![enter image description here](https://cdn.statically.io/img/i.sstatic.net/CPL7W.png)
Note that in the 3-dimensional complex space spanned by basis $\boldsymbol{\lbrace}\boldsymbol{u}\overline{\boldsymbol{u}},\boldsymbol{d}\overline{\boldsymbol{d}},\boldsymbol{s}\overline{\boldsymbol{s}}\boldsymbol{\rbrace}$, this basis is replaced by $\boldsymbol{\lbrace}\boldsymbol{\pi^{0},\boldsymbol{\eta},\boldsymbol{\eta}^{\prime}}\boldsymbol{\rbrace}$ through a special unitary transformation $\mathrm{V}\in SU(3)$,
\begin{equation}
\begin{bmatrix}
\boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\sqrt{\tfrac{1}{2}} & \boldsymbol{-} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\
\sqrt{\tfrac{1}{6}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & \boldsymbol{-}\sqrt{\tfrac{2}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\
\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
=\mathrm{V}
\begin{bmatrix}
\boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
\tag{1}\label{1}
\end{equation}
see Figure.
$
\newcommand{\FR}[2]{{\textstyle \frac{#1}{#2}}}
\newcommand{\BK}[3]{\left|{#1},{#2}\right\rangle_{#3}}
\newcommand{\BoldExp}[2]{{#1}^{\boldsymbol{#2}}}
\newcommand{\BoldSub}[2]{{#1}_{\boldsymbol{#2}}}
\newcommand{\MM}[4]
{\begin{bmatrix}
#1 & #2\\
#3 & #4\\
\end{bmatrix}}
\newcommand{\MMM}[9]
{\textstyle \begin{bmatrix}
#1 & #2 & #3 \\
#4 & #5 & #6 \\
#7 & #8 & #9 \\
\end{bmatrix}}
\newcommand{\CMRR}[2]
{\begin{bmatrix}
#1 \\
#2
\end{bmatrix}}
\newcommand{\CMRRR}[3]
{\begin{bmatrix}
#2 \\
#3
\end{bmatrix}}
\newcommand{\CMRRRR}[4]
{\begin{bmatrix}
#1 \\
#2 \\
#3 \\
#4
\end{bmatrix}}
\newcommand{\RMCC}[2]
{\begin{bmatrix}
#1 & #2
\end{bmatrix}}
\newcommand{\RMCCC}[3]
{\begin{bmatrix}
#1 & #2 & #3
\end{bmatrix}}
\newcommand{\RMCCCC}[4]
{\begin{bmatrix}
#1 & #2 & #3 & #4
\end{bmatrix}}
$
$\boldsymbol{\S\:}\textbf{A. Mesons from three quarks}$ $\boldsymbol{u},\boldsymbol{d},\boldsymbol{s} : \boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{8}$
Suppose we know the existence of three quarks only : $\boldsymbol{u}$, $\boldsymbol{d}$ and $\boldsymbol{s}$. Under full symmetry these are the basic states, let
\begin{equation}
\boldsymbol{u}=
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}
\qquad
\boldsymbol{d}=
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\qquad
\boldsymbol{s}=
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\tag{001}\label{001}
\end{equation}
of a 3-dimensional complex Hilbert space of quarks, say $\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$. A quark $\boldsymbol{\xi} \in \mathbf{Q}$ is expressed in terms of these basic states as
\begin{equation}
\boldsymbol{\xi}=\xi_u\boldsymbol{u}+\xi_d\boldsymbol{d}+\xi_s\boldsymbol{s}=
\begin{bmatrix}
\xi_u\\
\xi_d\\
\xi_s
\end{bmatrix}
\qquad \xi_u,\xi_d,\xi_s \in \mathbb{C}
\tag{002}\label{002}
\end{equation}
For a quark $\boldsymbol{\zeta} \in \mathbf{Q}$
\begin{equation}
\boldsymbol{\zeta}=\zeta_u\boldsymbol{u}+\zeta_d\boldsymbol{d}+\zeta_s\boldsymbol{s}=
\begin{bmatrix}
\zeta_u\\
\zeta_d\\
\zeta_s
\end{bmatrix}
\tag{003}\label{003}
\end{equation}
the respective antiquark $\overline{\boldsymbol{\zeta}}$ is expressed by the complex conjugates of the coordinates
\begin{equation}
\overline{\boldsymbol{\zeta}}=\overline{\zeta}_u \overline{\boldsymbol{u}}+\overline{\zeta}_d\overline{\boldsymbol{d}}+\overline{\zeta}_s\overline{\boldsymbol{s}}=
\begin{bmatrix}
\overline{\zeta}_u\\
\overline{\zeta}_d\\
\overline{\zeta}_s
\end{bmatrix}
\tag{004}\label{004}
\end{equation}
with respect to the basic states
\begin{equation}
\overline{\boldsymbol{u}}=
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}
\qquad
\overline{\boldsymbol{d}}=
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\qquad
\overline{\boldsymbol{s}}=
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\tag{005}\label{005}
\end{equation}
the antiquarks of $\boldsymbol{u},\boldsymbol{d}$ and $\boldsymbol{s}$ respectively. The antiquarks belong to a different space, the space of antiquarks $\overline{\mathbf{Q}}\equiv \mathbb{C}^{\boldsymbol{3}}$.
Since a meson is a quark-antiquark pair, we'll try to find the product space
\begin{equation}
\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\: \left(\equiv \mathbb{C}^{\boldsymbol{9}}\right)
\tag{006}\label{006}
\end{equation}
Using the expressions \eqref{002} and \eqref{004} of the quark $\boldsymbol{\xi} \in \mathbf{Q}$ and the antiquark $\overline{\boldsymbol{\zeta}} \in \overline{\mathbf{Q}}$ respectively, we have for the product meson state $ \mathrm{X} \in \mathbf{M}$
\begin{equation}
\begin{split}
\mathrm{X}=\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_u\overline{\zeta}_d \left( \boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_u\overline{\zeta}_s \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+ \\
&\xi_d\overline{\zeta}_u \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_d\overline{\zeta}_d \left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_d\overline{\zeta}_s \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+\\
&\xi_s\overline{\zeta}_u \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_s\overline{\zeta}_d \left( \boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_s\overline{\zeta}_s \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)
\end{split}
\tag{007}\label{007}
\end{equation}
In order to simplify the expressions, the product symbol $"\boldsymbol{\otimes}"$ is omitted and so
\begin{equation}
\begin{split}
\mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \boldsymbol{u}\overline{\boldsymbol{u}}+\xi_u\overline{\zeta}_d \boldsymbol{u}\overline{\boldsymbol{d}}+\xi_u\overline{\zeta}_s \boldsymbol{u}\overline{\boldsymbol{s}}+ \\
&\xi_d\overline{\zeta}_u \boldsymbol{d}\overline{\boldsymbol{u}}+\xi_d\overline{\zeta}_d \boldsymbol{d}\overline{\boldsymbol{d}}+\xi_d\overline{\zeta}_s \boldsymbol{d}\overline{\boldsymbol{s}}+\\
&\xi_s\overline{\zeta}_u \boldsymbol{s}\overline{\boldsymbol{u}}+\xi_s\overline{\zeta}_d \boldsymbol{s}\overline{\boldsymbol{d}}+\xi_s\overline{\zeta}_s \boldsymbol{s}\overline{\boldsymbol{s}}
\end{split}
\tag{008}\label{008}
\end{equation}
Due to the fact that $\mathbf{Q}$ and $\overline{\mathbf{Q}}$ are of the same dimension, it's convenient to represent the meson states in the product 9-dimensional complex space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ by square $3 \times 3$ matrices instead of row or column vectors
\begin{equation}
\mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=
\begin{bmatrix}
\xi_u\overline{\zeta}_u & \xi_u\overline{\zeta}_d & \xi_u\overline{\zeta}_s\\
\xi_d\overline{\zeta}_u & \xi_d\overline{\zeta}_d & \xi_d\overline{\zeta}_s\\
\xi_s\overline{\zeta}_u & \xi_s\overline{\zeta}_d & \xi_s\overline{\zeta}_s
\end{bmatrix}=
\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}
\begin{bmatrix}
\overline{\zeta}_u \\
\overline{\zeta}_d \\
\overline{\zeta}_s
\end{bmatrix}^{\mathsf{T}}
=
\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}
\begin{bmatrix}
\overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s
\end{bmatrix}
\tag{009}\label{009}
\end{equation}
The product space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ is created by completion of the set of states \eqref{008} with arbitrary complex coefficients
\begin{equation}
\begin{split}
\mathrm{X}=&\mathrm{x}_{_{11}}\boldsymbol{u}\overline{\boldsymbol{u}}+\mathrm{x}_{_{12}} \boldsymbol{u}\overline{\boldsymbol{d}}+\mathrm{x}_{_{13}} \boldsymbol{u}\overline{\boldsymbol{s}}+ \\
&\mathrm{x}_{_{21}}\boldsymbol{d}\overline{\boldsymbol{u}}+\mathrm{x}_{_{22}} \boldsymbol{d}\overline{\boldsymbol{d}}+\mathrm{x}_{_{23}} \boldsymbol{d}\overline{\boldsymbol{s}}+ \qquad \mathrm{x}_{_{ij}} \in \mathbb{C}\\
&\mathrm{x}_{_{31}} \boldsymbol{s}\overline{\boldsymbol{u}}+\mathrm{x}_{_{32}} \boldsymbol{s}\overline{\boldsymbol{d}}+\mathrm{x}_{_{33}} \boldsymbol{s}\overline{\boldsymbol{s}}
\end{split}
\tag{010}\label{010}
\end{equation}
that is
\begin{equation}
\mathrm{X}=
\begin{bmatrix}
\mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\
\mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\
\mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}}
\end{bmatrix} \:, \qquad \mathrm{x}_{_{ij}} \in \mathbb{C}
\tag{011}\label{011}
\end{equation}
So $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ is identical to $\mathbb{C}^{\boldsymbol{9}}$ with base states
\begin{align}
&\boldsymbol{u}\overline{\boldsymbol{u}}=
\begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\boldsymbol{u}\overline{\boldsymbol{d}}=
\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\boldsymbol{u}\overline{\boldsymbol{s}}=
\begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\tag{012a}\label{012a}\\
&\boldsymbol{d}\overline{\boldsymbol{u}}=
\begin{bmatrix}
0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\boldsymbol{d}\overline{\boldsymbol{d}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\:\boldsymbol{d}\overline{\boldsymbol{s}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix}
\tag{012b}\label{012b}\\
&\boldsymbol{s}\overline{\boldsymbol{u}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{bmatrix}
\quad
\:\boldsymbol{s}\overline{\boldsymbol{d}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{bmatrix}
\quad
\:\boldsymbol{s}\overline{\boldsymbol{s}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1
\end{bmatrix}
\tag{012c}\label{012c}
\end{align}
This basis is represented symbolically by a $3\times 3$ array
\begin{equation}
\mathcal{F}_{\mathbf{M}}=
\begin{bmatrix}
\boldsymbol{u}\overline{\boldsymbol{u}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\
\boldsymbol{d}\overline{\boldsymbol{u}} & \boldsymbol{d}\overline{\boldsymbol{d}} & \boldsymbol{d}\overline{\boldsymbol{s}}\\
\boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & \boldsymbol{s}\overline{\boldsymbol{s}}
\end{bmatrix}
\tag{013}\label{013}
\end{equation}
In this Hilbert space the usual inner product between states
\begin{equation}
\mathrm{X}=
\begin{bmatrix}
\mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\
\mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\
\mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}}
\end{bmatrix} \:, \qquad
\mathrm{Y}=
\begin{bmatrix}
\mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\
\mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\
\mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}}
\end{bmatrix}
\tag{014}\label{014}
\end{equation}
is
\begin{equation}
\begin{split}
\langle \mathrm{X},\mathrm{Y}\rangle \equiv &\mathrm{x}_{_{11}}\overline{\mathrm{y}}_{_{11}}+\mathrm{x}_{_{12}}\overline{\mathrm{y}}_{_{12}}+\mathrm{x}_{_{13}}\overline{\mathrm{y}}_{_{13}}+\\
&\mathrm{x}_{_{21}}\overline{\mathrm{y}}_{_{21}}+\mathrm{x}_{_{22}}\overline{\mathrm{y}}_{_{22}}+\mathrm{x}_{_{23}}\overline{\mathrm{y}}_{_{23}}+\\
&\mathrm{x}_{_{31}}\overline{\mathrm{y}}_{_{31}}+\mathrm{x}_{_{32}}\overline{\mathrm{y}}_{_{32}}+\mathrm{x}_{_{33}}\overline{\mathrm{y}}_{_{33}}
\end{split}
\tag{015}\label{015}
\end{equation}
which, using the $3\times 3$ matrix representation of states, is the trace of the matrix product $\mathrm{X}\BoldExp{\mathrm{Y}}{*}$
\begin{equation}
\langle \mathrm{X},\mathrm{Y}\rangle =\mathrm{Tr}\left[\mathrm{X}\BoldExp{\mathrm{Y}}{*}\right]
\tag{016}\label{016}
\end{equation}
given that $\BoldExp{\mathrm{Y}}{*}$ is the complex conjugate of the transpose of $\mathrm{Y}$
\begin{equation}
\BoldExp{\mathrm{Y}}{*}\equiv
\BoldExp{
\begin{bmatrix}
\mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\
\mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\
\mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}}
\end{bmatrix}}
{*}
=
\overline{\begin{bmatrix}
\mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\
\mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\
\mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}}
\end{bmatrix}^{\mathsf{T}}}
=
\begin{bmatrix}
\overline{\mathrm{y}}_{_{11}} & \overline{\mathrm{y}}_{_{21}} & \overline{\mathrm{y}}_{_{31}}\\
\overline{\mathrm{y}}_{_{12}} & \overline{\mathrm{y}}_{_{22}} & \overline{\mathrm{y}}_{_{32}}\\
\overline{\mathrm{y}}_{_{13}} & \overline{\mathrm{y}}_{_{23}} & \overline{\mathrm{y}}_{_{33}}
\end{bmatrix}
\tag{017}\label{017}
\end{equation}
Now, under a unitary transformation $\;W \in SU(3)\;$ in the 3-dimensional space of quarks $\;\mathbf{Q}\;$, we have
\begin{equation}
\BoldExp{\boldsymbol{\xi}}{'} = W\boldsymbol{\xi}
\tag{018}\label{018}
\end{equation}
so in the space of antiquarks $\overline{\mathbf{Q}}\;$, since $\;\BoldExp{\boldsymbol{\zeta}}{'}=W \boldsymbol{\zeta}\;$
\begin{equation}
\overline{\BoldExp{\boldsymbol{\zeta}}{'}}= \overline{W}\;\overline{\boldsymbol{\zeta}}
\tag{019}\label{019}
\end{equation}
and for the meson state
\begin{align}
\BoldExp{\mathrm{X}}{'} & =\BoldExp{\boldsymbol{\xi}}{'}\boldsymbol{\otimes}\overline{\BoldExp{\boldsymbol{\zeta}}{'}}=\left(W\boldsymbol{\xi}\vphantom{\overline{W}\overline{\boldsymbol{\zeta}} }\right)\left(\overline{W}\overline{\boldsymbol{\zeta}} \right)
=
\Biggl(W\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}\Biggr)
\Biggl(\overline{W}\begin{bmatrix}
\overline{\zeta}_u\\
\overline{\zeta}_d\\
\overline{\zeta}_s
\end{bmatrix}\Biggr)^{\mathsf{T}}
\nonumber\\
& = W\Biggl(\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}
\begin{bmatrix}
\overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s
\end{bmatrix}\Biggr)\overline{W}^{\mathsf{T}}
=W\left(\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}\right)\BoldExp{W}{*}=W\;\mathrm{X}\;\BoldExp{W}{*}
\nonumber
\tag{020}\label{020}
\end{align}
that is
\begin{equation}
\BoldExp{\mathrm{X}}{'} = W\;\mathrm{X}\;\BoldExp{W}{*}
\tag{021}\label{021}
\end{equation}
Above equation \eqref{021} is the transformation law of meson states in the 9-dimensional space $\;\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\;$ induced by a unitary
transformation $\;W \in SU(3)\;$ in the 3-dimensional space of quarks $\mathbf{Q}$.
Under this transformation law the inner product of two meson states is invariant because its relation with the trace, equation \eqref{016}, yields
\begin{equation}
\langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\mathrm{Tr}\left[\BoldExp{\mathrm{X}}{'}\BoldExp{\BoldExp{\mathrm{Y}}{'}}{*}\right]=\mathrm{Tr}\Bigl[\left(W\mathrm{X}\BoldExp{W}{*}\right) \BoldExp{\left(W\mathrm{Y}\BoldExp{W}{*}\right)}{*}\Bigr]=\mathrm{Tr}\Bigl[W \left( \mathrm{X}\BoldExp{Y}{*}\right)\BoldExp{W}{*}\Bigr]=\mathrm{Tr}\Bigl[\mathrm{X}\BoldExp{Y}{*}\Bigr]
\tag{022}\label{022}
\end{equation}
The last equality in above equation \eqref{022} is valid since under the transformation law \eqref{021} the trace remains invariant. More generally, for unitary $\;W \in SU(n)\;$ and $\;A\;$ a $\;n \times n\;$ complex matrix the transformation
\begin{equation}
\BoldExp{\mathrm{A}}{'} = W\;\mathrm{A}\;\BoldExp{W}{*}
\tag{023}\label{023}
\end{equation}
if expressed in terms of elements, yields (we use the Einstein summation convention)
\begin{equation}
\BoldExp{a_{ij}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma j}}{*}
\tag{024}\label{0242}
\end{equation}
so
\begin{equation}
\mathrm{Tr}\Bigl[\BoldExp{\mathrm{A}}{'}\Bigr]=\BoldExp{a_{ii}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma i}}{*}=(\BoldExp{w_{\sigma i}}{*}w_{i\rho})a_{\rho\sigma}=\delta_{\sigma\rho}a_{\rho\sigma}=a_{\rho\rho}=\mathrm{Tr}\Bigl[A\Bigr]
\tag{025}\label{025}
\end{equation}
proving the invariance of inner product under the transformation law \eqref{021}
\begin{equation}
\langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\langle \mathrm{X},\mathrm{Y}\rangle
\tag{026}\label{026}
\end{equation}
Now, obviously the meson state represented by the identity matrix
\begin{equation}
\mathrm{I}=
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
\tag{027}\label{027}
\end{equation}
is unchanged under the transformation \eqref{021} and if normalized yields
\begin{equation}
\BoldSub{\mathrm{F}}{0}=\sqrt{\tfrac{1}{3}}
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
=\sqrt{\tfrac{1}{3}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}+\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \BoldExp{\boldsymbol{\eta}}{\prime}
\tag{028}\label{028}
\end{equation}
that is, it represents the $\;\BoldExp{\boldsymbol{\eta}}{\prime}\;$ meson.
The 1-dimensional subspace $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$ spanned by this state is invariant. Note that $\;\BoldExp{\boldsymbol{\eta}}{\prime}=\sqrt{3}\cdot \mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]$.
Any meson state orthogonal to this space, $\mathrm{X}\perp\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace} $, remains orthogonal under the transformation. But
\begin{equation}
\mathrm{X}\perp \boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\Leftrightarrow\langle \mathrm{X},\BoldSub{\mathrm{F}}{0}\rangle =0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\BoldSub{\mathrm{F}}{0}^{\boldsymbol{*}}\right]=0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\right]=0
\tag{029}\label{029}
\end{equation}
So, the 8-dimensional linear subspace of all meson states with traceless matrix representation is the orthogonal complement of the 1-dimensional subspace $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$ and if $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}\;$ is any basis which spans this space then
\begin{equation}
\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}=\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}^{\boldsymbol{\perp}}=\Bigl\{ \mathrm{X} \in \mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\; :\; \mathrm{Tr}\left[X\right]=0 \; \Bigr\}
\tag{030}\label{030}
\end{equation}
This space is invariant under the transformation \eqref{021}.
There are arbitrary many choices of the basis $\;\left(\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\right)\;$ but a proper one must correspond to mesons in the real world and be orthonormal if possible.
So, the normalized traceless meson state
\begin{equation}
\BoldSub{\mathrm{F}}{3}=\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\
0 & \boldsymbol{-}1 & \hphantom{\boldsymbol{-}}0\\
0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0
\end{bmatrix}
=\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\equiv \BoldExp{\boldsymbol{\pi}}{0}
\tag{031}\label{031}
\end{equation}
represents of course the $\;\BoldExp{\boldsymbol{\pi}}{0}\;$ meson (pion).
The basis $\mathcal{F}_{\mathbf{M}}$ may be expressed symbolically as sum of a diagonal and a traceless component
\begin{equation}
\begin{split}
&\mathcal{F}_{\mathbf{M}}=\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}+\Bigl[\mathcal{F}_{\mathbf{M}}-\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}\Bigr]\\
&=\begin{bmatrix}
\dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}
\end{bmatrix}
+
\begin{bmatrix}
\dfrac{\left(2\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3}{\rule[0ex]{-10pt}{0ex}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\
\boldsymbol{d}\overline{\boldsymbol{u}} & \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}+2\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3} & \boldsymbol{d}\overline{\boldsymbol{s}} \\
\boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & {\rule[-2ex]{-10pt}{6ex}} \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}+2\boldsymbol{s}\overline{\boldsymbol{s}}\right)}{3}
\end{bmatrix}
\end{split}
\tag{032}\label{032}
\end{equation}
The 3rd diagonal element of the traceless component of $\mathcal{F}_{\mathbf{M}}$, if opposed and normalized, yields
\begin{equation}
\BoldSub{\mathrm{F}}{8}=\sqrt{\tfrac{1}{6}}
\begin{bmatrix}
1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\
0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\\
0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}2
\end{bmatrix}
=\sqrt{\tfrac{1}{6}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}-2\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \boldsymbol{\eta}
\tag{033}\label{033}
\end{equation}
that is, it represents the $\;\boldsymbol{\eta}\;$ meson.
(to be continued in $\boldsymbol{\S\:}\textbf{B}$)