*Preliminary remarks:*

In the comments the OP noted that he is primarly interested in the question whether the dual operator of an irreducible operator is irreducible, so I will focus on this aspect here.

The OP asked for a reference where this topic is treated. But to the best of my knowledge, all the standard references for Banach lattices and positive operators do not discuss characterisations of irreducible operators in detail. The topic might appear in some research papers or theses, but the only reference I could find is Gao's thesis quoted in Remark 6 below.

Below I discuss a few results (with proofs) which might be of interest for the OP. Several similar results are contained in Gao's thesis, but with a focus on *band* irreducibility rather than irreducibility of the dual operator (note that the term "irreducibility" is usually interpreted as *ideal* irreducibility unless otherwise specified).

*Assumptions:* Throughout, let $E$ be a (real or complex) Banach lattice and let $T$ be a positive (and thus bounded) linear operator on $E$. We denote the dual space of $E$ by $E'$ and the *dual* (or *adjoint*) operator of $T$ by $T'$; obviously, $T'$ is positive, too.

To rule out any misunderstandings, let us agree on the following terminology:

**Definition.** The operator $T$ is called irreducible if $0$ and $E$ are the only $T$-invariant closed ideals in $E$.

The following characterisation of irreducibility is quite useful for our purposes:

**Proposition 1.** The operator $T$ is irreducible if and only if for every $f \in E_+ \setminus \{0\}$ and every $\varphi \in E'_+ \setminus \{0\}$ there exists an integer $n \in \mathbb{N}_0 := \{0,1,2,\dots\}$ such that $\langle \varphi, T^n f\rangle > 0$.

**Proof.** "$\Rightarrow$" Assume that there exist $f \in E_+ \setminus \{0\}$ and $\varphi \in E'_+ \setminus \{0\}$ such that $\langle \varphi, T^nf\rangle = 0$ for all $n \in \mathbb{N}_0$. We define
\begin{align*}
I := \{g \in E: \; \langle \varphi, T^n|g|\rangle = 0 \text{ for all } n \in \mathbb{N}_0\}.
\end{align*}
Then $I$ is a closed and $T$-invariant ideal in $E$. We have $f \in I$, so $I$ is non-zero. Moreover, every positive vector in $I$ is contained in the kernel of $\varphi$. Since $\varphi$ is non-zero, we conclude that $I \not= E$. Hence, $T$ is not irreducible.

"$\Leftarrow$" Assume that $T$ is not irreducible. Then there exists a $T$-invariant closed ideal $I$ in $E$ such that $\{0\} \subsetneq I \subsetneq E$. The quotient space $E/I$ is non-zero, so there exists a non-zero positive functional $\psi \in (E/I)'$. Let $q: E \to E/I$ denote the quotient mapping and define $\varphi = \psi \circ q = q'\psi \in E'$. Then $\varphi \in E'_+ \setminus \{0\}$, and $\varphi$ vanishes on $I$. Since $I$ is non-zero, we can find a vector $0 \le f \in I \setminus \{0\}$. We have $T^nf \in I$ for all $n \in\mathbb{N}_0$ (since $I$ is $T$-invariant) and hence, $\langle \varphi, T^nf\rangle = 0$ for all $n \in \mathbb{N}_0$.

**Corollary 2.** If $T'$ is irreducible, then so is $T$.

**Corollary 3.** Assume that $E$ is reflexive. Then $T$ is irreducible if and only if $T'$ is irreducible.

We even have the following result which is a bit more general than Corollary 3:

**Corollary 4** Assume that both $E$ and $E'$ have order continuous norm. Then $T$ is irreducible if and only if $T'$ is irreducible.

**Proof.** We only have to show "$\Rightarrow$". Assume that $T$ is irreducible, let $\varphi \in E'_+ \setminus \{0\}$ and $\psi \in E''_+ \setminus \{0\}$. Since $E$ has order continuous norm, the band in $E''$ generated by $E$ coincides with the space of all order-continuous functionals on $E'$ (see [Schaefer: Banach Lattices and Positive Operators, 1974, Corollary 1 to Theorem II.5.10 on page 90]). However, $E'$ has also order continuous norm, so every element of $E''$ is an order-continuous functional on $E'$. Hence, the band generated in $E''$ by $E$ equals $E''$. Moreover, $E$ is an ideal in $E''$ since $E$ has order continuous norm. Thus, there exists a vector $f \in E_+ \setminus \{0\}$ such that $f \le \psi$. Since $T$ is irreducible, there exists (by Proposition 1) an integer $n \in \mathbb{N}_0$ such that $\langle \varphi, T^nf \rangle > 0$. Hence, $\langle \psi, (T')^n\varphi\rangle \ge \langle \varphi, T^nf\rangle > 0$, so $T'$ is irreducible (again by Proposition 1).

**Remark 5:** *The* classical example where Corollary 4 can be applied, while Corollary 3 cannot, is the space $E = c_0$ of null sequences whose dual space is given by $E' = \ell^1$.

**Remark 6:** One can prove a bit more general results if one also considers the notion of *band irreducibility* (in particular on the dual space $E'$). Details can for instance be found in the PhD thesis of Niushan Gao, in particular in Lemma 0.31 on page 20.

Finally, it is certainly worthwhile to point out a few negative results:

**Proposition 7.** Assume that $E$ is not reflexive, but has order continuous norm. Then the bi-dual operator $T''$ on $E''$ is *never* irreducible (no matter whether $T$ and $T'$ are irreducible).

**Proof.** As $E$ has order continuous norm, it is an ideal in $E''$. However, we have $E \not= \{0\}$ and $E \not= E''$ since $E$ is not reflexive. Hence, $E$ is a non-trivial closed $T$-invariant ideal in $E''$.

**Proposition 8.** Assume that $E_+$ does not contain a quasi-interior point. Then $T$ cannot be irreducible.

**Proof.** Let $\lambda$ be a real number which is larger than the spectral radius of (the complex extension of) $T$. Choose an arbitray vector $f \in E_+ \setminus \{0\}$ and set $g := (\lambda - T)^{-1}f = \sum_{n=0}^\infty \frac{T^n}{\lambda^{n+1}}$. Then it is easy to see that the closed ideal generated by $g$ is $T$-invariant. This ideal is not equal to $E$ since $g$ is, by assumption, not a quasi-interior point of $E_+$; moreover, the ideal is non-zero since it contains $g$. Hence, $T$ is not irreducible.

**Corollary 9.** Let $K$ be an uncountable compact Hausdorff space and let $E = C(K)$ by the space of all scalar-valued continuous functions on $K$ (endowed with the supremum norm). Then $T'$ is not irreducible (not matter whether $T$ is irreducible or not).

**Proof.** According to [Schaefer, op. cit., Example 4 on page 99], the dual space $E'$ does not contain quasi-interior points. Hence, the assertion follows from Proposition 8.

**Remark 10.** If consider Corollary 9 and Remark 6, it is worthwhile to point out that the dual operator $T'$ in Corollary 9 is not even band irreducible. Indeed, the dual space $E' = C(K)'$ is an AL-space, so it has order continuous norm, and therefore the notions *irreducible* and *band irreducible* coincide on $E'$.

irreducible". What I wrote instead clearly doesn't make any sense at all. $\endgroup$ – Jochen Glueck Feb 28 '18 at 10:254more comments