Is there an infinite dimensional analogue of the characterisation of irreducible matrices

Question

There is this result characterizing irreducible square matrices (e.g., L. Hogben, Handbook of Linear Algebra):

If $P$ is a non-negative square matrix, then $P$ is irreducible iff every eigenvector corresponding to the spectral radius $\rho(P)$ is a scalar multiple of a positive vector.

Is there any analogue of this theorem (especially an analogue of the If part) for irreducible operators (in the sens that $A$ defined on a Banach lattice is irreducible if there is no non-trivial closed ideal that is invariant for $A$)?

In particular, I am interested of such characterisation in the case of operators in $L^1(0,1)$.

Answer 1

The characterisation of irreducible matrices as suggested in the question is not correct. The matrix $$ P = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} $$ is a counterexample: it has spectral radius $1$ and the eigenspace for this eigenvalue is spanned by the vector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$; yet, $P$ is not irreducible.

However, if we impose an additional assumption on $P$, then the characterisation becomes true and can be generalised to infinite dimensions. To simplify the notation, we assume that $\rho(P) = 1$.

Let $E$ be a Banach lattice with positive cone $E_+$ and let $P: E \to E$ be a positive linear operator, meaning that $PE_+ \subseteq E_+$. We call a vector $x\in E$

• a fixed vector of $P$ if $x = Px$,
• a super fixed vector of $P$ if $0 \le x \le Px$,
• a quasi-interior point of $E_+$ if $x \in E_+$ and if the set $\bigcup_{c \ge 0} \{y \in E: \, |y| \le c x\}$ is dense in $E$.

Theorem. Suppose that $E$ has order continuous norm, that $\rho(P) = 1$ is an eigenvalue of $P$ and that every super fixed point of $P$ is actually a fixed point.
Then $P$ is irreducible if and only if the fixed space $\ker(1 - P)$ is one-dimensional and spanned by a quasi interior point of $E_+$.

Proof. The implication "$\Rightarrow$" can be shown in the same way as Theorem V.5.2(i) in H. H. Schaefer, Banach Lattices and Positive Operators (1974); for this implication we do not need the assumption that the norm on $E$ be order continuous.

To show "$\Leftarrow$", let $x$ be a quasi-interior point of $E_+$ which is a fixed vector of $P$. Assume for a contradiction that $I \subseteq E$ is a non-trivial closed $P$-invariant ideal. Since $E$ has order continuous norm, $I$ is a projection band in $E$, so there exists a band projection $Q_1: E \to E$ onto $I$. Let $Q_2 = 1 - Q_1$ be the complementary band projection. Both $Q_1$ and $Q_2$ are non-zero since $I$ is non-trivial. Since $x$ is a quasi-interior point of $E_+$, this implies that both $Q_1x$ and $Q_2x$ are non-zero.

We have $PQ_1x = Q_1PQ_1x \le Q_1Px = Q_1x$, so $Q_2x = x - Q_1x$ is a super fixed vector and thus a fixed vector of $P$. Therefore, $Q_1x = x - Q_2x$ is a fixed vector of $P$, too - which is a contradiction since $Q_1x$ and $Q_2x$ are linearly independent. This proves "$\Leftarrow$".

Remarks:

1. All $L^p$-spaces ($1 \le p < \infty$) are Banach lattices with order continuous norms; in particular, the Theorem applies to the space $L^1(0,1)$ mentioned in the question.

2. The assumption that every super fixed point of $P$ be a fixed point is, for instance, fulfilled in each of the following two cases:

   (a) If $\|P\| \le 1$ and if, in addition, the norm on $E$ is strictly increasing, meaning that $\|f\| < \|g\|$ whenever $0 \le f \le g$ but $f \not= g$.

   (b) If there exists a functional $0 \le \varphi \in E'$ which fulfils $P'\varphi = \varphi$ (where $P': E' \to E'$ denotes the dual operator of $P$) and which is strictly positive, meaning that $\langle \varphi, x \rangle > 0$ whenever $x \in E_+ \setminus \{0\}$.

3. The matrix $P = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$ does neither fulfil (a) nor (b). It is illustrative to note that $P$ is contractive with respect to the $\|\cdot\|_\infty$-norm, but that this norm is not strictly increasing. Also note that the fixed space of the dual operator $P'$ is spanned by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ which is not strictly positive.