I'm very confused about how to approach this question. The question is
If $\omega\in\Lambda^n(V)$ is the volume element determined by $T$ and $\mu$, and $w_1,\dots,w_n\in V$, show that $$|\omega(w_1,\dots,w_n)|=\sqrt{\det(g_{ij})}$$ where $g_{ij}=T(w_i,w_j)$.
Hint: If $v_1,\dots,v_n$ is an orthonormal basis and $w_i=\sum_{j=1}^{n}a_{ij}v_j$, show that $g_{ij}=\sum_{k=1}^{n}a_{ik}a_{kj}$.
Now there are two parts to my confusion.
The first part is, by definition of the inner product, isn't the most general form of $T(w_i,w_j)=\sum_{k=1}^{n}a_{ik}a_{jk}$, essentially $Av\cdot A^Tv$, so the order of the indices would be opposite of what he gives?
The second part is, how does that inner product factor into finding the size of the volume element? I've considered taking the tensor product of the volume element with itself, to try and use the definition $|x|=\sqrt{x^2}$, but this seems to be a dead end. My best guess to how this factors into the proof is that he then uses Gram-Schmidt to produce an orthonormal basis using $w_i$, but this still doesn't explain how the square root gets there.
Could I get a hint towards how to solve this problem (I still want to give this a go, I just think I'm going in the wrong direction)? I think I've exhausted all of my ideas, but none of them seem to actually use the algebraic ideas w.r.t. tensors that he has brought up in the chapter.
Edit: Some other things I noticed about the question are that the inner product only seems to hold if $A$ (for $w_i=Av$) is symmetric ($A=A^T$), which would essentially mean $|\omega(w_1,\dots,w_n)|=\sqrt{\det(A)^2}$, which is easily found from an earlier result from the book.
