If $R:SU(3)\to GL(V)$ is a complex representation then $V=\oplus {W}_{k,\mathrm{\ell}}$ where $${W}_{k,\mathrm{\ell}}=\{v\in V:R(D({\theta}_{1},{\theta}_{2}))v={e}^{i(k{\theta}_{1}+\mathrm{\ell}{\theta}_{2})}v\forall {\theta}_{1},{\theta}_{2}\},$$ where $D({\theta}_{1},{\theta}_{2})=\left(\begin{array}{ccc}\hfill {e}^{i{\theta}_{1}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{i{\theta}_{2}}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {e}^{i({\theta}_{1}+{\theta}_{2})}\hfill \end{array}\right)$ .
SU(3) overview
SU(3)
We now turn to the group $SU(3)$ , the group of 3by3 special unitary matrices. This is an instructive example: its representation theory shares many similarities to the representation theory of other semisimple groups.
When we studied $SU(2)$ , the key tool we introduced was the weight decomposition. In other words, given a representation of $SU(2)$ on a complex vector space $V$ , we decomposed $V$ as a direct sum of subspaces ${W}_{k}$ where ${W}_{k}$ was a simultaneous eigenspace (with eigenvalue ${e}^{ik\theta}$ ) for the action of the diagonal subgroup $T=\{\left(\begin{array}{cc}\hfill {e}^{i\theta}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{i\theta}\hfill \end{array}\right):{e}^{i\theta}\in U(1)\}$ . This was possible because $T\cong U(1)$ .
For $SU(3)$ , the subgroup of diagonal matrices is isomorphic to $U(1)\times U(1)$ : $$T=\{\left(\begin{array}{ccc}\hfill {e}^{i{\theta}_{1}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{i{\theta}_{2}}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {e}^{i({\theta}_{1}+{\theta}_{2})}\hfill \end{array}\right):({e}^{i{\theta}_{1}},{e}^{i{\theta}_{2}})\in U(1)\times U(1)\}.$$ $T$ here stands for torus: a 2dimensional torus has two circular coordinates (latitude and longitude) just like $U(1)\times U(1)$ . For compact matrix groups more generally, the analogue of $T$ will be a (possibly higherdimensional) maximal torus (more on that later).
For $SU(2)$ , our torus was just a copy of $U(1)$ , so we only needed one integer $k$ to index the weight spaces. Now we need two: one for each $U(1)$ factor in the torus.
We will postpone the proof until we have seen some examples of weight diagrams. The weight diagrams will now be patterns of dots in the plane.
Examples
The standard representation
Take $V={\mathbf{C}}^{3}$ , the standard representation of $SU(3)$ (i.e. considering a matrix in $SU(3)$ tautologically as an element of $GL(3,\mathbf{C})$ ). We pick the standard basis ${e}_{1},{e}_{2},{e}_{3}\in V$ . We have $$R(D({\theta}_{1},{\theta}_{1}))=\left(\begin{array}{ccc}\hfill {e}^{i{\theta}_{1}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{i{\theta}_{2}}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {e}^{i({\theta}_{1}+{\theta}_{2})}\hfill \end{array}\right),$$
so $$R(D({\theta}_{1},{\theta}_{2})){e}_{1}={e}^{i{\theta}_{1}}{e}_{1},R(D({\theta}_{1},{\theta}_{2})){e}_{2}={e}^{i{\theta}_{2}}{e}_{2},R(D({\theta}_{1},{\theta}_{2})){e}_{3}={e}^{i({\theta}_{1}+{\theta}_{2})}{e}_{3}$$ so

${e}_{1}$ spans the weight space ${W}_{1,0}$ ($k=1$ , $\mathrm{\ell}=0$ ).

${e}_{2}$ spans the weight space ${W}_{0,1}$ ($k=0$ , $\mathrm{\ell}=1$ ).

${e}_{3}$ spans the weight space ${W}_{1,1}$ ($k=1$ , $\mathrm{\ell}=1$ ).
The obvious way to represent these would be to simply draw the points $(k,\mathrm{\ell})=(1,0),(0,1)(1,1)$ in the plane. For reasons that will become clear when we discuss the Killing form, we will draw the $\mathrm{\ell}$ axis at 120 degrees to the $k$ axis. For now, the justification is just that this restores the symmetry we broke when we chose to express the bottomleft entry of $D({\theta}_{1},{\theta}_{2})$ in terms of the other two diagonal entries (instead of writing something equally valid like $\left(\begin{array}{ccc}\hfill {e}^{i{\theta}_{1}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{i({\theta}_{1}+{\theta}_{3})}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {e}^{i{\theta}_{3}}\hfill \end{array}\right)$ ).
We have also drawn the origin in grey in this picture: it is not a weight of the standard representation, the weights are red.
Sym^{2}(C^{3})
Consider ${\mathrm{Sym}}^{2}({\mathbf{C}}^{3})$ . A basis for this is given by $${e}_{1}^{2},{e}_{1}{e}_{2},{e}_{1}{e}_{3},{e}_{2}^{2},{e}_{2}{e}_{3},{e}_{3}^{2}.$$ How do these transform under ${\mathrm{Sym}}^{2}(D({\theta}_{1},{\theta}_{2}))$ ?
Since ${e}_{1}\mapsto {e}^{i{\theta}_{1}}{e}_{1}$ , we have $${e}_{1}^{2}\mapsto {e}^{i2{\theta}_{1}}{e}_{1}^{2},$$ so lives in the weight space ${W}_{2,0}$ . Similarly, we get $${e}_{1}^{2}\in {W}_{1,0},{e}_{1}{e}_{2}\in {W}_{1,1},{e}_{1}{e}_{3}\in {W}_{0,1},{e}_{2}^{2}\in {W}_{0,2},{e}_{2}{e}_{3}\in {W}_{1,0},{e}_{3}^{2}\in {W}_{2,2}.$$ For example, ${e}_{2}{e}_{3}\mapsto {e}^{i{\theta}_{2}}{e}^{i({\theta}_{1}+{\theta}_{2})}{e}_{2}{e}_{3}={e}^{i{\theta}_{1}}{e}_{2}{e}_{3}$ .
Plotting these weights in the plane (with respect to our tilted axes) we get:
We have also added:

the line $k=\mathrm{\ell}$ , to complete the symmetry of the diagram;

the points $(0,0)$ , $(1,0)$ , $(0,1)$ , $(1,1)$ in grey just to give a sense of the size of the diagram. Only the red dots are weights of the representation.
The kinds of diagrams we will get for weight diagrams of $SU(3)$ representations will be triangles and hexagons.
Proof of lemma
Consider the subgroup $${T}_{1}=\{D({\theta}_{1},0)=\left(\begin{array}{ccc}\hfill {e}^{i{\theta}_{1}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {e}^{{\theta}_{1}}\hfill \end{array}\right):{e}^{i{\theta}_{1}}\in U(1)\}\subset T$$ This is isomorphic to $U(1)$ and $V$ decomposes into weight spaces for the action of ${T}_{1}$ : $$V=\oplus {U}_{k},{U}_{k}=\{v\in V:R(D({\theta}_{1},0))v={e}^{ik{\theta}_{1}}v\}.$$
Define ${T}_{2}=\{D(0,{\theta}_{2}):{e}^{i{\theta}_{2}}\in U(1)\}$ .
The action of ${T}_{2}$ preserves each ${U}_{k}$ , i.e. $u\in {U}_{k}$ implies $R(D(0,{\theta}_{2}))u\in {U}_{k}$ .
This lemma will imply the result because now each ${U}_{k}$ decomposes under the action of ${T}_{2}$ : $${U}_{k}=\oplus {W}_{k,\mathrm{\ell}},{W}_{k,\mathrm{\ell}}=\{v\in {U}_{k}:R(D(0,{\theta}_{2}))v={e}^{i\mathrm{\ell}{\theta}_{2}}v\}.$$ In other words, elements of ${W}_{k,\mathrm{\ell}}$ transform as desired under both ${T}_{1}$ and ${T}_{2}$ and hence under $T$ : $$R(D({\theta}_{1},{\theta}_{2}))v=R(D({\theta}_{1},0))R(D(0,{\theta}_{2}))v={e}^{i(k{\theta}_{1}+\mathrm{\ell}{\theta}_{2})}v.$$
If $v\in {U}_{k}$ then $R(D({\theta}_{1},0))v={e}^{ik{\theta}_{1}}v$ . We want to show that $R(D(0,{\theta}_{2}))v\in {U}_{k}$ , i.e. that $$R(D({\theta}_{1},0))R(D(0,{\theta}_{2}))v={e}^{ik{\theta}_{1}}R(D(0,{\theta}_{2}))v.$$
Let's compute. Since $D({\theta}_{1},0)$ and $D(0,{\theta}_{2})$ commute with one another, the same is true of $R(D({\theta}_{1},0))$ and $R(D(0,{\theta}_{2}))$ . Therefore $$R(D({\theta}_{1},0))R(D(0,{\theta}_{2}))v=R(D(0,{\theta}_{2}))R(D({\theta}_{1},0))v.$$ Since $v\in {U}_{k}$ , we get $$R(D({\theta}_{1},0))R(D(0,{\theta}_{2}))v=R(D(0,{\theta}_{2})){e}^{ik{\theta}_{1}}v={e}^{ik{\theta}_{1}}R(D(0,{\theta}_{2}))v,$$ as required.
This argument works more generally for finding weight decompositions of representations of $U{(1)}^{n}$ : you inductively find the weight space decomposition with respect to $U{(1)}^{n1}$ and show that the final factor preserves the weight spaces (because the group is abelian), then decompose these weight spaces further with respect to the action of the final $U(1)$ factor. This would give us $n$ integer weights ${k}_{1},\mathrm{\dots},{k}_{n}$ and our weight diagrams would be collections of dots in $n$ dimensional space. For $SU(4)$ we get weight diagrams in ${\mathbf{R}}^{3}$ , for $SU(5)$ we get weight diagrams in ${\mathbf{R}}^{4}$ , etc.
Preclass exercises
If ${\mathbf{C}}^{3}$ denotes the standard representation of $SU(3)$ , what is the weight diagram of ${\mathrm{Sym}}^{3}({\mathbf{C}}^{3})$ ? Can you guess the weight diagram of ${\mathrm{Sym}}^{n}({\mathbf{C}}^{3})$ ?
What do you think the weight diagram of the standard 4dimensional representation of $SU(4)$ would look like?