Dual basis

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space $V$, its dual basis consists of linear functionals $e_1^{\ast },\dots ,e_n^{\ast }$ satisfyingThe Kronecker delta means that each rule gives one on its own basis vector and zero on the others. These rules extract coordinates:Applying $e_i^{\ast }$ to a linear combination removes every term except the $i$th. Consequently the dual basis is a basis of the dual space, and every linear functional has expansion $\phi ={\sum }_i\phi (e_i)e_i^{\ast }$. ^[ref].

Coordinate extraction works for a nonorthogonal basis too. With $b_1=(1,0)$ and $b_2=(1,1)$, the dual basis isThe checks $b_1^{\ast }(b_1)=1$, $b_1^{\ast }(b_2)=0$, $b_2^{\ast }(b_1)=0$, and $b_2^{\ast }(b_2)=1$ fix the two rules. For $v=(3,1)$ they give coordinates $(2,1)$, so $v=2b_1+b_2$. The usual dot products with $b_1,b_2$ instead give $3,4$ and do not extract these coordinates: simple projection formulas depend on an orthonormal basis.

The same idea explains polynomial interpolation. On polynomials of polynomial degree at most $n$, evaluation at distinct interpolation nodes $x_0,\dots ,x_n$ gives the dual basis to the Lagrange cardinal polynomials, because ${\ell }_i(x_j)={\delta }_{ij}$. Samples are coordinates in this basis, so the reconstruction $p={\sum }_{i}p(x_i){\ell }_i$ is a dual-basis expansion. Differentiation weights from nodal evaluations and factorial monomial basis for polynomial differentiation apply the same principle to different measurements.