_TZ_6393-periastron.dviMATH 3221 ADVANCED LINEAR ALGEBRADAILY ASSIGNMENT MARCH 15, 2023My apologies for posting this so late! I was trying to make the numbers work out on the“find the adjoint...

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Answer To: _TZ_6393-periastron.dviMATH 3221 ADVANCED LINEAR ALGEBRADAILY ASSIGNMENT MARCH 15, 2023My...

Dr. Saloni answered on Mar 16 2023
48 Votes
1
Math 3221
35.
To find the adjoint of TP , we need to find the operator T∗P such that hTP (A),
Bi = hA, T∗P (B)
for all A, B ∈ Mn(C). Using the definition of TP , we have:
hTP (A), Bi = hP−1AP, Bi = tr((P−1AP)∗B) = tr(P−1A∗PB)
= tr(A∗PBP−1) = hA, PBP−1i.
Thus, we need T∗P = PBP−1, where B is the adjoint of P−1, i.e., P−1B = BP−1 = adj(P−1).
If TP is self-adjoint, then we have T∗P = TP , which implies that PBP−1 = P−1AP for all A ∈ Mn(C).
Multiplying both sides by P on the left and P−1 on the right, we get BP−1 = P−1A(adj(P))P−1.
Taking adjoints on both sides, we get B = P−1(adj(P))P−1A∗. Thus, B is uniquely determined by
P and is...
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