# Is it possible that A(i, j) = A(u, v), or that A(i, j) > A(u, v) or that A(i, j) < A(u, v)?

Let A be a real m × n matrix, with no two entries equal. Let A(i, j) be the matrix entry obtained by selecting the least entry in each row, and then the greatest of these entries. Let A(u, v) be obtained by selecting the largest entry from each column, and then the least entry from these.

Q1.

Is it possible that A(i, j) = A(u, v), or that A(i, j) > A(u, v) or that A(i, j) < A(u, v)?

Q2.

What would the answers to these questions be if A(u, v) were chosen by taking the largest entry from each row, and then the least of these?

Ive established that each element is unique so cant be repeated, and we have the two equations

A(i, j) = MAX[{MIN(A(1, n), MIN(A(2, n),..., MIN(A(m, n)]

and

A(u, v) = MIN[{MAX(m, 1), MAX(m, 2),..., MAX(m, n)],

and feel all 3 equations are satisfied so am unsure how to approach this question, any help would be appreciated.

For the second question, I believe taking the least entry from each row will be only satisfied where A(i, j) = A(u, v) as both questions uses the minimum elements from the rows, and hence MAX=MIN.

Regarding Q1:

All three are possible:

sage: A
[ 1  2  3  4]
[ 5  6  7  8]
[ 9 10 11 12]
[13 14 15 16]
sage: max(map(min, A.rows()))
13
sage: min(map(max, A.columns()))
13

sage: B
[ -1   5  -1   0]
[  3   2  -3  15]
[ -7   3  -1   0]
[ -3   0   0 -15]
sage: max(map(min, B.rows()))
-1
sage: min(map(max, B.columns()))
0

sage: C
[  4 -12   1   0]
[  0   1  -1   0]
[  1   1   3 -11]
[  4  -1   8   2]
sage: min(map(max, C.columns()))
1
sage: max(map(min, C.rows()))
-1