Complex Projective Space as a $U(1)$ quotient As is well known, one can view $\mathbb{CP}^n$ as a quotient of the unit $(2n + 1)$-sphere in $\mathbb{C}^{n+1}$ under the action of $U(1)$, since every line in $\mathbb{C}^{n+1}$ intersects the unit sphere in a circle. 
Moreover, we have $S^{2n + 1} = SU(n+1)/SU(n)$, where $SU(n)$ embeds into the bottom right-hand corner (say).
My question is: Is there an embedding $j$ of $U(1)$ into $SU(n+1)/SU(n)$ that gives the $U(1)$ action as a left (right) multiplication, i.e. such that $A.e^{i \theta} = Aj(e^{i \theta})$, for all $A \in SU(n+1)/SU(n)$?
For $n=1$, it's easy: $SU(2) = S^3$, and we embed $e^{i\theta}$ as
$\left(
  \begin{array}{cc}
   e^{i \theta}  & 0  \\\\
   0             & e^{-i \theta}
  \end{array}
\right)$.
 A: The U(1) group is the torus in SU(N+1) which commutes with SU(N). In the example
given in the question where SU(N) is chosen as the bottom N-dimensional block. U(1) consists
of the diagonal matrices 
diag{exp(N*i*theta), exp(-i*theta), . . . . (N-times) exp(-i*theta)}
Please observe that the restriction to the bottom N-dimensional block is proportional to the unit matrix thus it commutes with the whole of SU(N), also it belongs to SU(N+1), since its has a unit determinant.
The reason that the U(1) and the SU(N) factors commute is due to a theorem by  A. Borel  which states that the denominator subgroup of homogeneous Kaehlerian spaces must be the centralizer of a torus. In our case the torus is the U(1) subgroup and the certralizer is SU(N)*U(1)
A: From wikipedia: $\mathbb{CP}^n$ is a Symmetric space of type AIII for $p=n$, $q=1$.  There are embeddings of both $U(1)$ and $SU(n)$ into $S(U(n) \times U(1)) \subset SU(n+1)$ which give you your quotient by right multiplication.
A: 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.
