= Solution
Use the <A5 root system> with <simple roots> $\alpha_i=e_i-e_{i+1}$ for $1\leq i\leq5$, in the hyperplane $\sum_{i=1}^6x_i=0$. The <Dynkin diagram> is the five-node chain
$$
\underset{1}{\circ}\!-\!\underset{2}{\circ}\!-\!\underset{3}{\circ}\!-\!\underset{4}{\circ}\!-\!\underset{5}{\circ}.
$$
Its <Cartan matrix> is
$$
\boxed{A_5=\begin{pmatrix}
2&-1&0&0&0\\-1&2&-1&0&0\\0&-1&2&-1&0\\0&0&-1&2&-1\\0&0&0&-1&2
\end{pmatrix}.}
$$
Deleting the final node leaves the <A4 root system>, with <simple roots> $\alpha_1,\ldots,\alpha_4$. Concretely, the <Lie algebra homomorphism> $\mathfrak{sl}_5(\mathbb C)\hookrightarrow\mathfrak{sl}_6(\mathbb C)$ sends $X$ to $\operatorname{diag}(X,0)$ and preserves the <Lie bracket>. This realizes the required <Lie subalgebra> and fixes the embedding used below.
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