Is the sum of a Wishart matrix and a deterministic psd matrix "almost Wishart"? Let $XX^T$ be a Wishart matrix, generated by taking the columns of $X$ to be i.i.d. standard $p$-variate normal vectors. Let $AA^T$ be a non-random positive definite matrix.
Though it is not possible to get an eigendecomposition for the sum $XX^T+AA^T$ merely from information about the eigenstructure of the summands, I have a more basic question:

Is it true that $XX^T+AA^T$ can be expressed as $YDY^T$, where the entries of $Y$ are i.i.d. variables (not necessarily normal?) and $D$
  is a self-adjoint non-negative matrix?  (or perhaps $D$ is even
  diagonal?)

This would be quite interesting, since one might be able to say something about the spacing of the eigenvectors of the sum without knowing its precise eigendecomposition.
Edit: My intuition is that the sum should be Wishart, since as I recall the sum of independent Wishart matrices is Wishart. The matrices are independent, and though the second matrix is not random, it seems that "from the perspective of $XX^T$, there is no difference between a Wishart matrix with expectation $AA^T$ and $AA^T$ itself. Can this be made rigorous?
EDIT 2 Please note that I am primarily asking for an understanding of the independence of the entries of $Y$.
 A: You can do this with $D$ as the identity matrix. You have $X\in\mathbb R^{p\times n}$ and $AA^T\in\mathbb R^{p\times p}$. The matrix $XX^T + AA^T$ is a symmetric matrix with entries that are real, and so can be diagonalized by some orthogonal matrix $W\in\mathbb R^{p\times p}$:
$$
XX^T + AA^T = W \begin{bmatrix} \lambda_1 \\ & \lambda_2 \\ & & \lambda_3 \\ & & & \ddots \\ & & & & \lambda_p \end{bmatrix} W^T
$$
where $\Lambda$ and $W$ depend on $X$ and $A$, and the $\lambda$s can be taken $\ge 0$ because $XX^T+AA^T$ is non-negative definite.  So now let
$$
\Lambda^{1/2} = \begin{bmatrix} \sqrt{\lambda_1} \\ & \sqrt{\lambda_2} \\ & & \sqrt{\lambda_3} \\ & & & \ddots \\ & & & & \sqrt{\lambda_p} \end{bmatrix}.
$$
Then we can let
$$
Y = W\Lambda^{1/2}
$$
and let $D$ be the $p\times p$ identity matrix, and then we have
$$
XX^T + AA^T = Y D Y^T.
$$
If you let $X$ vary while $A$ is fixed, you get a function $X\mapsto Y$.
All of the above can be said without saying anything at all about randomness or probability distributions.  But now if you impose some probability distribution on the random matrix $X$ while saying $\Pr(A=\text{one specified fixed matrix})$, then that determines a probability distribution of the matrix $Y$.
But $Y D Y^T$ has neither a Wishart distribution nor a noncentral Wishart distribution unless $A=0$.  The reason for that is as follows.  The set of all non-negative-definite $p\times p$ symmetric matrices with real entries has a natural partial order defined as follows: $P\le Q$ precisely if $Q-P$ is non-negative definite.  It is not hard to show that that satisfies the definition of a partial order relation.
Now observe that your random matrix $YY^T$ is supported entirely on the set of matrices $\ge AA^T$ in this partial order.  That does not happen with Wishart matrices.  With a Wishart matrix $V$ one would have $0 < \Pr(V \le AA^T) \le \Pr(V \not\ge AA^T)$.  Consider in particular what happens when $p=1$.  The support is $[a^2,\infty)$.  The support of either a central or a non-central chi-square distribution is $[0,\infty)$.
