\cup B(0,|-fa+eb+c+h|+1)\cup B(f+a,1)$.
ii) With the matrix $A$.
$B(a,|b|+|c|+|d|)\cup B(f,|g|+|h|+|e|)\cup B(0,1)$.
Remark 1. In both cases, we obtain $3$ or $1$ ball of radius $\geq 1$.
Remark 2. If $c=d=g=h=0$, then $spectrum(A)=\{0,0\}\cup spectrum(\begin{pmatrix}a&b\\e&f\end{pmatrix})$.
EDIT. If you want $\rho(A)<1$, then do as follows.
Let $U=\begin{pmatrix}a&b\\e&f\end{pmatrix},V=\begin{pmatrix}c&d\\g&h\end{pmatrix}$. Then $\det(A-\lambda I_4)=\det((U-\lambda I_2)(-\lambda I_2)-V)=\det(\lambda^2I_2-\lambda U -V)$.
If you want a small $\rho(A)$, then choose a small $\det(V)$ and a small $\rho(U)$, that is, small $a+f,af-be$.