Is this matrix block matrix Hurwitz for some $c_1, c_2, c_3 > 0$? Given the next well partitioned real squared matrix
$$
M = \begin{bmatrix}A & \frac{c_3}{c_1}BC^T \\
c_3c_2C & E- c_3^2\frac{c_2}{c_1}CC^T\end{bmatrix},
$$
where $A$ is Hurwitz (all the eigenvalues have negative real part), $CC^T$ positive definite, $E$ has its eigenvalues on the imaginary axis and is invertible (it represents an oscillator, in fact it is a block diagonal matrix with blocks similar to $\begin{bmatrix}0 & -\omega \\ \omega & 0 \end{bmatrix}$). Finally $c_1, c_2$ and $c_3$ are real positive constants.
I know that for whatever selection of $c_1, c_2$ and $c_3$ lower right block matrix is always Hurwitz.
If one of the non-diagonal blocks is equal to zero, then the eigenvalues of $M$ are the eigenvalues of the diagonal block matrices.
My question here is, if I do $\frac{c_3}{c_1}$ sufficiently small (or $c_3c_2$) then one of the non-diagonal blocks will approach to zero, but on the other hand I am also varying the eigenvalues of the lower right block (which is always Hurwitz).
Can I say that I can make $M$ Hurwitz?
 A: We claim that $M$ is weakly stable (i.e. all its eigenvalues have non-positive real parts) when $E$ is invertible and $k=c_3^2c_2/c_1(>0)$ is sufficiently small, and it is stable (such that all its eigenvalues have negative real parts) for any sufficiently small positive $k$ if $C\left[I+(A-i\omega I)^{-1}B\right]C^Ty\ne0$ for every eigenpair $(i\omega, y)$ of $E$.
Note that a matrix $M$ is (strictly) stable if and only if for any nonzero complex vector $x$, there exists a positive definite matrix $G$ (that may depend on $x$) such that $\operatorname{Re}(x^\ast GMx)<0$ (see Horn and Johnson, Topics in Matrix Analysis, p.107, theorem 2.4.11). Now, suppose $E$ is invertible. Let $\epsilon>0$. Perturb $E$ by $-\epsilon I$ in $M$ and call the resulting matrix $M_\epsilon$.
Note that $(A-wI)^{-1}$ exists and is bounded for every complex number $w$ with nonnegative real part, because $A$ is stable. Also, the eigenvalues of $E-\epsilon I-zI$ are bounded away from zero whenever $\Re(z)\le0$. Therefore, when $k$ is sufficiently small, the matrix
$$
E-\epsilon I-zI - kC\left[I+(A-wI)^{-1}B\right]C^T\tag{1}
$$
is invertible for all complex numbers $z$ and $w$ with nonnegative real parts. So, for such $z$ and $w$, the system of equations
\begin{align*}
y_1:= Ax_1 + \frac{c_3}{c_1}BC^Tx_2 &= wx_1,\tag{2}\\
y_2:=c_3c_2Cx_1 + (E-kCC^T)x_2 &= zx_2\tag{3}
\end{align*}
has only the trivial solution $x^\ast=(x_1^\ast,x_2^\ast)=0$. In other words, when $x\ne0$ and $p$ or $q$ is small, $y_1\ne wx_1$ or $y_2\ne zx_2$. Therefore, there exist positive definite matrices $G_1,G_2$ such that $\operatorname{Re}(x_1^\ast G_1y_1+x_2^\ast G_2y_2)<0$, i.e. $\operatorname{Re}(x^\ast GM_\epsilon x)<0$ for $G=G_1\oplus G_2$. Hence $M_\epsilon$ is (strictly) stable and by continuity, $M$ is weakly stable.
So, the remaining question is whether $M$ can possess a purely imaginary eigenvalue $\lambda=it$ with $t\in\mathbb R$. This occurs when
$$
E-itI - kC\left[I+(A-itI)^{-1}B\right]C^T\tag{4}
$$
is singular for some real $t$. Hence our assertion. Note that $(4)$ could really hold. For instance, when $A,B,C,E$ are $2\times2$, $C=I_2$ and $B=E-A$, if $y$ is an eigenvector of $E$ corresponding to the eigenvalue $i\omega$, then $\pmatrix{py\\ y}$ is an eigenvector of $M$ corresponding to the same eigenvalue.
