Coordinates are functionals in disguise
Fix a basis e1, ..., en of V. Every vector has a unique coordinate vector: v = c1·e1 + ... + cn·en. Now ask: what kind of object is 'the i-th coordinate'? The map e^i that sends v to ci is linear — so it is a linear functional, an element of V*. Reading off a coordinate IS a measurement.
These n functionals e^1, ..., e^n are the dual basis. They are pinned down by one clean rule: e^i(ej) = 1 if i = j, else 0. In words: the i-th measurement reads 1 on the i-th basis vector and ignores the rest. The superscripts (covectors) vs subscripts (vectors) bookkeeping is the heart of covariance and contravariance.
Same dimension, but no canonical link
Because the dual basis has exactly n members, V* has the same dimension as V (when V is finite-dimensional). So V and V* are isomorphic as abstract spaces. But here is the subtlety: the isomorphism v_i -> e^i depends on the basis you chose. Pick a different basis and the correspondence changes. There is no natural, basis-free isomorphism V -> V*. Hold that thought — it returns powerfully in Guide 5.
Computing a dual basis
Basis of R^2 (non-standard):
v1 = [1; 1], v2 = [1; -1]
Want covectors f1, f2 (rows) with f_i(v_j) = delta_ij.
Stack v1,v2 as columns: B = [1 1; 1 -1]
Dual basis rows are the ROWS of B^-1.
B^-1 = (1/-2) [ -1 -1; -1 1 ]
= [ 1/2 1/2 ; 1/2 -1/2 ]
So f1 = [1/2 1/2], f2 = [1/2 -1/2]
Check: f1(v1) = 1/2 + 1/2 = 1, f1(v2) = 1/2 - 1/2 = 0 ✓Notice the inverse appearing: when basis vectors transform by a matrix B, their dual functionals transform by B^-1 (transposed). Vectors and covectors react to a basis change in opposite ways — that is exactly why physicists distinguish covariant and contravariant indices.