Monday, April 29, 2019

linear algebra - How to calculate this determinant?




How to calculate this determinant?





A=[n1kkkkkn1kkkkkkkn1]n×n




where n,kN are fixed.



I tried for n=3 and got the characteristic polynomial as (x2k)2(x2+2k).




How to find it for general nN?


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 n1 and n with multiplicity 1. This can be seen by noting that J has n1 dimensional kernel and trace n.



Your matrix A is exactly kJ+(nk1)I where I denotes the n×n identity matrix. The eigenvalues of A are therefore nk1 with multiplicity n1 and nk+nk1 with multiplicity 1. The determinant of A is then (nk+nk1)(nk1)n1.


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