Determinant of $E^T H E$ where $E$ is semi-orthogonal and $H$ is positive definite Is there a way to simplify/obtain alternative forms of $\text{det} \left(E^T H E \right)$ where $E \in \mathbb{R}^{m \times n}$, $m > n$ is semi-orthogonal (meaning that its columns are orthonormal) and $H \in \mathbb{R}^{m \times m}$ is positive definite?
 A: The Cauchy interlacing theorem states that if $\lambda_1,\dots,\lambda_m$ are the eigenvalues of $H$ in decreasing order, and $\mu_1,\dots,\mu_n$ are the eigenvalues of $E^THE$, then we have the inequalities
$$
\lambda_j \geq \mu_j \geq \lambda_{j+(m-n)}
$$
for $j = 1,\dots,n$.  Conversely, for any tuple of real numbers $\mu_1,\dots,\mu_n$ satisfying the above inequalities for all $j$, there exists a semi-orthogonal $E$ such that the $\mu_j$ are the eigenvalues of $E^THE$.
Consequently: denote
$$
a= \prod_{j=m-n+1}^m \lambda_j, \qquad
b = \prod_{j={1}}^n \lambda_j 
$$ 
we can state that
$$
a \leq \det(E^THE) = \prod_{j=1}^n \mu_j \leq b
$$
and that any determinant on the interval $[a,b]$ can be attained with a suitable semi-orthogonal matrix $E$.

In terms of the exterior product (AKA wedge-product, AKA alternating tensor product), we can state that $a = \lambda_{\min}(\wedge^k H)$, and $b = \lambda_{\max}(\wedge^k H)$.  Note that $\wedge^k H$ will be a size-$\binom mk$ square matrix.
It might also be useful to observe that we have
$$
\lambda_m^n \leq a \leq \det(H)^{n/m}
 \leq b \leq \lambda_1^n$$
