An nxn

matrix A is called

idempotent if A

^{2} =A.

**Claim:** Each

eigenvalue of an idempotent matrix is either 0 or 1.

**Proof:** 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 p(x)=x

^{2} . Then p(A)=A

^{2} .

p(A)v = (A

^{2} )*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=A

^{2} , and λ v = Av = (A

^{2} )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.