- Considerations on Transpose Matrices

1. $[A^T]_{B^*} = ([A]_B)^T$ #

In a vector space $V$, the matrix $A$ satisfies the following relation. That is, $$ \sub{e’}{i} = Ae_i = A_{i}^j e_j $$ and this corresponds to $$ e^{i} = \sub{A}{j}^i e’^{\ j} (= A^{T} e’^{\ i}) $$ in the dual space $V^{\ast}$.

Since we can recognize that $A_{j}^{i}$ is an element of the transpose matrix where the rows and columns of $A_{i}^{j}$ are interchanged, we can rewrite it as $$ A_{j}^{i} e’^{\ j} = \sum_j (A^T)_i^{j} e’^{\ j} $$ which corresponds to $A^T e’^{\ i}$. That is, we can express it as $$ A^T e’^{\ i} = \sum_j (A^T)_i^{j} e’^{\ j} $$

When expressing this rigorously in matrix notation with respect to the basis, $$ [A^T]_{B^{\ast}} = ([A]_B)^T $$ holds true.

2. $(A^T f)(v) = f(Av)$ #

Meanwhile, $$ (A^T f)(v) = (A^T f_i e^i)(v) = (f_i A^T e^i)(v) $$

$$ = \left(f_i \sum_j (A^T)_i^j e^j\right)(v) = (f_i A_j^i e^j)(v) $$

$$ = (f_i A_j^i e^j)(v^k e_k) = f_i v^k A_j^i \delta_k^j $$

$$ = f_i v^k A_k^i = f_i A_k^i v^k = f_i (Av)^i = f(Av) $$

Through this process, we can see that $$ (A^T f)(v) = f(Av) $$ holds true, which corresponds to $$ (\sub{[A^T]}{B^{\ast}} \sub{[f]}{B^{\ast}})( \sub{[v]}{B} ) = \sub{[f]}{B^{\ast}} (\sub{[A]}{B} \sub{[v]}{B}) $$