 Solution

A centre $c$ turns position into a displacement $x-c$. If $v_1,\dots ,v_N$ are a basis of the $N$-dimensional vector space, every position has unique coordinatesThus $N$ 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.