How to show $ tr(AB^\intercal)=\sum\limits_{i=1}^n\sum\limits_{j=1}^n a_{ij}b_{ij}$

Question

Show that for any two matrixes $A = a_{ij}$ with $1\leqi,j\leq n , B = b_{ij}$ with $1\leqi,j\leq n$.

$tr(AB^\intercal)=\sum\limits_{i=1}^n\sum\limits_{j=1}^n a_{ij}b_{ij}$

I'm confused as to how to prove this? Why wouldn't the $b_{ij}$ be $b_{ji}$ if it is the transpose? Other than maybe the idea that since its the sum of the diagonals transpose doesn't change this?

Answer 1

• $(i,j)$-th entry of $AB^T$ is $\sum_{k=1}^n a_{ik}b_{jk}$

• $(i,i)$-th entry of $AB^T$ is $\sum_{k=1}^n a_{ik}b_{ik}=\sum_{j=1}^n a_{ij}b_{ij}$.

I will leave the last step to you.