The Lie algebra \(\mathfrak{su}(3)\)

When looking at the three \(\mathfrak{su}(2)\) generator matrices
\[
\begin{equation}
t_x = \frac{i}{2}\,\begin{pmatrix}
0 & -1 \\
-1 & 0 \end{pmatrix}\\
t_y = \frac{i}{2}\,\begin{pmatrix}
0  & i \\
-i & 0  \end{pmatrix}\\
t_z = \frac{i}{2}\,\begin{pmatrix}
-1 & 0 \\0 & 1
\end{pmatrix}
\label{lasu3:eq:repre1}
\end{equation}
\] (see this post) one could be tempted to assume that the nine matrices
\[
\begin{equation}
t_x = \frac{i}{2}\,\begin{pmatrix}
0 & -1 & 0\\
-1 & 0 & 0\\
0 & 0 & 0 \end{pmatrix}\qquad
t_y = \frac{i}{2}\,\begin{pmatrix}
0  & i & 0 \\
-i & 0 & 0 \\
0 & 0 & 0  \end{pmatrix}\\
t_z = \frac{i}{2}\,\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \end{pmatrix}\qquad
u_x = \frac{i}{2}\,\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & -1\\
0 & -1 & 0 \end{pmatrix}\\
u_y = \frac{i}{2}\,\begin{pmatrix}
0 & 0 & 0 \\
0 & 0  & i \\
0 & -i & 0 \end{pmatrix}\qquad
u_z = \frac{i}{2}\,\begin{pmatrix}
0 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \end{pmatrix}\\
v_x = \frac{i}{2}\,\begin{pmatrix}
0 & 0 & -1\\
0 & 0 & 0\\
-1 & 0 & 0 \end{pmatrix}\qquad
v_y = \frac{i}{2}\,\begin{pmatrix}
0  & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0  \end{pmatrix}\\
v_z = \frac{i}{2}\,\begin{pmatrix}
-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \end{pmatrix}
\label{lasu3:eq:repre2}
\end{equation}
\] form a Lie algebra as well. Calculation of all 81 commutators (Lie products) yields :

	\(t_x\)	\(t_y\)	\(t_z\)	\(u_x\)	\(u_y\)	\(u_z\)	\(v_x\)	\(v_y\)	\(v_z\)
\(t_x\)	\(0\)	\(t_z\)	\(-t_y\)	\({v_y\over 2}\)	\(-{v_x\over 2}\)	\({t_y\over 2}\)	\({u_y\over 2}\)	\(-{u_x\over 2}\)	\(-{t_y\over 2}\)
\(t_y\)	\(-t_z\)	\(0\)	\(t_x\)	\(-{v_x\over 2}\)	\({v_y\over 2}\)	\(-{t_x\over 2}\)	\(-{u_x\over 2}\)	\({u_y\over 2}\)	\({t_x\over 2}\)
\(t_z\)	\(t_y\)	\(-t_x\)	\(0\)	\(-{u_y\over 2}\)	\({u_x\over 2}\)	\(0\)	\({v_y\over 2}\)	\(-{v_x\over 2}\)	\(0\)
\(u_x\)	\(-{v_y\over 2}\)	\({v_x\over 2}\)	\({u_y\over 2}\)	\(0\)	\(u_z\)	\(-u_y\)	\(-{t_y\over 2}\)	\({t_x\over 2}\)	\(-{u_y\over 2}\)
\(u_y\)	\({v_x\over 2}\)	\(-{v_y\over 2}\)	\(-{u_x\over 2}\)	\(-u_z\)	\(0\)	\(u_x\)	\(-{t_x\over 2}\)	\({t_y\over 2}\)	\({u_x\over 2}\)
\(u_z\)	\(-{t_y\over 2}\)	\({t_x\over 2}\)	\(0\)	\(u_y\)	\(-u_x\)	\(0\)	\({v_y\over 2}\)	\(-{v_x\over 2}\)	\(0\)
\(v_x\)	\(-{u_y\over 2}\)	\({u_x\over 2}\)	\(-{v_y\over 2}\)	\({t_y\over 2}\)	\({t_x\over 2}\)	\(-{v_y\over 2}\)	\(0\)	\(v_z\)	\(-v_y\)
\(v_y\)	\({u_x\over 2}\)	\(-{u_y\over 2}\)	\({v_x\over 2}\)	\(-{t_x\over 2}\)	\(-{t_y\over 2}\)	\({v_x\over 2}\)	\(-v_z\)	\(0\)	\(v_x\)
\(v_z\)	\({t_y\over 2}\)	\(-{t_x\over 2}\)	\(0\)	\({u_y\over 2}\)	\(-{u_x\over 2}\)	\(0\)	\(v_y\)	\(-v_x\)	\(0\)

(The row value is the first entry, the column value the seconds entry in the commutator \([\cdot, \cdot]\).) The yellow-colored values mark the Lie product of \(t_x\), \(t_y\), \(t_z\) with themselves (green for \(u_x\), \(u_y\), \(u_z\), blue for \(v_x\), \(v_y\), \(v_z\)) and suggest that this algebra includes three copies of \(su(2)\).
 
Alas, the set \(t_x\), \(t_y\), \(t_z\), \(u_x\), \(u_y\), \(u_z\), \(v_x\), \(v_y\) and \(v_z\) doesn't form a basis since only eight of the nine vectors are linear independent. E.g., we find \(v_z = t_z + u_z\) and we could choose the eight elements \(t_x\), \(t_y\), \(t_z\), \(u_x\), \(u_y\), \(u_z\), \(v_x\), \(v_y\). Of course, instead of \(v_z\) we could just as well disregard \(t_z\) or \(u_z\). Either way, the multiplication table loses its nice symmetry and, for the set \(t_x\), \(t_y\), \(t_z\), \(u_x\), \(u_y\), \(u_z\), \(v_x\), \(v_y\), looks like this

	\(t_x\)	\(t_y\)	\(t_z\)	\(u_x\)	\(u_y\)	\(u_z\)	\(v_x\)	\(v_y\)
\(t_x\)	\(0\)	\(t_z\)	\(-t_y\)	\({v_y\over 2}\)	\(-{v_x\over 2}\)	\({t_y\over 2}\)	\({u_y\over 2}\)	\(-{u_x\over 2}\)
\(t_y\)	\(-t_z\)	\(0\)	\(t_x\)	\(-{v_x\over 2}\)	\({v_y\over 2}\)	\(-{t_x\over 2}\)	\(-{u_x\over 2}\)	\({u_y\over 2}\)
\(t_z\)	\(t_y\)	\(-t_x\)	\(0\)	\(-{u_y\over 2}\)	\({u_x\over 2}\)	\(0\)	\({v_y\over 2}\)	\(-{v_x\over 2}\)
\(u_x\)	\(-{v_y\over 2}\)	\({v_x\over 2}\)	\({u_y\over 2}\)	\(0\)	\(u_z\)	\(-u_y\)	\(-{t_y\over 2}\)	\({t_x\over 2}\)
\(u_y\)	\({v_x\over 2}\)	\(-{v_y\over 2}\)	\(-{u_x\over 2}\)	\(-u_z\)	\(0\)	\(u_x\)	\(-{t_x\over 2}\)	\({t_y\over 2}\)
\(u_z\)	\(-{t_y\over 2}\)	\({t_x\over 2}\)	\(0\)	\(u_y\)	\(-u_x\)	\(0\)	\({v_y\over 2}\)	\(-{v_x\over 2}\)
\(v_x\)	\(-{u_y\over 2}\)	\({u_x\over 2}\)	\(-{v_y\over 2}\)	\({t_y\over 2}\)	\({t_x\over 2}\)	\(-{v_y\over 2}\)	\(0\)	\(t_z+u_z\)
\(v_y\)	\({u_x\over 2}\)	\(-{u_y\over 2}\)	\({v_x\over 2}\)	\(-{t_x\over 2}\)	\(-{t_y\over 2}\)	\({v_x\over 2}\)	\(-(t_z+u_z)\)	\(0\)

In order to show that \(t_x\), \(t_y\), \(t_z\), \(u_x\), \(u_y\), \(u_z\), \(v_x\), \(v_y\) with the Lie multiplication table shown above is indeed a Lie algebra, it needs to be checked that
\[
\begin{equation*}
 [a, b] = -[b, a] \\
 [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0
\end{equation*}
\] The first equation (anti-symmetry) is valid since the entries in the table form an anti-symmetric matrix. The second (Jacobi identity) needs to be checked for all \(8\cdot 8\cdot 8 = 512\) possibilities, which did my computer for me. The Lie algebra of the eight vectors \(t_x\), \(t_y\), \(t_z\), \(u_x\), \(u_y\), \(u_z\), \(v_x\), \(v_y\) with the multiplication table above form the Lie algebra \(\mathfrak{su}(3) \).