$A=\begin{pmatrix}1 & 1\cr 1 & 4\end{pmatrix}$
$B=\begin{pmatrix}1 & 1\cr 1 & 5\end{pmatrix}$
the above two matrix is positive definte matrix, but $AB$ is not a symmetrical matrix. if we add $AB=BA$,then product of two positive definite matrices is positive definite.
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...
Best Answer
You could view it as the parabola $Kx^2=y, K>0 $ taken up to higher dimensions. In place of the positive constant $K$, a positive definite matrix $A$ would ensure that the high dimensional parabola (visualise it as a bowl) takes all positive values for all $x\in \mathbb{R}^n$.
See this question for why definiteness is needed when considering ordering among matrices.