An nxn matrix A is called idempotent if A2 =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)=x2 . 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.

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