Hi, here's the problem
Let A be an n x n matrix, where n >= 2, with the following properties:
1) a_ij > 0 for all i,j = 1,2,.....n
2) 2a_ii > Sum_j=1..n (a_ij)
Show that A is invertible.
The second condition can be rewritten as a_ii > Sum_j =! i (a_ij) . This means that an element on the main diagonal is greater than the sum of all other elements in the row.
Let's show that the determinant is not equal to 0 (i.e. A is invertible)
If we use column operations to get 0's everywhere in the first row except for the first element (a_11) the determinant will be written as
det A = a_11 * A_11 where A_11 is the matrix A where we removed the first row and column
the top left element of this matrix will be a_22 - ka_21 where k is such that a_12-k*a_11 = 0 . where k < 1 Due to condition 2)
the top left element of A_11 must therefore be positive (condition 2) )
We can continue doing this and show that det A> 0 but it gets messier as we go along and it's a bit harder to prove that the elements are positive
Does anyone know a better way of solving this ?