Question #24839

Prove that, if S is square, U is unitary, and U^(conjugate transpose)SU = T is upper triangular, then the eigenvalues of S and T are the same and S and T have the same trace. (Use the facts that det(AB) = det(A)det(B), and tr(ABC) = tr(CAB) = tr(BCA)).

Expert's answer

1

det(S-kI)=det(I) det(S-kI) = det(U*U) det(S-kI) = det(U*) det(S-kI) det(U) = det(

U*(S-kI)U )= det(U*SU-kU*U)=det(T-kI),

thus characteristic polynomials are equal , and therefore they have the same roots.

2

tr(T) = tr (U*SU) = tr (UU*S) = tr(S), so traces are the same.

det(S-kI)=det(I) det(S-kI) = det(U*U) det(S-kI) = det(U*) det(S-kI) det(U) = det(

U*(S-kI)U )= det(U*SU-kU*U)=det(T-kI),

thus characteristic polynomials are equal , and therefore they have the same roots.

2

tr(T) = tr (U*SU) = tr (UU*S) = tr(S), so traces are the same.

## Comments

## Leave a comment