# Spanning vectors of simple cubic lattice

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

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

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.