The matrix $\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ is orthogonal and indefinite. 
 $\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}$ is positive definite and not orthonormal.
and the Identity matrix $I$ is of course both orthogonal and positive definite. Let $S$ be the intersection of orthogonal matrices and positive definite matrices. We have seen that $S$ is non-empty. 


By a positive definite matrix, I mean either a symmetric or asymmetric matrix $M \in \mathbb{R}^{p^2}$ whose quadratic form satisfies $\forall x \in \mathbb{R}^p \setminus \{0\}: x^TMx > 0$. Let $P$ denote the set of all positive definite matrices. Then $P$ is a convex cone.


$S$ is not a convex cone, unlike $P$. Also unlike $P$, $S$ is closed under multiplication The product of any two matrices in $S$ is guaranteed to be at least PSD.[1]


I am interested in complete characterizations of this space $S$, which globally behaves more like the space of orthogonal matrices. *My real motivation is that* I want to know whether there are efficient procedures for testing whether a matrix $M $ is in $S$ that are faster than testing for positive definiteness which requires the calculation of eigenvalues? E.g. such procedures might take only $O(p^2)$ computations. I tried to google for resources but nothing relevant came up.


[1] Orthogonal matrices are of course closed under multiplication. The product of two PD matrices is PD PSD matrices is PSD iff their product itself is normal, which is true in $S$.


Reference: On a product of positive semidefinite matrices, A.R. Meenakshi, C. Rajian, Linear Algebra and its Applications, Volume 295, Issues 1–3, 1 July 1999, Pages 3–6.


*In case someone is wondering, the real real reason, why* I am interested in this question is because I want to "efficiently" find a "reasonably good" positive definite approximation of a matrix whose $QR$ decomposition is known to me. The details of this part are best left for future.



The space of positive definite orthogonal matrices
You may find the Cayley transform to be useful here:


As is well-known and easy to prove, every orthogonal $n$-by-$n$ matrix $R$ that does not have $-1$ as an eigenvalue can be written uniquely in the form
$$
R = (I-A)(I+A)^{-1}
$$
for some anti-symmetric matrix $A$ for which $I+A$ is invertible, and, conversely, if $A$ is any anti-symmetric matrix such that $(I+A)$ is invertible, the matrix $R$ in the above formula is orthogonal and $I+R$ is invertible.


In fact, $A = (I-R)(I+R)^{-1}$, so $A$ is easy to find. The two matrices $R$ and $A$ are said to be *Cayley transforms* of each other, and this provides a 'rational parametrization' of the orthogonal group minus the hypersurface consisting of the orthogonal matrices that have $-1$ as an eigenvalue. (Note that $R$ and $A$ commute.)


Now $R$ satisfies the stronger condition that $x^TRx>0$ for all nonzero $x\in\mathbb{R}^n$ if and only if (setting $y=(I+A)^{-1}x$ or, equivalently $x = (I+A)y$), we have
$$
0 < x^TRx = y^T(I+A)(I-A)y = y^T(I-A^2)y = |y|^2-|Ay|^2 
$$
for all $y\in \mathbb{R}^n$. Equivalently, the matrix norm of $A$, i.e., $\|A\|$, should be strictly less than $1$.


Thus, via the Cayley transform, your set $S$ is parametrized by the open convex set in the vector space of anti-symmetric $n$-by-$n$ matrices consisting of those anti-symmetric matrices $A$ whose matrix norm is less than $1$.


