This question is related to Computing the volume of a fundamental domain of a lattice. I have been trying to understand similar concepts as in this question, but still unclear about couple points.
Let $v_1, ..., v_k$ be linearly independent vectors in $\mathbb{R}^n$ and $k < n$. Let $\sigma$ be the parallelpiped formed by these $k$ vectors inside $\mathbb{R}^n$. Then as described in one of the answers $$ vol(\sigma)=\sqrt{\mid \det A A^T\mid} $$ with matrix $A = [v_1 ... v_k]$.
Question 1. What does the volume mean precisely in this context? It makes sense to me when $k = n$, it is the Lebesgue measure on $\mathbb{R}^n,$ but was not sure when $k < n$. Any clarification is appreciated!
Question 2. I was wondering if someone could explain how the two answers in Computing the volume of a fundamental domain of a lattice are the same?
I'm just going to answer the first question and say why the formula should be clear.
Question 1:
Let's say you have a $k$-dimensional real vector space $V$. In order to define a Lebesgue measure on $V$, it's enough to select a basis. You transport Lebesgue measure on $R^k$ to $V$ via the coordinate map $V \to R^k$ associated with that basis.
Now if you replace the basis you have with a second one, the measure on $V$ will remain unchanged as long as the transition matrix between the two bases has determinant $\pm 1$. So in order to define a measure on $V$, instead of singling out one basis, it's enough to single out a class of bases which are all related to each other in this way.
Returning now to your problem, let $V$ be the subspace of $R^n$ spanned by $v_1, \dots, v_k$. It so happens that we can distinguish such a class of bases of $V$, namely, all bases which are orthonormal with respect to the standard inner product of $R^n$. Since the transition matrices between these bases are orthogonal, they have determinant $\pm 1$, as required. This defines the measure on $V$.
Proof of the formula:
The first answer given in the link is, I think, not much more helpful than what I wrote above, so I'll just say why the formula in the second answer should be obvious.
I imagine you already agree with the formula $(\det AA^T)^{1/2}$ in the case $k = n$, as this is just $|\det A|$. Clearly, $AA^T = (<v_i,v_j>)_{i,j}$ (this equality also being valid when $k < n$). Thus when $k = n$, we have an expression for the volume of the parallelepiped in terms of the inner products of the spanning vectors.
But as I explained in answering Question 1, in the case $k < n$ the concept of volume in $V$ depends only on the inner product induced on $V$ by the one on $R^n$. So the very same formula for the volume of the parallelipiped must be valid in the case of $V$.