we used row-operations and column operations to evaluate det(A - λI).
the matrix we wound up with is NOT A-λI, it just has the same determinant.
it is clear that with your original matrix A(e1) = (1,1,1,.....,1) ≠ n(e1) = (n,0,0,.....,0).
we used row-operations and column operations to evaluate det(A - λI).
the matrix we wound up with is NOT A-λI, it just has the same determinant.
it is clear that with your original matrix A(e1) = (1,1,1,.....,1) ≠ n(e1) = (n,0,0,.....,0).
you're trying to find the eigenvectors of A = (a_ij) ,where every entry is 1, right?
if v is an eigenvector, then Av = λv, so (A-λI)v = 0.
when λ = n, this is: (A-nI)v = 0, a homogeneous linear system of equations.
e1 is not a solution to that system.