# How can I show that 0 is not an eigenvalue of the Hamiltonian matrix?

I have the following Hamiltonian matrix

\begin{align} H = \begin{pmatrix}A&-BR^{-1}B^{\intercal}\\-C^{\intercal}C&-A^{\intercal}\end{pmatrix} \end{align}

that admits eigenvalues in pairs, i.e. if $$\lambda$$ is an eigenvalue, then so is $$-\lambda$$. Now, $$H$$ is clearly a block-matrix and has has the dimension $$H\in\mathbb{R}^{2n\times2n}$$ (each of the blocks is in $$\mathbb{R}^{n\times n}$$). For this case we have that $$R$$ is positive definite and the special relations that

\begin{align} \Gamma = \begin{bmatrix}B&AB&A^{2}B&A^{3}B&\dots&A^{n-1}B\end{bmatrix}\quad \text{and}\quad \Omega =\begin{bmatrix}C\\CA\\CA^{2}\\CA^{3}\\\vdots\\CA^{n-1}\end{bmatrix} \end{align}

are of full rank. In control theory, or systems theory, we say that $$(A,B)$$ is reachable and $$(C,A)$$ is observable when $$\Gamma$$ and $$\Omega$$ are of full rank, respectively (just a side note). Since my knowledge of matrix algebra and linear algebra is weak, I have yet to come up with any fruitful approach that would solve this problem. I have tried assuming, for contradiction, that 0 is an eigenvalue, which implies that

\begin{align} Hx = 0 \end{align}

where $$x$$ is the corresponding eigenvector, but that is about as far as that takes me. Can I use, somehow, that the eigenvalues come in pairs? Is it not the case that $$H^{\intercal}H$$ shares eigenvalues with $$H$$, although they are called singular values?

Edit:

Proof that eigenvalues come in pair:

\begin{align*} \lambda\begin{bmatrix}v\\w\end{bmatrix} &= \begin{bmatrix}A&-BR^{-1}B^{\intercal}\\-C^{\intercal}C&-A^{\intercal}\end{bmatrix}\begin{bmatrix}v\\w\end{bmatrix} = \begin{bmatrix}Av-BR^{-1}B^{\intercal}w\\-C^{\intercal}Cv-A^{\intercal}w\end{bmatrix} \\ & \implies \left\{\begin{matrix}Av-BR^{-1}B^{\intercal}w=\lambda v\\-C^{\intercal}Cv-A^{\intercal}w=\lambda w \end{matrix}\right. \end{align*}

We investigate what happens when we right-multiply $$H^{\intercal}$$ by the vector $$\begin{bmatrix}w&-v\end{bmatrix}^{\intercal}$$:

\begin{align*} H^{\intercal}\begin{bmatrix}w\\-v\end{bmatrix}&= \begin{bmatrix}A^{\intercal}&-C^{\intercal}C\\-BR^{-1}B^{\intercal}&-A\end{bmatrix}\begin{bmatrix}w\\-v\end{bmatrix} = \begin{bmatrix}A^{\intercal}w + C^{\intercal}Cv\\-BR^{-1}B^{\intercal}w +Av\end{bmatrix} \\ & = \begin{bmatrix}-\lambda w\\\lambda v\end{bmatrix}= -\lambda \begin{bmatrix} w\\- v\end{bmatrix} \end{align*}

Since $$H$$ and $$H^{\intercal}$$ share eigenvalues (same characteristic polynomial) we have showed what we wanted.

Edit:

The Hamiltonian matrix is associated with the algebraic Riccati equation:

\begin{align*} A^{\intercal}P+ PA - PBR^{-1}B^{\intercal}P + C^{\intercal}C = 0. \end{align*}

Here $$P\in\mathbb{R}^{n\times n}$$ is a symmetric positive definite solution that can be shown to exist and be unique when $$\Gamma$$ and $$\Omega$$ are of full rank. Note that $$C^{\intercal}C \geq 0$$ (symmetric positive semidefinite).

• Comments are not for extended discussion; this conversation has been moved to chat. Dec 20, 2020 at 12:57

Suppose that $$P$$ is positive definite and solves the CARE. It is known that the feedback control law $$u = -R^{-1}B^TP$$ minimizes the cost function $$\int_0^\infty (x^TC^TCx + u^TRu)\,dt,$$ and that this minimal cost is finite. Define $$K = A - BR^{-1}B^TP$$. The feedback input/state system is governed by $$\frac {dx}{dt} = Kx.$$ Because the cost-function integral converges, it must be the case that $$\lim_{t \to \infty} x(t) = 0$$ for any initial state $$x(0)$$. This can only be the case if $$K$$ is stable, which is to say that the eigenvalues of $$K$$ have negative real part.
On the other hand, we find that $$PK = -PBR^{-1}B^TP + PA = -C^TC - A^TP \implies\\ H \pmatrix{I\\P} = \pmatrix{A & -BR^{-1}B^T\\-C^TC & -A^T} \pmatrix{I \\ P} = \pmatrix{K\\PK} = \pmatrix{I\\P} K.$$ Let $$D$$ denote the Jordan form of $$K$$, and let $$X$$ be such that $$K = XDX^{-1}$$. We have $$H \pmatrix{I\\P} = \pmatrix{I\\P} XJX^{-1} \implies H \pmatrix{X\\PX} = \pmatrix{X\\PX} D.$$ In other words, each eigenvalue of $$K$$ is also an eigenvalue of $$H$$, and the associated eigenvectors are the columns of $$\pmatrix{X\\PX}$$.
Thus, $$H$$ has $$n$$ eigenvalues with negative real part. Because the eigenvalues of $$H$$ come in $$\pm$$ pairs, it must hold that $$H$$ has $$n$$ more eigenvalues with positive real part. Thus, all eigenvalues of $$H$$ have either negative or positive real part, which means that none of them can have zero real part.
In particular, we conclude that none of the eigenvalues of $$H$$ can be $$0$$.