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The simplest set of vectors spanning a simple cubic lattice is $\textbf{x}$, $\textbf{y}$, $\textbf{z}$, where they are all mutually perpendicular and of the same length.

I wanted to find another set of vectors that also spans the lattice.

So would $\textbf{x}$, $\textbf{y}$, $\textbf{x+y+z}$ work?

The first two span the $x-y$ plane and the third takes a starting point to the point in the opposite corner of the cube, so it should work.

I wanted to then show that the volume of the primitive cell, given by $V = | \textbf{a} \cdot \textbf{b} \times\textbf{c}|$, remains the same for both sets of vectors, but I don't even know where to begin with this, or if it's even correct.

EDIT: also any other spanning sets would be welcome if you could explain how they generate the lattice

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  • $\begingroup$ Any integer matrix with determinant 1 would do. $\endgroup$ Commented Mar 9, 2021 at 14:26

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Let us consider the matrix $B = [x,y,z]$ and $B' = [x,y,x+y+z]$ where $x,y,z$ are column vectors. As you have mentioned, they span the same lattice. Also, $$B' = B\ \underbrace{\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}}_{U}$$ We know that the volume of the parallelogram defined by vectors in a matrix is equal to the determinant of the matrix. So, we get $det(B')=det(BU) = det(B)det(U) = det(B)$.

To generate other spanning set (or basis) for the lattice, as mentioned by @Ivan Neretin, it is enough to multiply $B$ with any integer matrix $U$ whose determinant is $\pm1$. This is because if $C = BU$, then $B = CU^{-1}$ where $U^{-1}$ is again an integer matrix with determinant $\pm1$. Observe that this exactly the argument you have used to show that $B'$ is also a generating set for the lattice spanned by $B$. These integer matrices with determinant $\pm1$ are called unimodular matrices.

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