(A B)^{T} = B^{T} A^{T} .

For the requested two-dimensional check, put

A = (a c b d)

and

B = (e g f h)

. Both sides are

(a e + b g a f + bh ce + d g c f + d h) .

More generally, the entry formula

(A B)_{ji} = \sum_{k} A_{jk} B_{ki}

is exactly

(B^{T} A^{T})_{ij}

ii

Solution

The conjugate transpose also reverses factor order. If

U

and

V

(U V)^{†} (U V) = V^{†} U^{†} U V = V^{†} I V = I .

Thus their product is unitary. This is closure under multiplication in the unitary group.

H

H^{†} = H

. Reversing the product of

n

identical factors has no effect, so

(H^{n})^{†} = (H^{†})^{n} = H^{n} .

Hence every positive integer power is Hermitian. The identical-factor condition matters; a product of different Hermitian matrices need not be Hermitian.

A

and

B

(A B)^{†} = B^{†} A^{†} = B A

. Therefore

A B is Hermitian ⟺ A B = B A ⟺ [A, B] = 0.

The required condition is that the matrices commute.

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