Textbook Problem 4.4.17:

This problem asks for the maximum value of a real 5x5 determinant,
subject to the constraint that all entires have magnitudes <= 1.
Gil Strang comments that he doesn't know the "best" upper bound.

Let A be such a matrix, and factor A = QR.  Then since |Q| = 1,
it follows that |A| = |R|.  Furthermore, since A^T A = R^ R,
it also follows that the sum of the squares of any column of
A is the same as the sum of the squares of the same column of R.
Hence, the sum of squares in any column of R is <= 5, whence
each diagonal element of R has magnitude <= SQRT(5).  And since
R is upper triangular,  we have |A| = |R| <= SQRT(5^5) ~ 55.9...

The discussion in the "solutions" in the back of the book also
pursues some other, much weaker bounds, such as |A| <= 120.

More interesting, in my view, are lower bounds.  My favorite
candidate is B = J - 2I,  where I is the 5x5 identity, and
J is the whose 25 entires all of the same value, +1.
This particular matrix as |B| = 48.


CANDIDATES WITH ONLY INTEGER COMPONENTS

It is not hard to show that this is the maximum possible value
of the determinant if all entries have magnitude equal to 1.
In any such matrix, the only possible entries are +1 and -1.
The determinant is unchanged by elementary row operations,
so we can begin by adding or subtracting one copy of the first
row, as appropriate, to each of the following rows in order to
zero out all entries in the first column below the diagonal.
Once this is done, all entries in all rows except the first
must be EVEN integers.  So we can one factor of two out of
each such row, and thus conclude that the value of the
determinant is an integer multiple of 16.  But since 16 time 4
is 64, which exceeds the prior bound of 55.9..., we are able
to conclude that any 5x5 determinant whose entries are all
+1 or -1 must have a determinant no greater than 48.

I can also show that if all entries of A with magnitude <1
all lie on the same row of column, then |A| < 48.
To this end, write a cofactor expansion of A, in terms of
the relevant row or column.  Each cofactor must have magnitude
which is a multiple of 8.  If these magnitudes summed to
anything >=56, then by appropriately changing each entry
in the relevant row to +1 or -1, we could find a new A
whose determinant was >=56, contradicting the bound of
55.9.

This argument also shows that the matrix B = J - 2I is a local
maximum.  Decreasing the magnitude of any single entry of B
simply reduces the value of its determinant.  This statement
is also true for any other B which has all entries of
magnitude = 1, and |B| = 48.

So there appears considerable evidence to suggest that the
best possible bound is |A| <= 48.  However, I have not yet
found any proof of that assertion.

=======

QR FACTORIZATION

Let A be a candidate matrix, and let A = QR.  Then since Q
is orthonormal, valumes of parallelpipeds with vertices at 0
and at any number of columns of Q will be the same as volumes
of parallelpipeds with vertices at 0 and at the corresponding
columns of R.  So it may be advantageous to factor A and then
study the related "R".

Consider the QR factorization of

B = J - 2I = Q R

The orthogonal matrix Q can be written in this form:

-2   3   3   3   3    SQRT(1/20)
 2  -3   3   3   3            SQRT(1/30)
 2   2  -4   3   3                    SQRT(1/42)
 2   2   2  -5   3                            SQRT(1/56)
 2   2   2   2  -6                                    SQRT(1/72)

Here, the first matrix has only "3" above the diagonal, only "2"
below the diagonal, and diagonal entries of -2, -3, -4, -5, and -6.
The second matrix is diagonal, and its entires are the reciprocal
roots of 4x5, 5x6, 6x7, 7x8, and 8x9.

The upper triangular matrix R can be written as 2I times this:

SQRT(5/4 ) SQRT(1/20) SQRT(1/20) SQRT(1/20) SQRT(1/20)
           SQRT(6/5 ) SQRT(1/30) SQRT(1/30) SQRT(1/30)
                      SQRT(7/6 ) SQRT(1/42) SQRT(1/42)
                                 SQRT(8/7 ) SQRT(1/56)
                                            SQRT(9/8 )



The product of its diagonal entries telescopes to
SQRT(1/4) SQRT(9) = 3/2, which, when multiplied by the
determinant of 2 I (which is 32) gives 48, as was known.

I suspect this result may be best-possible, in the following
sense:

Conjecture: Given any n real 5-dimensional vectors, each of
which has all components bounded between -1 and 1, inclusive,
then the maximum volume that can be enclosed in the parallelpiped
generated by 0 and these n vectors is as follows:

n    max Vn

1      SQRT(5) 
2    2 SQRT(6)
3    4 SQRT(7)
4    8 SQRT(8)
5   16 SQRT(9) = 48