Volume of a parallelepiped Suppose $\Lambda$ is a lattice in $\mathbb{R}^n$ of rank $r$
and $\mathbf{b}_1, ..., \mathbf{b}_r \subseteq \mathbb{R}^n$ its basis.
I know that if we pick any orthonormal vectors $\mathbf{e}_{r+1}, ....,\mathbf{e}_{n} \subseteq \mathbb{R}^n$
so that $\mathbf{b}_j \cdot \mathbf{e}_k = 0$ for each $j,k$ then
the volume of the parallelepiped is given by
$$
d(\Lambda) = |\det [ \mathbf{b}_1, ..., \mathbf{b}_r, \mathbf{e}_{r+1}, ..., \mathbf{e}_n ]|.
$$
How can I show that this is independent of the choice of $\mathbf{b}_j$
and $\mathbf{e}_k$? Thanks!
 A: Let $B = [b_1 , \ldots , b_r , e_{r+1} , \ldots , e_n]$ and $A = [a_1 , \ldots , a_r ,  e'_{r+1} , \ldots , e'_n]$. Since $b_1 , \ldots , b_r$ and $a_1 , \ldots , a_r$ generated the same lattice, there is a $r \times r$ unimodular transformation, $U$, such that $ [b_1 , \ldots , b_r , e'_{r+1} , \ldots , e'_n] = A[\begin{array}{cc} U & 0 \\ 0 & I \end{array}]$. So we only have to show that the choice of $e_i$ does not change $\det(B)$.
Apply Graham-Schmidt to $b_1 , \ldots , b_r$, which gives $$v_j = 
\sum_{i=1}^r c_{ij} b_j,$$ and the $v_j$ are pairwise orthogonal. Using that all the $v_j$ and $e_i$ are pairwise orthogonal, we get $$
\det(B) = \det([v_1 , \ldots , v_r , e_{r+1} , \ldots , e_n]) = |v_1||v_2| \ldots |v_r| |e_{r+1}| \ldots |e_n| = |v_1||v_2| \ldots |v_r| .$$ This shows that $\det(B)$ is independent of the choice of $e_i$. 
A: Observe that (the absolute value of) the determinant of a square matrix $A$ is equal to the square root of the determinant of $A^tA$. Apply this to your matrix and the result will be a block diagonal matrix with one block equal to the identity matrix and the $b_i$ have disappeared!
