We defined projections in a vector space V(C) with a positive definite Hermitian product >> as linear operators P, such that P=P=P*. Let's consider the case of the finite-dimensional vector space V'(C). a) Show that if a linear operator P on V(C) is a projection, then it has all eigenvalues exclusively in the set {0,1}, b) Show that a Hermitian (self-adjoint) operator P=P* which has all eigenvalues in the set {0,1} is a projection. (Hint: If you want to use the appropriate Spectral Theorem explain why you can use it. Also, remember to justify the claim that the square of a diagonal matrix is a diagonal matrix with squared elements.] We defined projections in a vector space V(C) with a positive definite Hermitian product >> as linear operators P, such that P=P=P*. Let's consider the case of the finite-dimensional vector space V'(C). a) Show that if a linear operator P on V(C) is a projection, then it has all eigenvalues exclusively in the set {0,1}, b) Show that a Hermitian (self-adjoint) operator P=P* which has all eigenvalues in the set {0,1} is a projection. (Hint: If you want to use the appropriate Spectral Theorem explain why you can use it. Also, remember to justify the claim that the square of a diagonal matrix is a diagonal matrix with squared elements.]