# Purely imaginary generalized eigenvalues

Consider real block matrices $$M_1=\left[\begin{matrix}A & 0 \\ 0 & B\end{matrix}\right]$$ $$M_2=\left[\begin{matrix}0 & C \\ -C^T & 0\end{matrix}\right]$$ with $A,B$ square symmetric positive definite (of the same size).

We're interested in the solutions of $|M_2-\lambda M_1|=0$. It is clear that the solutions have a Hamiltonian structure:

• $|M_2-\lambda M_1|$ is a polynomial in $\lambda$ with real coefficients, so its solutions are symmetric with respect to the real axis.

• Also, $|M_2-\lambda M_1|=\pm |M_2+\lambda M_1|$, so the solutions are symmetric with respect to the imaginary axis.

But it doesn't look like this is the whole story: $\lambda$ appears to always be purely imaginary. For example $$\left| \begin{array}{cccc} -17 \lambda & -7 \lambda & 8 & 3 \\ -7 \lambda & -34 \lambda & 4 & 4 \\ -8 & -4 & -5 \lambda & -20 \lambda \\ -3 & -4 & -20 \lambda & -100 \lambda \\ \end{array} \right|=0$$ $$1058 \lambda ^4+3417 \lambda ^2+8=0$$ has four purely imaginary solutions.

When $A,B$ are merely symmetric (not positive definite), the $\lambda$ are not always purely imaginary.

Is it actually true that the solutions of $|M_2-\lambda M_1|=0$ are purely imaginary? I'm looking for a proof, or hints on how to construct one.

• I don't see a contradiction between 1) being symmetrical with respect to coordinate axes and 2) being pure imaginary. – Jean Marie Sep 27 '17 at 5:54
• I agree, it's not a contradiction. I want to understand why they are purely imaginary. – Wouter Sep 27 '17 at 6:12

On further thought, if $M_1$ is indeed symmetric, then it is diagonalized by an orthogonal matrix $Q$: $$M_1=Q D Q^T$$ Then $$|M_2-\lambda M_1|=|M_2-\lambda Q D Q^T|=|Q^T M_2 Q-\lambda D|$$ where $Q^T M_2 Q$ is antisymmetric.

If $M_1$ is positive definite, the diagonal entries of the diagonal matrix $D$ (the eigenvalues of $M_1$) are positive and its square root $D^{1/2}$ is simply the entrywise square root. $$|Q^T M_2 Q-D^{1/2}\lambda D^{1/2}|=|D||D^{-1/2}Q^T M_2 Q D^{-1/2}-\lambda I|$$ $|D|$ is nonzero, so $$|M_2-\lambda M_1|=0\implies |D^{-1/2}Q^T M_2 Q D^{-1/2}-\lambda I|=0$$ and $D^{-1/2}Q^T M_2 Q D^{-1/2}$ is real anti-symmetric, so $$|D^{-1/2}Q^T M_2 Q D^{-1/2}-\lambda I|=0$$ is the eigenvalue equation for a real anti-symmetric (anti-Hermitian) matrix, so $\lambda$ is purely imaginary.

Your problem is interesting and your solution is good. I have a different (half) solution that, despite being not completed, has the interest to connect the question to an eigenvalue issue in dimension 2 instead of 4 (more generally $n$ instead of $2n$).

Consider for a short while your example with polynomial :

$$\tag{1}\varphi(\lambda):=1058 \lambda ^4+3417 \lambda ^2+8=0$$

I propose the interpretation of $\varphi$ as the result of the substitution of $\lambda$ by $\lambda^2$ in

$$\tag{2}\chi(\lambda):=1058 \lambda ^2+3417 \lambda +8=0$$

1) The interest of considering $\chi$ is that it is the characteristic polynomial of the $2 \times 2$ matrix:

$$D:=-C^TA^{-1}CB^{-1}$$

with all the interest that using "pure-flesh" eigenvalues can have.

2) Moreover, it would suffice to show that all roots of (2), i.e., all eigenvalues of $D$, are negative, to prove that all the roots of (1) are pure imaginary.

Let us now prove 1) in the general case. Due to the fact that a positive-definite matrix is invertible:

$$\det(M_2-\lambda M_1)=\det(M_2M_1^{-1}-\lambda M_1M_1^{-1})det(M_1)$$ is zero if and only if

$$\det(E)=0 \ \ \text{with} \ \ E:=M_2M_1^{-1}-\lambda I_4$$

As an easy computation shows that $$E=\left[\begin{matrix}-\lambda I_2 & CB^{-1} \\ -C^TA^{-1} & -\lambda I_2\end{matrix}\right]=-\lambda\left[\begin{matrix}I_2 & -\tfrac{1}{\lambda}CB^{-1} \\ -\tfrac{1}{\lambda}C^T & I_2\end{matrix}\right];$$

it suffices now to apply Property 3 in (https://en.wikipedia.org/wiki/Schur_complement) which is a consequence of Aitken factorization (see as well(https://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.AppP.d/IFEM.AppP.pdf) property P7), to obtain :

$$\det(E)=\lambda^4\det(I_2-\dfrac{1}{\lambda}C^TA^{-1}I_2^{-1}\dfrac{1}{\lambda}CB^{-1})$$

Let us assume $\lambda \neq 0$. $\det(E)$ will be zero if and only if:

$$\det(\lambda^2I_2+\underbrace{(-C^TA^{-1}CB^{-1})}_D)=0$$

establishing part 1).

Unfortunately, I have still not been able to establish part 2. I have done extensive random testing showing : in all cases, the eigenvalues of $D$ where negative, but...