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 $\mathfrak{su}(3)$.