An nxn matrix
A is called idempotent
of an idempotent matrix is either 0 or 1.
Let A be an nxn matrix, and let λ be an eigenvalue of A, with corresponding eigenvector
v. Then by definition of eigenvalue and eigenvector, Av= λ v.
Consider the polynomial
. Then p(A)=A2
p(A)v = (A2
)*v = A*A*v = A*(A*v) = A*( λ *v) = A* λ *v
Note at this point that λ is a scalar
, and so commutes
with the matrix A.
A* λ *v = λ *A*v = λ *(A*v) = λ * λ *v = ( λ 2
)*v = p( λ )v
In other words, p(A)v = p( λ )v.
Since A is idempotent, A=A2
, and λ v = Av = (A2
)v = p(A)v = p( λ )v. We may then conclude that since λ *v = ( λ 2
)*v, λ must be equal to λ 2
In any scalar field we are working with (real or complex), the only scalar solutions to this equation are 0 and 1.
Taken from homework assignment from a class titled "Matrix Theory" at the University of Iowa, 2002.