transpose

Conjugate transpose

The conjugate transpose is the transpose of a matrix with each element being the complex conjugate:

$$\begin{array}{r}{A}^{\ast }=(\bar{A}{)}^T=\overline{A^T}\end{array}$$

For real-valued matrices, this is just the transpose:

$$A_{ij}^T=A_{ji}$$

This has nothing to do with adjunct/adjugate matrix — which annoyingly is also sometimes called “adjoint”.

The conjugate transpose is the real adjoint, i.e. this operator applied to a matrix: https://en.wikipedia.org/wiki/Hermitian_adjoint

$(AB{)}^T=B^T{A}^T[⟹\forall A\in {\mathbb{R}}^{n\times n}:(A{A}^T{)}^T=A{A}^T]$

Let $m,n,p\in \mathbb{N},A\in {\mathbb{R}}^{m\times p}$, and $B\in {\mathbb{R}}^{p\times n}$. Then,

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

Note the order of multiplication reverses.

Proof:

$$(AB{)}_{ij}^T=(AB{)}_{ji}=\sum _k^p{A}_{jk}{B}_{ki}=\sum _k^p{B}_{ik}^T{A}_{kj}^T=(B^T{A}^T{)}_{ij}$$

$(A^{-1}{)}^T{A}^T=I_n⟹(A^{-1}{)}^T=(A^T{)}^{-1}$

Transpose of the inverse ($A{A}^{-1}=I_n$):
$⟹(A^{-1}{)}^T{A}^T=(A{A}^{-1}{)}^T=I_n^T=I_n$
$⟹(A^{-1}{)}^T=(A^T{)}^{-1}$

Sometimes denoted as $A^{-T}$