I am starting to learn tensors but i'm already stuck in the first question of my problem set, which says:
Show that the inverse transformation of $$g'_{i} = \sum_{j=1}^{3} a_{ij} g_{j} $$
is $$ g_{i} = \sum_{j=1}^{3} a_{ji} g'_{j}$$
I have tried using the definition of inverse transformation and applying it to both sides of the first equation but didn't manage to arrive at anything useful. I've also tried using the definitions $g'_{i} = \mathbf{ê'_{i}}\cdot \mathbf{g}$ and $\mathbf{g} = \sum_{j=1}^{3} g_{j} \cdot \mathbf{ê_{j}}$ and manipulating them into the equation but it also didn't work.
I assume $a_{ij}$ is an orthogonal transformation, which means
$$\delta_{jk}=\sum_{i=1}^3a_{ij}a_{ik}\tag{1}$$
Using the Einstein summation convention equation $(1)$ is $$\delta_{jk}=a_{ij}a_{ik}\tag{2}$$ Now we wish to show that $$g_i=a_{ji}g_j^\prime\tag{3}$$ is the inverse transformation of $$g_i^\prime=a_{ij}g_j\tag{4}$$ Well, lets multiply both sides of $(4)$ with $a_{ik}$ $$a_{ik}g_i^\prime=a_{ik}a_{ij}g_j=\delta_{kj}g_j=g_k$$ Finally we rename the indices $i\to j$ and $k\to i$ $$g_i=a_{ji}g_j^\prime$$