The product of non-zero eigen values of the matrix is..... 1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 1
Dear user,
Let, the given matrix be A.
1 0 0 0 1
0 1 1 1 0
A = 0 1 1 1 0
0 1 1 1 0
1 0 0 0 1
To find eigen value,
Put, A = (lambda) (I) 5
=> A – (lambda) X (I) 5 = 0
|1 0 0 0 1| |1 0 0 0 0|
|0 1 1 1 0| |0 1 0 0 0|
|0 1 1 1 0| - (X) |0 0 1 0 0| = 0
|0 1 1 1 0| |0 0 0 1 0|
|1 0 0 0 1| |0 0 0 0 1|
Thus,
1-X 0 0 0 1
0 1-X 1 1 0
0 1 1-X 1 0 = 0
0 1 1 1-X 0
1 0 0 0 1-X
Now, take determinant value
=> -x 5 + 5x 4 – 6x 3 = 0
=> -x 3 (x 2 - 5x + 6) = 0
=> x 3 (x 2 – 3x - 2x + 6) = 0
=> x 3 ((x - 3) -2(x - 3)) = 0
Thus,
=> x 3 (x-3)(x-2) = 0
=> x = 0, x = 3, x=2
Now, since x is not = 0 as per question
Thus, valid eigen values are: 2,3.