Bounds on determinant

Question

I was just curious about the bounds on the determinant of a 3x3 matrix whose elements take values between 0 and 5. I believe this bound is around +-1040

Answer 1

This is a simulation of 1,000,000 random matrices for which the entries $A_{ij} \in (0,5)$ were uniformly distributed. The mean value of the determinant is 0.003774572, with standard deviation 23.31994. The minimum determinant was -160.43715 and the maximum 175.49375 . The following image shows the distribution adjusted by Epanechnikov kernel:

I think your bounds are a bit off...

Answer 2

Hint: You can expand the determinant. Here's one way to approach it (informally)...

For integers $a,\ldots,i\in[0,5]$, we define the matrix determinant $$\left|\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right|=a(ei-fh)+b(fg-di)+c(dh-eg).$$

Since all the terms are non-negative, the product $P$ of any two entries is $0\le P \le 25$. Then, what can you say about the difference $ei-fh$ and the other two cofactor expressions?

From here, it should be clearer how to derive the bounds of the determinant. [feel free to comment if you have any questions or remarks]

EDIT: Of course, the entries need not be integers, as you mentioned. We can let $a,\ldots,i$ be any real numbers in the interval $[0,5]$, though the method above and its result are the same.

Answer 3

The answer is 250.

Here is a brute force approach.

Consider the function $f(x_1,x_2,x_3) = \det \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}$. The function $f$ is continuous and the set $[0,5]^{3 \times 3}$ compact, so there is some maximiser of $f$, say $(\hat{x}_1, \hat{x}_2, \hat{x}_3)$.

Now consider $f(\hat{x}_1+ \lambda e_k, \hat{x}_2, \hat{x}_3)= f(\hat{x}_1, \hat{x}_2, \hat{x}_3) + \lambda f( e_k, \hat{x}_2, \hat{x}_3)$.

Let $\alpha = f( e_k, \hat{x}_2, \hat{x}_3)$. If $\alpha = 0$, then we can replace $\hat{x}_1$ by $\hat{x}_1+ \lambda e_k$ for any $\lambda$, hence we can choose $\lambda$ so that the $k$th entry of $\hat{x}_1+ \lambda e_k$ is $0$ or $5$.

If $\alpha>0$, we can choose $\lambda$ so that the $k$th entry of $\hat{x}_1+ \lambda e_k$ is $5$.

If $\alpha<0$, we can choose $\lambda$ so that the $k$th entry of $\hat{x}_1+ \lambda e_k$ is $0$.

This analysis can be repeated for each column. Hence we can restrict our search to the space $\{0,5\}^{3 \times 3}$. A quick examination of the $2^9$ matrices shows that the maximum is 250 and this is attained by, for example, $\begin{bmatrix} 0 & 5 & 5 \\ 5 & 0 & 5 \\ 5 & 5 & 0 \end{bmatrix}$.

Here is the Python program that examined the $2^9$ matrices:

import numpy
import itertools

cols = list(itertools.product((0 ,5), (0 ,5), (0 ,5)))
mats = itertools.product(cols, cols, cols)
sup = 0
sup_p = None
for p in mats:
    d = numpy.linalg.det(p)
    if d > sup:
        sup = d
        sup_p = p

print sup
print sup_p