Transpose of a matrix

[nicoc]

I need help to demonstrate this equality. I dont know how to start. Thanks in advance.

PD: A and B are conformable

 

[ksdhart2]

The greatest trick in all of algebra is that if we don't know some value, we can give it a name so we can talk about it and work with it. That applies to variables as well as matrices. Let $A$ and $B$ be the most generic matrices possible:

$\displaystyle A = \begin{bmatrix} A_1 & A_2 \\ A_3 & A_4 \end{bmatrix}, \: B = \begin{bmatrix} B_1 & B_2 \\ B_3 & B_4 \end{bmatrix}$

Then do the matrix multiplication in the usual way:

$\displaystyle AB = \begin{bmatrix} A_1 & A_2 \\ A_3 & A_4 \end{bmatrix} \begin{bmatrix} B_1 & B_2 \\ B_3 & B_4 \end{bmatrix} = \begin{bmatrix} \text{???} & \text{???} \\ \text{???} & \text{???} \end{bmatrix}$

What do you get when you transpose this result? And how does that compare with the product of the two individual matrices transposed? Be careful to note that the order of the matrices swapped, as matrix multiplication is not commutative.

 

[Romsek]

I'd write out $\displaystyle (AB)^T_{i,j}$ and do a bit of rearranging of indices by noting for example $\displaystyle A^T_{i,j}=A_{j,k}$

and see where that takes you


$\displaystyle (AB)_{i,j} = \sum \limits_k A_{i,k}B_{k,j}$
\(\displaystyle (AB)^T_{i,j} = (AB)_{j,i} =

\sum \limits_k A_{j,k}B_{k,i} =

\sum \limits_k B_{k,i}A_{j,k}=

\sum \limits_k B^T_{i,k}A^T_{k,j} =

B^TA^T
\)

 

[Steven G]

As Romsek noted, just use the definition of matrix multiplication

 

[Romsek]

I'd write out $\displaystyle (AB)^T_{i,j}$ and do a bit of rearranging of indices by noting for example $\displaystyle A^T_{i,j}=A_{j,k}$

Small error above. The final equation should be
$\displaystyle A^T_{i,j}=A_{j,i}$