Doubt about linear invariant term on paper "On theory of phase transitions", Landau 1937 In the  paper of Landau , "On theory of phase transitions " , page 3 http://www.ujp.bitp.kiev.ua/files/journals/53/si/53SI08p.pdf

It's is not clear to me what is meant by  "linear invariant"  in this context and why theses terms(linear invariants) cannot  be constructed in terms of "quantities transforming as an irreducible representation".
 A: Main ideas and definitions
In simple terms, a representation $\mathcal{R}$ is a set of matrices $r \in \mathcal{R}$ that act on a vector space $V$. An invariant subspace $W \subseteq V$ is a space of vectors which transform only among themselves.
A representation is irreducible if and only if the only invariant subspaces $W \subseteq V$ are $V$ itself, and the zero point $\{0\}$. Said in  a different way, there can be no similarity transformation which makes $\mathcal{R}$ consist of block diagonal matrices -- in that case, each block would define an invariant subspace.
Let a representation $\mathcal{R}$ be given, acting on a vector space $V$. Now, let a vector $\gamma \in V$ be given. Written in terms of a basis $\{\phi_i\}$ of $V$,
$$
\gamma = \sum_i c_i \phi_i.
$$
Evidently $\gamma$ is linear in the $\phi_i$. We will show that if $\gamma$ is also an invariant, then $\mathcal{R}$ is reducible, because it implies that there exists a nontrivial invariant subspace $W$. (It'll happen fast, so don't blink!)
So, suppose that $\gamma$ is an invariant vector. By the definition of invariance, for any matrix $r \in \mathcal{R}$,
$$
r \circ \gamma = \gamma.
$$
This means that $\gamma$ defines an invariant subspace $W$ -- namely the subspace spanned by scalar multiples of $\gamma$ itself, since muliplying $\gamma$ by any scalar gets you another invariant vector in $W$. (This is a subspace $W \subset V$.) Therefore $\mathcal{R}$ is reducible.
Going the other way, this implies that if $\mathcal{R}$ is irreducible, there can be no linear invariants.
Intuition
The intuition is that for an irreducible representation $\mathcal{R}$, all the vectors $v \in V$ are "maximally mixed" among themselves; there's no way to find any linearly independent subset of vectors of $V$ which doesn't mix with the others under one or more transformations belonging to $\mathcal{R}$.
When you find a linear invariant, you find a subspace which doesn't mix with the others under the transformations in $\mathcal{R}$, contradicting irreducibility.
