10
$\begingroup$

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 $n!$.

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ Hadamard inequality gives $n^{n/2}$, which is better than $n!$ but probably much worse than the optimal bound. $\endgroup$
    – Fedor Petrov
    Commented Nov 28, 2016 at 15:38
  • 2
    $\begingroup$ Perhaps it's worth mentioning that if $C(n)$ is the best possible constant (so Fedor has shown $C(n)\leq n^{n/2}$), then $C(n)\leq C(nk)^{1/k}$, $\forall k$--as can be seen by creating a block diagonal matrix with $k$ copies of $A$ down the diagonal. This doesn't help much with Fedor's bound, but it shows that either $C(n)$ grows rapidly in $n$ or $C(n)=1$ for all $n$. $\endgroup$
    – Brian Street
    Commented Nov 28, 2016 at 16:00
  • $\begingroup$ No matter what the bound, it can not escape the dependence on $n$. I guess. $\endgroup$
    – T. Amdeberhan
    Commented Nov 28, 2016 at 16:37

2 Answers 2

Reset to default
12
$\begingroup$

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 $3/2>1$.

Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ I checked this by explicitly solving everything, and it's right. The equations are solved by $a=\pi/6$, $b=5\pi/6$, $c=-\pi/6$, $d=\pi/6$, for what it's worth. $\endgroup$
    – Brian Street
    Commented Nov 28, 2016 at 16:20
  • $\begingroup$ It appears (from numerical optimization) that $3/2$ is the maximum value for $n=2$. For $n=3$ the maximum value I get is $2.25$, e.g. for matrix $$\left[ \begin {array}{ccc} -1&\exp \left( 5/6\,i\pi \right) &\exp \left( 4/ 3\,i\pi \right) \\ \exp \left( 5/3\,i\pi \right) &\exp \left( 7/6\,i\pi \right) &1\\ \exp \left( 7/6\,i\pi \right) &0&\exp \left( 11/{6}\; i \pi \right) \end {array} \right] $$ $\endgroup$
    – Robert Israel
    Commented Nov 28, 2016 at 21:25
  • $\begingroup$ Is this hasty? For $n+1$, maximum is $n+\frac1{2^n}$. $\endgroup$
    – T. Amdeberhan
    Commented Nov 28, 2016 at 23:10
  • 2
    $\begingroup$ @T.Amdeberhan See the comment on the OP's question on block diagonal matrices. Your formula predicts that $C(2)=3/2$ and $C(12) = 11+2^{-11}$, which violates the inequality with $n=2$ and $k=6$ because $(11+2^{-11})^{1/6} \approx 1.491 < 3/2$. Perhaps a better guess would be $C(n) = (3/2)^{n-1}$, which also matches the given numerical data for $n=1,2,3$ and does not seem to be in conflict with any of the bounds. $\endgroup$
    – Logan M
    Commented Nov 29, 2016 at 7:24
6
$\begingroup$

As a complement to Fedor's answer, let me note the following special case, where the desired claim does hold.

Prop. Let $A$ be a complex matrix and $A=X+iY$ be its Cartesian decomposition, i.e., $X=\frac12(A+A^*)$ and $Y=\frac{1}{2i}(A-A^*)$. If $X>0$ (i.e., it is positive definite), then in fact $|\det A| \ge \det X$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .