Proof of Identity for Transpose of a Product of Matrices
If
\[A\]
and \[B\]
are matrices that can be multiplied, then \[(AB)^{T} = B^T A^T\]
Proof
Let
\[A=(a_{ij})\]
be an \[n \times m\]
matrix and \[B=(b_{ij})\]
be an \[n \times r\]
matrix.Then
\[C=(c_{ij})\]
where\[c_{ij} = a_{i1}b_{1j}+ a_{i2}b_{2j}+ ...+ a_{in}b_{nj} =\sum_{k=1}^{k=m} a_{ik}b_{kj} \]
\[c_{ij}\]
is the element in the \[i\]
^{th} row and \[j\]
^{th} column of \[C\]
and the element in the \[j\]
^{th} row and \[i\]
^{th} column of \[C^T =(AB)^T\]
.The elements
\[b_{1j}, \: b_{2j} ,..., \: b_{mj}\]
form the \[j\]
^{th} column of \[B\]
and the \[j\]
^{th} row of \[B^T\]
.The elements
\[a_{i1}, \: a_{i2} ,..., \: a_{in}\]
form the \[i\]
^{th} row of \[A\]
and the \[i\]
^{th} column of \[A^T\]
.Thus the element in the
\[j\]
^{th} row and \[i\]
^{th} column of \[B^T A^T\]
is\[b_{1j} a_{i1} +b_{2j} a_{i2}+...+ b_{mj} a_{m1}= \sum_{k=1}^{k=m} a_{ik}b_{kj}\]
.Hence
\[(AB)^{T} = B^T A^T\]
