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
It's actually defined more generally for Hermitian matrices (matrices equal to the complex conjugate of their transpose). If a Hermitian matrix has at least one entry with a (nonzero) imaginary part, then the matrix is not symmetric, but it could still be positive (semi-)definite.
There are two reasons I can think of why we wouldn't extend the definitions to non-Hermitian matrices. The first is that if the matrix is not Hermitian, then we could have complex eigenvalues. In that context, "positive" has no meaning.
Another reason why I think we wouldn't has to do with quadratic forms. If $A$ is a real symmetric matrix, and $x$ is a vector of variables, then $f(x)=x^TAx$ is called a quadratic form. $f(x)>0$ ($f(x)\geq 0$) for all nonzero $x$ if and only if $A$ is positive definite (semi-definite). We could relax the condition that $A$ is symmetric, and define quadratic forms in the same way for a general real matrix. But the resulting quadratic form is the same as the one defined by the real symmetric matrix $(A+A^T)/2$. The quadratic form is positive definite if and only if $(A+A^T)/2$ is, so positive definiteness is not a property of $A$ as much as it is a property $(A+A^T)/2.$ In other words, we would end up studying symmetric matrices anyway, so wouldn't get anything new by extending the definition to non-symmetric real matrices.