# Hermitian Conjugate

1. Dec 29, 2007

### 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!

2. Dec 29, 2007

### raintrek

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

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

rather than

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

3. Dec 29, 2007

### 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

4. Dec 29, 2007

### HallsofIvy

Staff Emeritus
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.