Show that the eigenvalues of a non-symmetric matrix built from positive matrices have positive real parts Let $A, B, C \in \mathbb{R}^{n\times n}$ such that $N = \begin{bmatrix} A & B\\ B^{\top} & C\end{bmatrix}$ is a symmetric positive definite matrix. I'm trying to show that the following matrix
$$
M = \begin{bmatrix} A & B & 0\\ B^{\top} & C & -B^{\top} \\ - A &  -B & A\end{bmatrix}
$$
has eigenvalues with positive real part. Numerical tests suggest this is true, but I cannot prove it.
Edit It is not true that $M + M^\top$ has positive eigenvalues, i.e. that $\langle x, Mx\rangle \geq 0$ for all $x$, which is (?) a sufficient condition for $M$ to have eigenvalues with positive real parts.
Edit 2 After further numerical tests, letting $\lambda_{\min}$ the smallest eigenvalue of $N$, and $v_{\min}$ the smallest real part of the eigenvalues of $M$, it seems like we should have a bound like
$$
v_{\min} \geq \rho \cdot \lambda_{\min}
$$
where the constant $\rho$ is $\simeq 0.963$.
Edit 3 I expect that we can answer the question by finding the right factorization for $M$. For instance it is easy to show that
$$
\begin{bmatrix} A & B & -A\\ B^{\top} & C & -B^{\top} \\ - A &  -B & A\end{bmatrix}
$$
has positive eigenvalues since it equals $ \begin{bmatrix} I & 0 & -I \\ 0& I & 0\end{bmatrix}^{\top}N  \begin{bmatrix} I & 0 & -I \\ 0& I & 0\end{bmatrix}$
 A: A simple brute force method worked (even though I'm not happy with this).
Let $\zeta=\xi+i\eta$ be a non-positive eigenvalue of $M$ and
$\left[\begin{matrix} x & y & z \end{matrix}\right]^T$ be
a corresponding eigenvector.
This gives equations
\begin{align} 
Ax + By \qquad &= \zeta x \\
B^Tx + Cy - B^Tz &= \zeta y\\
-Ax -By + Az &= \zeta z
\end{align}
From the first and the third, one obtains
$$z = -\zeta(\zeta-A)^{-1}x 
\mbox{ and } x-z = (1+\zeta(\zeta-A)^{-1})x.$$
To ease the notation, put $A(\zeta):=1+\zeta(\zeta-A)^{-1}=(2\zeta-A)(\zeta-A)^{-1}$.
From the second,
$$y = (\zeta-C)^{-1}B^T(x-z) = (\zeta-C)^{-1}B^T A(\zeta) x.$$
By combining with the first, one obtains
$$(\zeta-A)x = By =  B(\zeta-C)^{-1}B^T A(\zeta)x.$$
Note that the imaginary part of $(\zeta- C)^{-1}$ is
$$\Im\frac{1}{\zeta- C}=\Im\frac{\bar{\zeta}-C}{|\zeta-C|^2}
=-\eta\frac{1}{|\zeta-C|^2}.$$
Take the inner product with $A(\zeta)x$ and look at the imaginary part:
$$\Im \langle (\zeta-A)x,A(\zeta)x\rangle 
= -\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$
Now, since
\begin{align*}
A(\bar{\zeta})(\zeta-A) 
&= \frac{(2\bar{\zeta}-A)(\zeta-A)^2}{|\zeta-A|^2} \\
&=\frac{2|\zeta|^2\zeta-4|\zeta|^2A+2\bar{\zeta}A^2-\zeta^2A+2\zeta A^2-A^3}{|\zeta-A|^2}, 
\end{align*}
one obtains
$$2\eta\langle \frac{|\zeta|^2-\xi A }{|\zeta-A|^2}x,x\rangle 
=-\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$
Hence, unless $\eta=0$,
$$\xi\geq\frac{|\zeta|^2}{\|A\|}>0.$$
Let's deal with the case $\eta=0$.
Suppose for a contradiction that $\xi\le0$. Then
\begin{align*}
\langle A(\xi)(-\xi+A)x,x\rangle 
&= \langle B(-\xi+C)^{-1}B^T A(\xi)x,A(\xi)x\rangle\\
&\le \langle BC^{-1}B^T A(\xi)x,A(\xi)x\rangle\\
&< \langle A A(\xi)x,A(\xi)x\rangle,
\end{align*}
but this is in contradiction with the fact that
$A(\xi)\succ0$ and $-\xi + A\succeq A(\xi)A$.
ADDED:
We assume $\left[\begin{smallmatrix} 
A_0 & B_0 & \\ 
B_0^T & C_0\end{smallmatrix}\right]\succ 2\epsilon$ and will show
that any eigenvalue of
$\left[\begin{smallmatrix} 
A_0 & B_0 & \\ 
B_0^T & C_0 & -B_0^T \\ 
-A_0 & -B_0 & A_0 
\end{smallmatrix}\right]$ has its real part at least $\epsilon$.
To ease notation, we consider $A=A_0-2\epsilon$ and $C=C_0-2\epsilon$
instead of $A_0$ and $C_0$.
By the IVT trick, it suffices to show
there is no solution for
$$\left[\begin{matrix} 
A+2\epsilon & B & \\ 
B^T & C+2\epsilon & -B^T \\ 
-(A+2\epsilon) & -B & A+2\epsilon 
\end{matrix}\right]
\left[\begin{matrix} 
x \\ y \\ z 
\end{matrix}\right]
 = (\epsilon+i\eta) \left[\begin{matrix} 
x \\ y \\ z 
\end{matrix}\right],$$
with $\left[\begin{smallmatrix} 
A & B & \\ B^T & C\end{smallmatrix}\right]\succ 0$,
$\epsilon>0$, $\eta\in\mathbb{R}$, and
$\left[\begin{smallmatrix} x & y & z \end{smallmatrix}\right]\neq 0$.
The first row + the third row: $-\epsilon x + (A+\epsilon) z = i\eta (x+z).$
Hence
$$x-z=(1-\frac{\epsilon+i\eta}{A+\epsilon-i\eta})x
=\frac{A-2i\eta}{A+\epsilon-i\eta}x=:A(\eta)x.$$
Together with the second row:
$$y=-(C+\epsilon-i\eta)^{-1}B^T(x-z)=-(C+\epsilon-i\eta)^{-1}B^TA(\eta)x.$$
It follows that $x\neq0$.
Together with the first row:
\begin{align*}
(A+\epsilon -i\eta) x
 = -By 
 = B(C+\epsilon-i\eta)^{-1}B^TA(\eta)x
\end{align*}
and so
\begin{equation}\tag{$\ast$} \langle (A+\epsilon -i\eta) x,A(\eta)x\rangle 
= \langle B(C+\epsilon-i\eta)^{-1}B^TA(\eta)x, A(\eta)x\rangle.\end{equation}
Do some calculations:
$$A(\eta)^*(A+\epsilon -i\eta) 
= |A+\epsilon-i\eta|^{-2}(A+\epsilon-i\eta)^2(A+2i\eta),$$
$$(A+\epsilon-i\eta)^2(A+2i\eta)
 = A^3 + 2 \epsilon A^2 + \epsilon^2 A + 3 \eta^2 A
  + 4\epsilon \eta^2 + 2i\eta (\epsilon A + \epsilon^2 - \eta^2),$$
$$A(\eta)^*B(C+\epsilon-i\eta)^{-1}B^TA(\eta)
 =A(\eta)^*B\frac{C+\epsilon +i\eta}{|C+\epsilon-i\eta|^2}B^TA(\eta),$$
$$A(\eta)^*B\frac{C+\epsilon}{|C+\epsilon-i\eta|^2}B^TA(\eta)
 \preceq A(\eta)^*B C^{-1} B^T A(\eta) \prec A|A(\eta)|^2,$$
$$A|A(\eta)|^2 = |A+\epsilon-i\eta|^{-2}(A^3+4\eta^2A).$$
Look at the real part of $(\ast)$: for $w=|A+\epsilon-i\eta|^{-1}x$,
$$\langle(2\epsilon A^2+\epsilon^2A+4\epsilon\eta^2)w,w\rangle
 < \langle \eta^2Aw,w\rangle.$$
Look at the imaginary part of $(\ast)$:
as $\eta\neq0$ from the previous inequality,
$$\langle(\epsilon A + \epsilon^2 - \eta^2)w,w\rangle
 =\langle A(\eta)^*B\frac{1}{|C+\epsilon-i\eta|^2}B^TA(\eta)x,x\rangle/2 
 \geq0.$$
Combine the last two:
\begin{align*}
2\epsilon \|Aw\|^2+\epsilon^2\langle Aw,w\rangle +4\epsilon\eta^2\|w\|^2 
&< \eta^2\langle Aw,w\rangle\\
&\le (\epsilon \frac{\langle Aw,w\rangle}{\|w\|^2} +\epsilon^2)\langle Aw,w\rangle\\
&\le\epsilon\|Aw\|^2 +\epsilon^2\langle Aw,w\rangle.
\end{align*}
We arrive at a contradiction.
A: This post is the solution to the limit case, as suggested by Denis Serre,
where the Schur complement $S=C-B^TA^{-1}B$ is zero.
This means that $W:=A^{-1/2}BC^{-1/2}$ is an orthogonal matrix.
Now put $S=\operatorname{diag}(A^{1/2}, C^{1/2}W^T, A^{1/2})$ and consider
$$N:=S^{-1}MS
=\left[\begin{matrix}
 A & D & 0 \\
 A & D & -A \\
 -A & -D & A
\end{matrix}\right],
$$
where $D:=WCW^T\succ0$.
Note that the second and the third rows are the same modulo the negative sign.
Suppose $\zeta=\xi+i\eta$ is a nonzero eigenvalue of $M$.
Then the corresponding eigenvector for $N$ is of the
form $\left[\begin{matrix} x & y & -y \end{matrix}\right]^T$
with nonzero vectors $x$ and $y$.
This gives equations
$$Ax + Dy = \zeta x$$
and
$$Ax + Dy  + Ay = \zeta y.$$
It follows that $\zeta x = (\zeta-A)y$ and
$$\zeta Dy=\zeta (\zeta-A)x = (\zeta-A)^2y=(\zeta^2-2\zeta A +A^2)y.$$
Take the inner product with $y$ and look at the imaginary part of it:
$$\eta \langle Dy,y\rangle = 2\xi\eta \langle y,y\rangle 
 - 2\eta \langle Ay,y\rangle.$$
Thus, one has $\xi>0$, provided that $\eta\neq0$.
One still has the same conclusion for $\eta=0$ because $\zeta Dy=(\zeta-A)^2y$.
A: This is only for the case n=1. Clearly the claim is true if B=0. Hence it suffices to show that there is never a zero or purely imaginary eigenvalue. The case of a zero eigenvalue leads to $a(ac-b^2)=0$, but this quantity is positive by assumption. The case of a purely imaginary eigenvalue $iy$ leads to the pair of equations $$A(AC-B^2)-y^2(2A+C)=0,$$ $$-A^2+2(B^2-AC)+y^2=0.$$ By elimination we conclude that $$(y^2-A^2)/2=y^2(2A+C)/A.$$ This is a contradiction, since the left hand side is less than $y^2/2$, and the right hand side is greater than $2y^2$.
A: Since there is not much progress on this question, let me give a partial result and a direct consequence.
Denote $S=C-B^TA^{-1}B$ the Schur complement of $A$ in $N$, which is positive definite. Then $M$ factorizes $UL$ with
$$U=\begin{pmatrix} A & B & 0 \\ 0 & S & -B^T \\ 0 & 0 & A \end{pmatrix},\qquad L=\begin{pmatrix} I & 0 & 0 \\ 0 & I & 0 \\ -I & 0 & I \end{pmatrix}.$$
We derive the formula $\det M=\det U\det L=(\det A)^2\det S$, that is
$$\det M=\det A\cdot\det N.$$
I have also been interested in the limit case where $S=0$, because if the OP's statement is true, then it remains true for this limit. Denoting $K=A^{-1}B$, one finds the characteristic polynomial
$$\chi_M(X)=X^n\det((XI-A)^2-XKK^TA).$$
This raises the sub-question:

Assume that $A,H$ are symmetric and positive definite. Is it true that the roots of $\det((XI-A)^2-XHA)$ have a positive real part ?

The answer is obviously positive when $H=h^2I$. Could this polynomial be the characteristic polynomial of some matrix $Q$ such that $Q+Q^T$ is positive definite ?
