Proving $(AB)^T=B^TA^T$: Can you find the flaw in the reasoning?

Question

Proving $(AB)^T=B^TA^T$ isn't difficult. An example of such a proof can be found here: http://www.math.ucdenver.edu/~esulliva/LinearAlgebra/ABT.pdf

However, if I take a slightly different approach, it doesn't work:

If $B=(b_{ij})$, then $B^T=(b_{ji})$, and similarly, $A^T=(a_{ji})$. Thus $B^TA^T=(b_{ji})(a_{ji}) = (\sum_k b_{jk}*a_{ki}).$ But the $(i,j) $ element of $(AB)^T$ is $(\sum_k b_{ki}*a_{jk})=(\sum_k a_{jk}*b_{ki})$, which is generally not equal to $(\sum_k b_{jk}*a_{ki})$.

What is the flaw in the reasoning? Any help is appreciated.

Answer 1

One may also address this problem in the context of inner products, which I find simpler since it avoids the necessity of index bookkeeping, viz:

For any matrix/operator $C$, $C^T$ is the unique operator satisfying

$\langle x, Cy \rangle = \langle C^Tx, y \rangle; \tag 1$

then

$\langle x, ABy \rangle = \langle (AB)^T x, y \rangle; \tag 2$

but

$\langle x, ABy \rangle = \langle A^Tx, By \rangle = \langle B^TA^Tx, y \rangle; \tag 3$

combining (2) and (3) yields

$\langle (AB)^T x, y \rangle = \langle x, ABy \rangle = \langle B^TA^Tx, y \rangle, \tag 4$

which holds for all $x$, $y$; thus

$(AB)^T x = B^TA^Tx \tag 5$

binds for every $x$, whence

$(AB)^T = B^TA^T. \tag 6$

$OE\Delta$.

Answer 2

When you write $A^\top = (a_{ji})$, you have to remember that you're writing down the $ij$-entry. So, yes, the $ij$-entry of $B^\top A^\top$ is $\sum (B^\top)_{ik}(A^\top)_{kj} = \sum b_{ki}a_{jk}$. You just wrote symbols and didn't think about actual entries of actual matrices.