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 $L^1(0,1)$.



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


