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Hermitian and skew-Hermitian matrices

Short answer

Conjugate transpose gives (BᶿAB)ᶿ = BᶿAᶿB. If Aᶿ = A, result equals BᶿAB (Hermitian). If Aᶿ = −A, result equals −BᶿAB (skew-Hermitian).

Show that the matrix Bᶿ A B is Hermitian or skew-Hermitian according as A is Hermitian or skew-Hermitian (Bᶿ denotes the conjugate transpose of B).

Let Bᶿ denote the conjugate transpose of B. We use the rules (XY)ᶿ = YᶿXᶿ and (Xᶿ)ᶿ = X. A matrix H is Hermitian if Hᶿ = H and skew-Hermitian if Hᶿ = −H.

Take the conjugate transpose of BᶿAB

(BᶿAB)ᶿ = Bᶿ Aᶿ (Bᶿ)ᶿ = Bᶿ Aᶿ B

(the order is reversed, and (Bᶿ)ᶿ = B).

Case 1: A is Hermitian (Aᶿ = A)

(BᶿAB)ᶿ = Bᶿ A B

So BᶿAB equals its own conjugate transpose, i.e. BᶿAB is Hermitian.

Case 2: A is skew-Hermitian (Aᶿ = −A)

(BᶿAB)ᶿ = Bᶿ(−A)B = −(BᶿAB)

So BᶿAB is skew-Hermitian. ∎

Hence BᶿAB is Hermitian or skew-Hermitian according as A is Hermitian or skew-Hermitian. (B can be any matrix for which the product is defined; no condition on B is needed.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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