How to calculate this determinant?
A=[n−1kkk…kkn−1kk…k…………kkkk…n−1]n×n
where n,k∈N are fixed.
I tried for n=3 and got the characteristic polynomial as (x−2−k)2(x−2+2k).
How to find it for general n∈N?
Answer
Here I've followed the same initial step as K. Miller. Instead of using a determinant identity I examine the eigenvalues A and consider their product.
If J denotes the n×n matrix of all 1's, then then eigenvalues of J are 0 with multiplicity n−1 and n with multiplicity 1. This can be seen by noting that J has n−1 dimensional kernel and trace n.
Your matrix A is exactly kJ+(n−k−1)I where I denotes the n×n identity matrix. The eigenvalues of A are therefore n−k−1 with multiplicity n−1 and nk+n−k−1 with multiplicity 1. The determinant of A is then (nk+n−k−1)(n−k−1)n−1.
No comments:
Post a Comment