Help with proving that the transpose of the product of any number of matrices is equal to the product of their transposes in reverse

Solution 1:

The fact that $(AB)^T=B^TA^T$ follows from the formulas $$ \begin{align} (ab)^T_{ki} &=ab_{ik}\\ &=\sum_{j=1}^na_{ij}b_{jk}\tag{1} \end{align} $$ and $$ \begin{align} (b^Ta^T)_{ki} &=\sum_{j=1}^nb^T_{kj}a^T_{ji}\\ &=\sum_{j=1}^nb_{jk}a_{ij}\\ &=\sum_{j=1}^na_{ij}b_{jk}\tag{2} \end{align} $$ To extend this to more than two matrices, use induction:

Suppose that for some $n$, we have $$ (A_1A_2\dots A_n)^T=A_n^T\dots A_2^TA_1^T\tag{3} $$ Note that we have already shown $(3)$ for $n=2$ using $(1)$ and $(2)$.

Then, using the two matrix result and $(3)$, we have $$ \begin{align} (A_1A_2\dots A_nA_{n+1})^T &=((A_1A_2\dots A_n)A_{n+1})^T\\ &=A_{n+1}^T(A_1A_2\dots A_n)^T\\ &=A_{n+1}^TA_n^T\dots A_2^TA_1^T\tag{4} \end{align} $$ Thus, the $n$ matrix result and the two matrix result imply the $n+1$ matrix result. Therefore, $(3)$ is true for two or more matrices.