# Kronecker delta indexing rules: How the left side indices are calculated?

Excerpt from Linear Algebra Done Right, 3rd Ed, Page 110:

3.114 The matrix of T' is the transpose of the matrix of T

Suppose T $\in$ L(V,W). Then M (T') = $(M(T))^t$.

Proof Let A = M(T) and C = M(T'). Suppose 1 $\leq$ j $\leq$ m and 1 $\leq$ k $\leq$ n.

From the definition of M(T') we have T'($\psi_j$) = $\sum_{r=1}^n$$C_{r,j}$$\phi_r$

The left side of the equation above equals $\psi \circ T$. Thus applying both sides of the equation above to $v_k$ gives ($\psi_j$ $\circ$ T)($v_k$) = $\sum_{r=1}^n$$C_{r,j}$$\phi_r$($v_k$)

We also have

($\psi_j$ $\circ$ T)($v_k$) = $\psi_j$(T$v_k$)

($\psi_j$ $\circ$ T)($v_k$) = $\psi_j$)( $\sum_{r=1}^n$$A_{r,k}$$w_r$)

($\psi_j$ $\circ$ T)($v_k$) = ( $\sum_{r=1}^n$$A_{r,k}$$\psi_j$($w_r$))

($\psi_j$ $\circ$ T)($v_k$) = $A_{j,k}$

I could not understand how finally we arrived at the transpose. I checked Hoffman & Kunze (page 115) and could see that the Kronecker delta gets used in the last step.

= $\sum_{k=1}^m$$A_{k,i}$$\delta_{j,k}$ =$A_{ji}$

I am expecting $\sum_{k=1}^m$$A_{k,i}$$\delta_{j,k}$ to replace elements of A to produce the transpose. That is, $A_{2,3}$ should get replaced by $A_{3,2}$. Had some misconception about the indices and later found out the index which appears more than once is to be summed-up. For the below expansion, I choose i =2 and j =3.

$A_{k,2}$$\delta_{3,k} = A_{1,2}$$\delta_{3,1}$ + $A_{2,2}$$\delta_{3,2} + A_{3,2}$$\delta_{3,3}$

So, that gives $A_{3,2}$

My question is, how does this decide to which position this goes to? Of course it goes into $A_{2,3}$, but from the notation I am finding it difficult to understand. We just chose i = 2 and j = 3, but did not say which one is row index and which one is column index. How does this get set?

$$\sum_{k=1}^m A_{ki}\delta_{jk}= \sum_{\substack{k=1\\k\neq j}} A_{ki}\delta_{jk}+ A_{ji}\delta_{jj}=\sum_{\substack{k=1\\k\neq j}} A_{ki} \, 0+A_{ji} \, 1=A_{ji}$$