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 $\leq 1$).

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 $n!$.

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 $n!$.