Building the meson octet and singlet

It is not arbitrary, but a result of the representation theory of $SU(3)$. Quark colors form a vector space $C^3$, and a quark antiquark-pair gives a tensor product $C^3 otimes (C^3)^* sim C^{3times 3}$, hence is represented by 3 by 3 matrices, on which $SU(3)$ acts by conjugation. This is a 9-dimensional vector space which, as a representation space of $SU(3)$, is reducible.

Indeed, the space of 3 by 3 matrices is a direct sum of the 1-dimensional space of multiples of the identity, on which $SU(3)$ acts trivially, and the 8-dimensional space of matrices of trace zero, on which $SU(3)$ acts (as conjugation preserves tracelessness). It is not very difficult to show that this representation is in fact irreducibly.

Thus the decomposition $3 otimes 3^* = 1 + 8$ into a singlet and an octet is not arbitrary, but determined by the properties of $SU(3)$.

It is a complete analogue of the decomposition $2 otimes 2^* = 1 + 3$ for $SU(2)$ (i.e., spin).

1. The Lie group $G$ behind the meson octet is the $SU(3)_{rm Flavor}$ Lie group over the three lightest quark flavors $u$, $d$, and $s$. More precisely, the fundamental $SU(3)$ representation $$V~=~text{Fund} ~=~{bf 3}$$ is a linear span of the $u$, $d$, and $s$ quarks. The three complex coefficients are collected in a $3times 1$ column vector $vin V$.

2. How does the $G=SU(3)_{rm Flavor}$ act on $V$? It acts from left by multiplying $vin V$ with a $3times3$ special unitary group matrix $gin G$, which results in a new column vector $v'=gvin V$. Similarly, $G$ acts on the complex conjugated representation $bar{V}$ as $bar{v}'=bar{g},bar{v}$. In particular, if we write $bar{v}$ as a $1times 3$ row vector $v^{dagger}$, the group acts as $v^{dagger prime}=v^{dagger}g^{dagger}=v^{dagger}g^{-1}$.

3. Lie group representation theory explains how the meson nonet decomposes in irreps, $$begin{align} {bf 3} otimes overline{bf 3}~=~&V otimes bar{V}cr ~=~&text{Fund} otimes overline{text{Fund}}cr ~cong~& text{Adj}oplus text{Sing}cr ~=~& {bf 8}oplus {bf 1}. end{align}$$

4. We may identify $$begin{align}V otimes bar{V}~cong~&V otimes V^daggercr ~cong~&{rm Mat}_{3 times 3}(mathbb{C}).end{align}tag{*}$$ Notice that $G$ acts on both spaces $V otimes V^dagger$ and ${rm Mat}_{3 times 3}(mathbb{C})$ by similarity transformations.

5. The $SU(3)$ singlet is the eta prime meson $$eta'~=~frac{uotimes bar{u}+ dotimes bar{d}+ sotimes bar{s}}{sqrt{3}}.$$ Under the identification (*), the $eta'$ corresponds to a $3 times 3$ matrix proportional to the ${bf1}_{3 times 3}$ unit matrix. How do we know this is the unique flavor singlet? On one hand, the singlet is characterized by being invariant under the group action, i.e. similarity transformations. On the other hand, we know from Schur Lemma, that the only matrices that are invariant under all similarity transformations are the ones proportional to the unit matrix. Equivalently, in terms of the corresponding Lie algebra $su(3)$, the only matrices that commute with all the Lie algebra generators are the ones proportional to the unit matrix. This latter condition may be viewed as the constraint that OP is requesting.

6. Finally let us mention, that the $SU(3)_{rm Flavor}$ symmetry is only an approximate symmetry in the standard model, as is evident from the different meson masses. The $SU(3)_{rm Flavor}$ symmetry can be decomposed via branching rules in (strong) $SU(2)$ isospin symmetry, see e.g. chapter 10 of 't Hooft's lecture notes. The pdf file is available here.