MHB Determinant of matrix with Aij = min(i, j)

Click For Summary
The determinant of an n x n matrix defined by Aij = min(i, j) is consistently 1. To prove this, one effective method is to perform elementary row operations, specifically subtracting the row above from each subsequent row. This operation preserves the determinant, allowing the transformed matrix to maintain the same determinant as the original. The resulting upper triangular matrix confirms that the determinant remains 1, as the product of the diagonal entries is 1. Thus, the determinant of the original matrix can be established as 1 through these operations.
nedf
Messages
4
Reaction score
0
Given a n x n matrix whose (i,j)-th entry is i or j, whichever smaller, eg.
[1, 1, 1, 1]
[1, 2, 2, 2]
[1, 2, 3, 3]
[1, 2, 3, 4]
The determinant of any such matrix is 1.
How do I prove this? Tried induction but the assumption would only help me to compute the term for Ann mirror.
 
Physics news on Phys.org
From each line subtract the line immediately above it.
 
I will get this?
[1 1 1 1]
[0 1 1 1]
[0 0 1 1]
[0 0 0 1]
How do I prove that this matrix has same determinant as the original one?
 
nedf said:
I will get this?
[1 1 1 1]
[0 1 1 1]
[0 0 1 1]
[0 0 0 1]
Yes.

nedf said:
How do I prove that this matrix has same determinant as the original one?
Subtracting another line is one of the elementary row operations and is known to preserve the determinant.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K