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**Isomorphism** If transpose was defined to you as $\displaystyle [A^T]_{ij} = [A]_{ji}$, then "summation sign" method as explained in the prev. post is most likely the only thing you could do.

However if you are doing linear algebra, and you are familiar with the notion of a linear functional $\displaystyle [x,y]$(Halmos' notation), then one could define $\displaystyle A'$, adjoint of linear transformation A as $\displaystyle [Ax,y] = [x,A'y]$.If $\displaystyle [A]_{\cal{X}}$ is the matrix of linear transformation A w.r.t the basis $\displaystyle \cal{X}$, then its easy to verify that the matrix of linear transformation A' is $\displaystyle [A]^T_{\cal{X}}$.

Now with all this setup, its extremely easy to prove $\displaystyle (AB)^T = B^TA^T$. Note that it suffices to prove $\displaystyle (AB)' = B'A'$.

But $\displaystyle \forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]$

Thus $\displaystyle (AB)' = B'A'$