My original answer was unsalvegable so I've deleted it and am posting a new "answer". As with the first one, I don't rate this as particularly an answer but more just trying to understand what's going on.

I was initially having trouble understanding Scott's answer, but now I think I do and I think it gives the matrix representation wanted which isn't quite what David wrote.

We have $SU(n+1)$ and inside this we have $SU(n)$ and quotient out to get $S^{2n+1}$. We also have a slightly larger subgroup which is $S(U(n) \times U(1))$, which contains $SU(n)$, such that the quotient is $\mathbb{CP}^n$.

Now, $S(U(n) \times U(1))$ is $U(n)$ via $A \mapsto (\det A^{-1},A)$ and the inclusion $SU(n) \to S(U(n) \times U(1))$ goes over to the standard inclusion. Here, $SU(n)$ is a normal subgroup and $U(n)$ is the semi-direct product of $SU(n)$ and $U(1)$ with the map $U(1) \to U(n)$ given by $\lambda \mapsto (\lambda, 1,\dots,1)$ (diagonal matrix). When taken over to $S(U(n) \times U(1))$ this becomes $\lambda \mapsto (\lambda^{-1},\lambda,1,\dots,1)$.

So then $SU(n+1)/S(U(n) \times U(1)) \cong (SU(n+1)/SU(n))/U(1)$ where $U(1) \to SU(n+1)$ is the map $\lambda \to (\lambda^{-1},\lambda,1,\dots,1)$.

This isn't the same as David's, I know, so it may not be what you want (since that answer's been accepted). Presumably only one satisfies the condition that you want and presumably it's David's since that answer's been accepted. Still, I was confused and I think I've straightened myself out now.