I am unsure how we should properly resolve this question.
As we have discussed at length in the comments, only square matrices can be positive semidefinite. Therefore, if $p\neq n$, there is no way that the product matrix $W=UV$ can be positive semidefinite, because it will also be non-square.
For grins, let's assume that $p=n$. Under what conditions is $W=UV$ positive semidefinite? This requires that $x^HUVx\geq 0$ for all complex vectors $x$. Alternatively, this is true if and only if $UV+VU^H=Q$ where $Q=W+W^H$ has nonnegative eigenvalues (and is therefore PSD itself). If $U$, $V$ are real, then you can relax the Hermitian transposes to real tranposes, and consider only reall vectors $x$.
Now for one special case, I know the answer. If $V$ is positive definite---i.e., not just PSD but nonsingular---and we require $W$ to be positive definite as well, then the answer follows from Lyapunov's theorem applied to linear systems:
- The eigenvalues of $U$ must have positive real part.
What about the more relaxed cases? That is, what if $V$ is only positive semidefinite? What if $W$ is only required to be positive semidefinite? I am afraid I do not know. I'm sure people who study Lyapunov's theorem for linear systems in some depth know...
First, one can argue that non-symmetric positive definite matrices are pathological, in the sense that when you move to the complex case all positive definite matrices are hermitian.
For a non-symmetric positive definite matrix you can say little more than the fact that it has positive eigenvalues. You don't have many of the nice properties that symmetry adds. For instance, without symmetry you don't even have that the singular values agree with the eigenvalues, nor diagonalizability.
Edit: here is why in the complex case, positive semidefinite implies hermitian. Actually, the proof implies that in the complex case $A$ is hermitian if and only if $x^*Ax\in\mathbb R$ for all $x$.
Assume $x^*Ax\in\mathbb R$ for all $x$. then
$$
\mathbb R\ni(y+\alpha x)^*A(y+\alpha x)=y^*Ay+\overline\alpha\,x^*Ay+\alpha\,y^*Ax+|\alpha|^2\,x^*Ax.
$$
As this expression is real, it equals its complex conjugate
$$
y^*Ay+\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y+|\alpha|^2\,x^*Ax.
$$
So
$$
\overline\alpha\,x^*Ay+\alpha\,y^*Ax=\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y.
$$
Taking first $\alpha=1$ and then $\alpha=i$, we get
$$
x^*Ay+y^*Ax=y^*A^*x+x^*A^*y,
$$
$$
-i\,x^*Ay+i\,y^*Ax=i\,y^*A^*x-i\,x^*A^*y.
$$
Multiplying the first equation by $i$ and adding, we get
$$
2i\,y^*Ax=2i\,y^*A^*x.
$$
As this works for any $x,y$, we deduce that $A=A^*$.
Best Answer
http://en.wikipedia.org/wiki/Positive-semidefinite_matrix
So, given symmetric $A,$ we have $A^2 = A A^T$ is symmetric positive semidefinite and has just one p.s.d. square root. So your projection is $$ A \mapsto \sqrt{A^2} $$ Meanwhile, if $A$ is already p.s.d., already in the cone, then $A \mapsto A,$ which is what you want for something called a projection.