<s>[INST] Is this a theorem (is it correct?)? My instruction notes have specified a theorem of matrix transpose that be there two compatible matrices $A$ and $B$ in respect of their sums and products, then:
$(AB)^T =A^TB^T$
So I set on to verify if indeed this is correct:
Let $A = \begin{pmatrix}2 & 2 \\ 6 & 4\end{pmatrix}  $
Let $B = \begin{pmatrix}1 & 0 \\ 7 & 3\end{pmatrix}  $
$A^T=\begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}  $  
$B^T=\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}  $  
$(AB) = \begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}$=$\begin{pmatrix}16 & 6 \\ 34 & 12\end{pmatrix}  $ 
hence $(AB)^T=\begin{pmatrix}16 & 34 \\ 6 & 12\end{pmatrix}$
and....
$A^TB^T=\begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}=\begin{pmatrix}2 & 32 \\ 2 & 26\end{pmatrix}$
$\therefore (AB)^T\neq A^TB^T $
Is the theorem wrong or am having a rough time understanding?
 [/INST] The assertion is false; what is true is that
$$
(AB)^T=B^TA^T
$$
which can be easily proved. Actually, the product $A^TB^T$ need not be defined when the product $AB$ is defined, so the equality doesn't make sense unless both matrices are square. However, the equality doesn't generally hold in this case.
Let's see what happens if you also have, for two square matrices, $(AB)^T=A^TB^T$:
\begin{gather}
A^TB^T=B^TA^T\\
(A^TB^T)^T=(B^TA^T)^T\\
(B^T)^T(A^T)^T=(A^T)^T(B^T)^T\\
BA=AB
\end{gather}
Conversely, it's obvious that from $AB=BA$ you can deduce $(AB)^T=A^TB^T$.
So the equality you read in your instructions is wrong.
</s>