Principal Solutions at Infinity of Given Ranks for Nonoscillatory Linear Hamiltonian Systems

Abstract

In this paper we study the existence and properties of the principal solutions at infinity of nonoscillatory linear Hamiltonian systems without any controllability assumption. As our main results we prove that the principal solutions can be classified according to the rank of their first component and that the principal solutions exist for any rank in the range between explicitly given minimal and maximal values. The minimal rank then corresponds to the minimal principal solution at infinity introduced by the authors in their previous paper, while the maximal rank corresponds to the principal solution at infinity developed by W. T. Reid, P. Hartman or W. A. Coppel. We also derive a classification of the principal solutions, which have eventually the same image. The proofs are based on a detailed analysis of conjoined bases with a given rank and their construction from the minimal conjoined bases. We illustrate our new theory by several examples.

Appendix 1: Auxiliary Results About Orthogonal Projectors

In this section we present some results from matrix analysis, in particular about orthogonal projectors. We also derive a representation of the set ${\mathcal B}(P_{\!**},P_{\!*},P)$ introduced in (2.2).

Lemma 9.1

Let ${P_{\!*}}\in {\mathbb R}^{n\times n}$ be an orthogonal projector and $p_*:=\mathrm{rank} \, {P_{\!*}}$. Furthermore, let ${V_{\!*}}\in {\mathbb R}^{n\times n}$ be the corresponding orthogonal matrix from (2.1), i.e.

$$\begin{aligned} {P_{\!*}}={V_{\!*}}\ \mathrm{diag}\{I_{p_*},\,0_{n-p_*}\}\,{V_{\!*}}^T. \end{aligned}$$

(9.1)

Let $p\in {\mathbb N}$ satisfy $p_*\le p \le n$. Then a matrix $P\in {\mathbb R}^{n\times n}$ is an orthogonal projector with

$$\begin{aligned} \mathrm{Im} \, {P_{\!*}}\subseteq \mathrm{Im} \, P\quad \text {and} \quad \mathrm{rank} \, P=p \end{aligned}$$

(9.2)

if and only if $P$ has the form

$$\begin{aligned} P={V_{\!*}}\ \mathrm{diag}\{I_{p_*},\,{R_{*}}\}\,{V_{\!*}}^T, \end{aligned}$$

(9.3)

where ${R_{*}}\in {\mathbb R}^{(n-p_*)\times (n-p_*)}$ is an orthogonal projector with rank equal to $p-p_*$.

Proof

It is easy to see that every matrix $P$ of the form (9.3) is symmetric and idempotent (i.e., it is an orthogonal projector) and (9.2) holds. Conversely, suppose that $P\in {\mathbb R}^{n\times n}$ is an orthogonal projector satisfying (9.2). Then we may write

$$\begin{aligned} P={V_{\!*}}\left( \begin{matrix} K_* &{} L_*\\ L_*^T &{} {R_{*}} \end{matrix}\right) {V_{\!*}}^T, \end{aligned}$$

(9.4)

where $K_*\in {\mathbb R}^{p_*\times p_*}$ is symmetric, $L_*\in {\mathbb R}^{p_*\times (n-p_*)}$, and ${R_{*}}\in {\mathbb R}^{(n-p_*)\times (n-p_*)}$ is symmetric. The first condition in (9.2) is equivalent with $P{P_{\!*}}={P_{\!*}}$, from which we get by using the representations in (9.1) and (9.4) that $K_*=I_{p_*}$ and $L_*=0_{p_*\times (n-p_*)}$. Thus, $P={V_{\!*}}\mathrm{\text {diag} }\{I_{p_*},\,{R_{*}}\}\,{V_{\!*}}^T$, where $\mathrm{\text {rank} }\,{R_{*}}=\mathrm{\text {rank} }P-p_*=p-p_*$ according to (9.2). Finally, the idempotence of $P$ now implies the idempotence of ${R_{*}}$, showing that ${R_{*}}$ is an orthogonal projector. $\square$

Theorem 9.2

Let ${P_{\!*}},P,{\tilde{P}}\in {\mathbb R}^{n\times n}$ be orthogonal projectors satisfying

$$\begin{aligned} \mathrm{Im} \, {P_{\!*}}\subseteq \mathrm{Im} \, P, \quad \mathrm{Im} \, {P_{\!*}}\subseteq \mathrm{Im} \, {\tilde{P}}, \quad \mathrm{rank} \, P=\mathrm{rank} {\tilde{P}}. \end{aligned}$$

(9.5)

Then there exists an invertible matrix $E\in {\mathbb R}^{n\times n}$ such that $E{P_{\!*}}={P_{\!*}}$ and  $\mathrm{Im} \, EP=\mathrm{Im} \,{\tilde{P}}$.

Proof

Let $p_*:=\mathrm{\text {rank} }\, {P_{\!*}}$ and $p:=\mathrm{\text {rank} }\, P=\mathrm{\text {rank} }{\tilde{P}}$. Then obviously $p\ge p_*$. Let ${V_{\!*}}\in {\mathbb R}^{n\times n}$ be the orthogonal matrix in (2.1) associated with projector ${P_{\!*}}$, that is, (9.1) holds. According to Lemma 9.1 there exist orthogonal projectors ${R_{*}},\tilde{R_{*}}\in {\mathbb R}^{(n-p_*)\times (n-p_*)}$ such that

$$\begin{aligned} P={V_{\!*}}\mathrm{\text {diag} }\{I_{p_*},\,{R_{*}}\}\,{V_{\!*}}^T, \quad {\tilde{P}}={V_{\!*}}\mathrm{\text {diag} }\{I_{p_*},\,\tilde{R_{*}}\}\,{V_{\!*}}^T, \end{aligned}$$

(9.6)

and $\mathrm{\text {rank} }\, R_*=\mathrm{\text {rank} }\, \tilde{R_{*}}=p-p_*$. Let $Z_*,\tilde{Z}_*\in {\mathbb R}^{(n-p_*)\times (n-p_*)}$ be orthogonal matrices in (2.1) associated with the projectors ${R_{*}}$ and $\tilde{R_{*}}$, that is, ${R_{*}}=Z_*\mathrm{\text {diag} }\{I_{p-p_*},\,0_{n-p}\}\,Z_*^T$ and $\tilde{R_{*}}=\tilde{Z}_*\mathrm{\text {diag} }\{I_{p-p_*},\,0_{n-p}\}\,\tilde{Z}_*^T$. It follows that

$$\begin{aligned} \tilde{Z}_*Z_*^T{R_{*}}=\tilde{Z}_*\mathrm{\text {diag} }\{I_{p-p_*},\,0_{n-p}\}\,Z_*^T=\tilde{R_{*}}\tilde{Z}_*Z_*^T. \end{aligned}$$

(9.7)

We set $E:={V_{\!*}}\mathrm{\text {diag} }\{I_{p_*},\,\tilde{Z}_*Z_*^T\}\,{V_{\!*}}^T\in {\mathbb R}^{n\times n}$. Then $E$ is nonsingular and from (9.1) it follows that $E{P_{\!*}}={P_{\!*}}$. Finally, by (9.6) and (9.7) we obtain

$$\begin{aligned} EP={V_{\!*}}\mathrm{\text {diag} }\{I_{p_*},\,\tilde{Z}_*Z_*^T{R_{*}}\}\,{V_{\!*}}^T={V_{\!*}}\mathrm{\text {diag} }\{I_{p_*},\,\tilde{R_{*}}\tilde{Z}_*Z_*^T\}\,{V_{\!*}}^T={\tilde{P}}E, \end{aligned}$$

which shows that $\mathrm{\text {Im} }\, EP=\mathrm{\text {Im} }\,{\tilde{P}}E=\mathrm{\text {Im} }\, {\tilde{P}}$. The proof is complete. $\square$

In the following (see Theorem 9.5) we provide a certain representation of the set ${\mathcal B}(P_{\!**},P_{\!*},P)$ defined in (2.2), where $P_{\!**}$, $P_{\!*}$, and $P$ are $n\times n$ orthogonal projectors such that

$$\begin{aligned} \mathrm{\text {Im} }\, P_{\!**}\subseteq \mathrm{\text {Im} }\, P_{\!*}\subseteq \mathrm{\text {Im} }\, P. \end{aligned}$$

(9.8)

First we consider the special case with $P=I$. In this case the elements of ${\mathcal B}(P_{\!**},P_{\!*},I)$ are characterized by (2.3). For ${\bar{G}}\in {\mathbb R}^{n\times n}$ we consider the matrices ${\bar{G}_{\!\!\perp }}$ and ${\bar{G}_{\!\parallel }}$ defined by

$$\begin{aligned} {\bar{G}_{\!\!\perp }}\!:={\bar{G}}\,(I-{P_{\!*}})\quad \text {and} \quad {\bar{G}_{\!\parallel }}\!:={\bar{G}}{P_{\!*}}. \end{aligned}$$

(9.9)

Then ${\bar{G}}={\bar{G}_{\!\!\perp }}+{\bar{G}_{\!\parallel }}$ holds. The next theorem provides a characterization of the elements ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$ via the matrices ${\bar{G}_{\!\!\perp }}$ and ${\bar{G}_{\!\parallel }}$ associated with ${\bar{G}}$ through (9.9). For convenience we will use the notation

$$\begin{aligned} {\bar{G}_{\!\!\perp }}{}^{\!\!T}\!:=({\bar{G}_{\!\!\perp }})^T\!, \quad {\bar{G}_{\!\parallel }}{}^{\!T}\!:=({\bar{G}_{\!\parallel }})^T\!, \quad {\bar{G}_{\!\!\perp }}{}^{\!\!\dagger }\!:=({\bar{G}_{\!\!\perp }})^\dagger \!, \quad {\bar{G}_{\!\!\perp }}{}^{\!\!\dagger T}\!:=({\bar{G}_{\!\!\perp }})^{\dagger T}\!. \end{aligned}$$

(9.10)

Theorem 9.3

Let ${P_{\!**}}$ and ${P_{\!*}}$ be orthogonal projectors with $\mathrm{Im} \, P_{\!**}\subseteq \mathrm{Im} \, P_{\!*}$. Then ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$ if and only if its corresponding matrices ${\bar{G}_{\!\!\perp }}$ and ${\bar{G}_{\!\parallel }}$ satisfy

$$\begin{aligned} \mathrm{Im} \, {\bar{G}_{\!\!\perp }}\!=\mathrm{Im} \,\,(I-{P_{\!*}}), \quad {P_{\!**}}\,{\bar{G}_{\!\parallel }}\!=0, \quad {P_{\!*}}\,{\bar{G}_{\!\parallel }}\!={\bar{G}_{\!\parallel }}{}^{\!T}{P_{\!*}}. \end{aligned}$$

(9.11)

Proof

Let ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$. First we show that $\mathrm{\text {Im} }\, {\bar{G}_{\!\!\perp }}\!=\mathrm{\text {Im} }\,\,(I-{P_{\!*}})=\mathrm{\text {Ker} }\,{P_{\!*}}$. Since by (2.3) the matrix ${P_{\!*}}\,{\bar{G}}$ is symmetric, we have ${P_{\!*}}\,{\bar{G}_{\!\!\perp }}={P_{\!*}}\,{\bar{G}}\,(I-{P_{\!*}})={\bar{G}}^T{P_{\!*}}\,(I-{P_{\!*}})=0$. This shows that $\mathrm{\text {Im} }\,{\bar{G}_{\!\!\perp }}\!\subseteq \mathrm{\text {Ker} }\,{P_{\!*}}$. Moreover, $\mathrm{\text {rank} }\,\,({\bar{G}_{\!\!\perp }}{}^{\!\!T}\!,\,{P_{\!*}})=\mathrm{\text {rank} }\,\,((I-{P_{\!*}})\,{\bar{G}}^T\!\!,\,{P_{\!*}})=\mathrm{\text {rank} }\,\,({\bar{G}}^T\!\!,\,{P_{\!*}})=n$, by the last condition in (2.3). But the definition of ${\bar{G}_{\!\!\perp }}$ in (9.9) yields that ${P_{\!*}}\,{\bar{G}_{\!\!\perp }}{}^{\!\!T}=0$, so that $\mathrm{\text {Im} }\,{\bar{G}_{\!\!\perp }}{}^{\!\!T}\subseteq \mathrm{\text {Ker} }\,{P_{\!*}}$. It follows that $\mathrm{\text {rank} }\,{\bar{G}_{\!\!\perp }}=\mathrm{\text {rank} }\,{\bar{G}_{\!\!\perp }}{}^{\!\!T}=\mathrm{\text {def} }{P_{\!*}}$ and thus, $\mathrm{\text {Im} }\, {\bar{G}_{\!\!\perp }}\!=\mathrm{\text {Im} }\,\,(I-{P_{\!*}})$. Furthermore, from (2.3) we get ${P_{\!**}}{\bar{G}_{\!\parallel }}={P_{\!**}}{\bar{G}}{P_{\!*}}=0$. The last condition in (9.11) follows again from (2.3) and from the idempotence of ${P_{\!*}}$, because ${P_{\!*}}\,{\bar{G}_{\!\parallel }}={P_{\!*}}\,{\bar{G}}{P_{\!*}}={P_{\!*}}{P_{\!*}}\,{\bar{G}}{P_{\!*}}={P_{\!*}}\,{\bar{G}}^T{P_{\!*}}{P_{\!*}}={\bar{G}_{\!\parallel }}{}^{\!T}{P_{\!*}}$. Conversely, let ${\bar{G}}\in {\mathbb R}^{n\times n}$ be such that the corresponding matrices ${\bar{G}_{\!\!\perp }}$ and ${\bar{G}_{\!\parallel }}$ defined in (9.9) satisfy (9.11). We shall prove that (2.3) holds. The first equality in (2.3) follows from (9.11) and from the identity ${P_{\!**}}={P_{\!**}}{P_{\!*}}$, since ${P_{\!**}}{\bar{G}}={P_{\!**}}({\bar{G}_{\!\!\perp }}+{\bar{G}_{\!\parallel }})={P_{\!**}}{P_{\!*}}\,{\bar{G}_{\!\!\perp }}=0$. The symmetry of ${P_{\!*}}\,{\bar{G}}$ follows from ${P_{\!*}}\,{\bar{G}}={P_{\!*}}\,({\bar{G}_{\!\!\perp }}+{\bar{G}_{\!\parallel }})={P_{\!*}}\,{\bar{G}_{\!\parallel }}$ and from the symmetry of ${P_{\!*}}\,{\bar{G}_{\!\parallel }}$. Finally, ${P_{\!*}}\,{\bar{G}_{\!\!\perp }}{}^{\!\!T}\!=0$ by (9.9) and $\mathrm{\text {rank} }\,{\bar{G}_{\!\!\perp }}{}^{\!\!T}=\mathrm{\text {rank} }\,{\bar{G}_{\!\!\perp }}=\mathrm{\text {def} }{P_{\!*}}$ by the first condition in (9.11). Hence, $\mathrm{\text {rank} }\,\,({\bar{G}}^T\!\!,\,{P_{\!*}})=\mathrm{\text {rank} }\,\,((I-{P_{\!*}})\,{\bar{G}}^T\!\!,\,{P_{\!*}})=\mathrm{\text {rank} }\,\,({\bar{G}_{\!\!\perp }}{}^{\!\!T}\!,\,{P_{\!*}})=n$, which completes the proof.$\square$

Remark 9.4

If ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$, then the first conditions in (9.9) and (9.11) imply that

$$\begin{aligned} {\bar{G}_{\!\!\perp }}\,{\bar{G}_{\!\!\perp }}{}^{\!\!\dagger }\!=I-{P_{\!*}}={\bar{G}_{\!\!\perp }}{}^{\!\!\dagger }\,{\bar{G}_{\!\!\perp }}. \end{aligned}$$

(9.12)

In the last result of this section we provide a representation of the set ${\mathcal B}(P_{\!**},P_{\!*},P)$ in (2.2) in terms of the elements ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$.

Theorem 9.5

Let ${P_{\!**}}$, ${P_{\!*}}$, and $P$ be orthogonal projectors satisfying (9.8). Then $(G,H)\in {\mathcal B}(P_{\!**},P_{\!*},P)$ if and only if for some ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$ we have

$$\begin{aligned} G=P\,{\bar{G}}\quad \text {and} \quad H=(I-P)\,{\bar{G}}. \end{aligned}$$

(9.13)

In this case, the matrix ${\bar{G}}$ is uniquely determined by $(G,H)$ with ${\bar{G}}=G+H$.

Proof

Let $(G,H)\in {\mathcal B}(P_{\!**},P_{\!*},P)$ and set ${\bar{G}}:=G+H$. Then clearly $G=P{\bar{G}}$ and $H=(I-P)\,{\bar{G}}$, by the third and fifth properties in (2.2). We show that the matrix ${\bar{G}}$ satisfies (2.3). Using the second and fifth equalities in (2.2) together with the identity ${P_{\!**}}={P_{\!**}}P$ we get ${P_{\!**}}{\bar{G}}={P_{\!**}}G+{P_{\!**}}H={P_{\!**}}PH=0$. Moreover, the identities ${P_{\!*}}={P_{\!*}}P$ and $PH=0$ and the symmetry of ${P_{\!*}}\,G$, i.e. the fourth equality in (2.2), imply the symmetry of ${P_{\!*}}\,{\bar{G}}$, because ${P_{\!*}}\,{\bar{G}}={P_{\!*}}\,G+{P_{\!*}}H={P_{\!*}}\,G+{P_{\!*}}PH={P_{\!*}}\,G$. For the last equality in (2.3) it suffices to prove that $\mathrm{\text {Ker} }\,{\bar{G}}\cap \mathrm{\text {Ker} }\,{P_{\!*}}=\{0\}$. Let $v\in {\mathbb R}^n$ be such that ${\bar{G}}v=0$ and ${P_{\!*}}v=0$. Then also $Gv=P{\bar{G}}v=0$ and $Hv=(I-P)\,{\bar{G}}v=0$. Thus, $v\in \mathrm{\text {Ker} }\, G\cap \mathrm{\text {Ker} }\, H\cap \mathrm{\text {Ker} }\, {P_{\!*}}$ and consequently, $v=0$ by the first equality in (2.2). Therefore, the matrix ${\bar{G}}$ belongs to the set ${\mathcal B}(P_{\!**},P_{\!*},I)$. Conversely, for any ${\bar{G}}\in {\mathcal B}(P_{\!**},P_{\!*},I)$ the pair $(G,H):=(P\,{\bar{G}},(I-P)\,{\bar{G}})$ satisfies conditions (2.2), as can be directly verified by (2.3). Hence, $(G,H)\in {\mathcal B}(P_{\!**},P_{\!*},P)$. Finally, we have $G+H={\bar{G}}$ and so the matrix ${\bar{G}}$ is unique. The proof is complete. $\square$