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Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C, A, B) = 0

User Drew Moore
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1 Answer

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Answer:

Three linear operators A,B, and C will satisfy the condition
[[A, B],C] + [[B,C), A] + [[C, A], B] = 0.

Step-by-step explanation:

According to the question we have to prove.


[[A, B],C] + [[B,C), A] + [[C, A], B] = 0

Now taking Left hand side of the equation and solve.


[[A, B],C] + [[B,C), A] + [[C, A], B]

Now use commutator property on it as,


=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0

Therefore, it is proved that
[[A, B],C] + [[B,C), A] + [[C, A], B] = 0.

User Tbhartman
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