Let $A$ be an $n\times n$ complex matrix, and write $A=X+iY$, where $X$ and $Y$ are real $n\times n$ matricies. Suppose that for every square submatrix $S$ of $A$, $|\mathrm{det}(S)|\leq 1$ (i.e., all minors of $A$ are complex numbers with modulus $\leq 1$). This includes the assumption that $|\mathrm{det}(A)|\leq 1$.


**Question:** Is $|\mathrm{det}(X)|\leq 1$?


**Note:** It is easy to see $|\mathrm{det}(X)|\leq n!$ by a simple induction (since every component of $A$ has modulus $\leq 1$--and therefore the same is true for $X$). However, computer simulations make me wonder if the above question might be true. I'd be up for hearing about any sort of bound which is better than $n!$.



Determinant of the "real part" of a matrix
Bound 1 does not hold already for $n=2$. Take a matrix $A=\pmatrix{e^{ia}&e^{ib}\\e^{ic}&e^{id}}$, it satisfies your conditions if $|a+d-b-c|\leqslant \pi/3$. On the other hand, $X=\pmatrix{\cos a&\cos b\\\cos c&\cos d}$ and $$\det X=\cos a\cos d-\cos b\cos c=\frac12\left(\cos(a+d)+\cos(a-d)-\cos(b+c)-\cos(b-c)\right)\\=
\frac12\left(2\sin\frac{b+c-a-d}2\sin\frac{a+d+b+c}2+\cos(a-d)-\cos(b-c)\right),
$$
thus if $a-d=0$, $b-c=\pi$, $a+d+b+c=\pi$, $b+c-a-d=\pi/3$ (I am lazy to solve this explicitly), this expression is equal to $3/2>1$.


