Suppose $K$ is an $n\times n$ Hermitian matrix and $0\leq K\leq I$, which means that $I-K$ is positive semidefinite. Let $E\subset \{1,2,\dotsc,n\}$. I wonder how to show that det$(M^{E})\geq 0$, where $M^{E}$ is defined as follows:
\begin{align\*}
 M^{E}\left(i,j\right)=\begin{cases}
 \delta\_{i,j}-K(i,j),&i\in E^{\complement}\\
 K(i,j),&\text{otherwise}
 \end{cases}.
\end{align\*}
My approach is that suppose the cardinality of $E$ is $k$, then I can do an induction on $n-k$. If $n-k=0$, then the determinant is nonnegative since $K$ is Hermitian and positive semidefinite. If $n-k=1$, then I can still get the desired result by Cauchy Interlace Theorem. But when $n-k\geq 2$, I don't know how to derive the result. Thanks for any hints.



Determinant of a matrix with entries specified by a set
We may assume that $E=\{1,2,\dots,k\}$. Then
$$
 M^E=\begin{bmatrix}
 A& C\\
 -C^\*& B
\end{bmatrix},
$$
where $A$ and $B$ are PSD (and Hermitian), and $C$ is some rectangular matrix. We claim that, under these constraints, the determinant is always non-negative.


We may assume that $A$ is non-singular, the other cases are obtained via taking a limit. Then, adding to the second multi-row $C^\*A^{-1}$ times the first multi-row, we get
$$
 \det M^E
 =\det \begin{bmatrix}
 A& C\\
 -C^\*+C^\*& B+C^\*A^{-1}C
\end{bmatrix}
=\det A\cdot \det(B+C^\*A^{-1}C).
$$
But both $B$ and $C^\*A^{-1}C$ are PSD (since $A^{-1}$ is PSD), hence their sum is also such. Therefore, the factors are both non-negative.


