Hermitian Conjugate of Matrix Explained

[raintrek]

Simple question, and pretty sure I already know the answer - I just wanted confirmation,

Considering the Hermitian Conjugate of a matrix, I understand that

A^{+} = A where A^{+} = (A^{T})^{*}

Explicitly,

(A_{nm})^{*} = A_{mn}

Would this mean that for a matrix of A, where A is

a b
c d

that

a b
c d

=

a* c*
b* d*

=

A11 A12
A21 A22

=

A11* A21*
A12* A22*

Thanks for the clarification!

 

[raintrek]

And can I also ask why this seems to be a general property of the Hermitian Conjugate?

(AB)^{+} = B^{+} A^{+}

rather than

(AB)^{+} = A^{+} B^{+}

 

[malawi_glenn]

for your first post, you have done correct.

a b
c d

becomes

a* c*
b* d*

when you do hermitian conjugate of it.

And
(AB)^{\dagger} = B^{\dagger} A^{\dagger}

Follows from
(AB)^{T} = B^{T} A^{T}

Very easy to prove

 

[HallsofIvy]

As for
(AB)^{\dagger} = B^{\dagger} A^{\dagger}
and
(AB)^{T} = B^{T} A^{T}

remember that multiplication of matrices is NOT commutative.
With (AB)^{T} = B^{T} A^{T} we have (AB)^T(AB)= (A^T)(B^T B)(A)= A^T A= I. If we tried, instead, (A^TB^T)(AB) we would have (A^T)(B^T A)(B) and we can't do anything with that.