= Solution
The projection deletes the third <Dynkin label>, so the four <weights> $v_i$ of <4/b/solution> restrict to $(1,0),(-1,1),(0,-1),(0,0)$. The first three are the defining <SU(3)> <weight system>, and the fourth is a singlet. Negating them gives the <dual representation>. Hence
$$
\boxed{\mathbf4\downarrow_{SU(3)}=\mathbf3\oplus\mathbf1,\qquad
\overline{\mathbf4}\downarrow_{SU(3)}=\overline{\mathbf3}\oplus\mathbf1.}
$$
For the <Adjoint representation>, the projected nonzero <weights> consist of the six <A2 root system> <roots>, the three defining <weights> and their negatives. There are also three zero <weights>. Two zero <weights> belong to the <SU(3)> <Adjoint representation>, leaving one singlet. Therefore the <SU4 adjoint branching to SU3> is
$$
\boxed{\mathbf{15}\downarrow_{SU(3)}=\mathbf8\oplus\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1.}
$$
One can see all four <invariant subspaces> directly by writing a <traceless matrix> as
$$
M=\begin{pmatrix}A&b\\c&d\end{pmatrix},\qquad \operatorname{tr}A+d=0.
$$
The traceless part of $A$ is $\mathbf8$, the column $b$ is $\mathbf3$, the row $c$ is $\overline{\mathbf3}$, and the remaining diagonal <matrix> is proportional to $\operatorname{diag}(1,1,1,-3)$ and is fixed by <SU(3)>. These have <dimensions> $8+3+3+1=15$. Equivalently restrict $\mathbf4\otimes\overline{\mathbf4}-\mathbf1$ and use $\mathbf3\otimes\overline{\mathbf3}=\mathbf8\oplus\mathbf1$.
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