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

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SUMMARY

The determinant of an n x n matrix defined by Aij = min(i, j) is conclusively 1 for all n. This is demonstrated through elementary row operations, specifically by subtracting the row immediately above from each row, which preserves the determinant. The resulting matrix transforms into an upper triangular form, confirming that the determinant remains unchanged. Therefore, the determinant of the original matrix is validated as 1.

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nedf
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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.
 
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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.
 

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