It is easy to see that Problem 02.1.2 holds more generally in that the result is valid for B Hermitian complex (rather than “symmetric real”), for m = 0,1,…,q (rather than just for “m < q”), for r ≠ 0 (rather than just for “integer r > 1”), and for tr(B^k) = tr(C^k), k = 1,2,3,4 (rather than for “k = 1,2,…”), as shown by Shanbhag (1970).

To see this let A = (1/r)B, D = (1/r)C, and α₁,α₂,…,α_q denote the eigenvalues of A. Then

because we know that if λ₁,…,λ_q are the (not necessarily distinct) eigenvalues of

, then λ₁ⁿ,…,λ_qⁿ are the eigenvalues of Mⁿ, irrespective of

(see, e.g., Lancaster, 1969, Theorem 2.5.2). A fundamental result on characteristic polynomials tells us that

(cf. Lancaster, 1969, p. 55). Hence

and so the α_i = 0 or 1. Because

it follows that precisely m of the α_i are equal to 1 and q − m are equal to 0. A Hermitian matrix is unitarily similar to the diagonal matrix of its eigenvalues, and the eigenvalues are all real (cf. e.g., Lancaster, 1969, Theorem 2.9.5 and Theorem 2.9.1). It follows then, as is well known, that a Hermitian complex matrix with all eigenvalues equal to 0 or 1 is idempotent. Our proof is complete.

One might be tempted to believe that a complex square matrix M, not Hermitian, with all eigenvalues equal to 0 and/or 1 is always idempotent. This, however, is erroneous, as the nilpotent matrix

shows. Although this matrix fails to be idempotent, it has the eigenvalue 0 with (algebraic) multiplicity 2.

When A and B are both Hermitian nonnegative definite, then the eigenvalues α_i ≥ 0, and, as Shanbhag (1970) points out, the condition tr(B^k) = tr(C^k), k = 1,2,3,4 may be replaced by tr(B^k) = tr(C^k), k = 1,2,3. This leads to

, and because the α_i are now all nonnegative we see immediately that α_i = 0 or 1.