Question #24837

Shur's Lemma: For any square matrix S therer is a unitary matrix U such that
U^(conjugate transpose)SU = T is an upper triangular matrix.
Use Shur's Lemma to prove that
(a) If S^(conjugate transpose)=S then there is a unitary matrix U such that U^(conjugate transpose)SU is a real diagonal matrix.
(b) If S is real and and S^T = S then there is an orthogonal matrix O such that O^TSO is a real diagonal matrix.

Expert's answer

1

S* = S . Then as U*SU=T, and T* = (U*SU)*=U*S*U**=U*SU, then T=T*, but T -

upper triangular, and thus it have to be diagonal, and elements on its diagonal

are conjugate. Thus T - real diagonal.

2

S^T=S.

Let D = U*SU, by part 1. Then D=D^T and thus (U*SU)^T=U^T S (U*)^T

= U^T S U', where U' means conjugate to U.

So, U*SU = U^T S U' and then U* S U = (U')*T S U', but Shur decomposition

is unique, and thus U=U', so U is real Hermitian, i.e. U - orthogonal.

S* = S . Then as U*SU=T, and T* = (U*SU)*=U*S*U**=U*SU, then T=T*, but T -

upper triangular, and thus it have to be diagonal, and elements on its diagonal

are conjugate. Thus T - real diagonal.

2

S^T=S.

Let D = U*SU, by part 1. Then D=D^T and thus (U*SU)^T=U^T S (U*)^T

= U^T S U', where U' means conjugate to U.

So, U*SU = U^T S U' and then U* S U = (U')*T S U', but Shur decomposition

is unique, and thus U=U', so U is real Hermitian, i.e. U - orthogonal.

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