= Solution
A centre $c$ turns position into a displacement $x-c$. If $v_1,\ldots,v_N$ are a basis of the $N$-dimensional <vector space>, every position has unique coordinates
$$
\boxed{x=c+\sum_{j=1}^N x_jv_j.}
$$
Thus \b[$N$ linearly independent vectors are necessary and sufficient]. Fewer vectors span a proper subspace and cannot represent arbitrary displacements. The coordinates use the field defining the space. For qubit tomography, the relevant parameter space is the real three-dimensional space of <Bloch vectors>, rather than the complex two-dimensional space of state amplitudes.
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