Does the Lie-Algebra of $SU(2)$ always have 3 generators? Because I'm reading about different representations of $SU(2)$ as $2\times2$, $3\times3$, $4\times4$ matrices etc. But I guess that even if we are looking at $7\times7$ matrix representation of $SU(2)$ we only need $3$ such matrices corresponding to the $3$ generators right?
And if we have a representation by say $7\times 7$ matrices, is it not possible by using similarity transformations to find an $3-$dimensional invariant subspace corresponding to these $3$ linear operators? Then what point is it to look for higher representations of $SU(2)$ then just $3\times 3$ real matrices?