Suppose R from SU(3) to G L U and S from SU(3) to G L V are irreducible representations of SU(3) with highest weight \lambda (with respect to some fixed choice of line of irrational slope). Then U is isomorphic to V.
This video is optional.
We have now proved that for any SU(3) representation, if we pick a highest weight then we get a highest weight subrepresentation. We now need to prove that there is only one irrep with a given highest weight.
Consider the representation R direct sum S on U direct sum V. This is not irreducible as both U and V are subrepresentations. If u in U and v in V are weight vectors with weight lambda then (u,v) in U direct sum V is a weight vector with weight lambda: H_theta (u,v) equals (H_theta u, H_theta v), which equals lambda of theta times (u, v)
Take the highest weight irreducible subrepresentation W inside U direct sum V of R direct sum S generated by (u, v) (coming from the theorem we proved in the previous video).
The projections maps pr_U from W to U and pr_V from W to V are morphisms of representations (exercise!).
There's a nice fact (Schur's lemma) which says that a morphism of irreducible representations like pr_U from W to U (respectively pr_V from W to V) is either an isomorphism or it's zero. To see this, note that the kernel of a morphism is a subrepresentation of W, and is therefore either 0 or W (because W is irreducible); the morphism is injective if the kernel is zero, and zero if the kernel is everything. Similarly, the image is a subrepresentation of U (respectively V), so is either 0 or U (respectively 0 or V). The morphism is surjective if the image is everything, and zero if the image is zero. The only possibilities are for the morphism to be an isomorphism or to be zero.
But pr_U of (u, v) equals u and pr_V of (u, v) equals v, so neither map is zero. Therefore U is isomorphic to W which is isomorphic to V.