= Solution
Embed the two off-diagonal <Pauli matrices> in each of the $(u,d)$, $(u,s)$ and $(d,s)$ two-dimensional subspaces. Together with two independent embedded diagonal <Pauli matrices>, this produces the <Gell-Mann matrices>:
$$
\begin{aligned}
\lambda_1&=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},&
\lambda_2&=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix},&
\lambda_3&=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix},\\
\lambda_4&=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix},&
\lambda_5&=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix},&
\lambda_6&=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix},\\
\lambda_7&=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix},&
\lambda_8&=\frac1{\sqrt3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}.&&
\end{aligned}
$$
To construct the last matrix explicitly, the $(u,s)$ diagonal embedding is $D_{us}=\operatorname{diag}(1,0,-1)$ and the $(d,s)$ embedding is $D_{ds}=\operatorname{diag}(0,1,-1)$. Their difference is $\lambda_3$, and their sum divided by $\sqrt3$ is $\lambda_8$. There are only two independent diagonal generators.
Each matrix is <Hermitian> and traceless. Their inner products satisfy $\operatorname{tr}(\lambda_a\lambda_b)=2\delta_{ab}$, proving <linear independence>. With $T_a=\lambda_a/2$, arbitrary transformations near the identity have $U=\exp(i\theta_aT_a)$. The eight matrices span the generators; they are not a vector-space basis of group elements themselves.
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