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Problem:

Given $N\in\mathbb{Z}^+$ and $\mathbf{A}\in[0,1]^{N\times N}$, is $\det(\mathbf{A})$ bounded? What are $\max(\det(\mathbf{A}))$ and $\min(\det(\mathbf{A}))$?

My Efforts:

I may guess that $\det(\mathbf{A})$ is bounded since it is the summation of some multiplications of $\mathbf{A}_{ij}$s and each element is bounded by $[0,1]$. But I do not know how to calculate $\max(\det(\mathbf{A}))$ and $\min(\det(\mathbf{A}))$ for any $N$.

I see that there is a post discussing the case of $N=3$ with $\mathbf{A}_{ij}\in \{0,1\}$ Reference.

I may know from Matrix Cookbook that $\frac{\partial}{\partial \mathbf{A}}\det(\mathbf{A})=\det(\mathbf{A})(\mathbf{A}^{-1})^\top$, but this formula may be not very helpful for this problem.

I also perform some simulation experiments by writing a toy Python program and find that $\det(\mathbf{A})$ may be NOT go to $+\infty$ or $-\infty$.

But how to exactly describe and analyse this?

BinChen
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1 Answers1

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It is bounded since $[0, 1]^{N \times N}$ is compact and its image under the continuous map $\det$ should also be a compact subset of $\mathbb{R}$. Also, for $N>1$, we have $\min(\det(A)) = -\max(\det(A))$, since for any $A$, the matrix $A'$ obtained by swapping first two rows of $A$ has a determinant $\det(A') = -\det(A)$. At last, it is enough to consider the matrices with entries in $\{0, 1\}$ instead of whole $[0, 1]$ by following the argument in the answer you linked ($3 \times 3$ case).

However, it seems that finding the closed formula for maximum/minimum in $N$ is open - see here.

Seewoo Lee
  • 15,670