Question

proof: if three matrices \[A_{n \times 1}, B_{n \times n} , C_{n \times 1} \], then
\[A^TBC=\sum_{i=1}^{n}B_{ij}[AC^T]_{ji}\]

Answer 1

proof: if three matrices \[A_{n \times 1}, B_{n \times n} , C_{n \times 1} \], then
\[A^TBC=\sum_{i=1}^{n}B_{ij}[AC^T]_{ji}\]

Answer 2

@ganeshie8 @TuringTest @iambatman @Abhisar @iGreen

Answer 3

The left hand side is a scalar, but the right hand side B and AC' are nxn.
and the indices of AC' should probably be ij (instead of ji).
Would the statement be:
\(A^TBC=\sum_{i=1}^n\sum_{j=1}^{n}B_{ij}[AC^T]_{ij}\)
or simply
\(A^TBC=B_{ij}[AC^T]_{ij}\)
using repeated indices

Answer 4

\[A ^{t}BC=a _{i}*b _{ij}*c _{j}=b _{ij}*(a _{i}*c _{j})=\]
\[=b _{ij}*(A ^{t} C)_{ij}=b _{ij}*(C ^{t}A)_{ji}=B(C ^{t}A)\]

Answer 5

where I used the summation rule on repeated indices