Finding coefficients using dual basis

[Supreme_BiH]

Let V be the space of polynomials of degree 3 or less over \Re. For every \lambda\in\Re the evaluation at \lambda is the map ev_{\lambda} such that V \rightarrow \Re is linear. How do we find the coefficients of ev_{2} in the basis dual to \{1,x,x^2,x^3\}?

 

[Rasalhague]

Suppose you have any finite-dimentional vector space, V, with dual space V^*. Suppose, for the sake of definiteness, it's 4-dimensional, like the one in your example. Let \mathbf{e}_\nu be the nu'th vector of some basis for V. Let \pmb{\varepsilon}_\mu be the mu'th vector of the dual basis for V^* corresponding to the basis { \mathbf{e}_\nu }. Let v be an arbitrary vector of V, and let v subscript nu be the nu'th coefficient of v in this basis. Let alpha be an arbitrary linear map from V to its field (In your case, that's the real numbers.), and let alpha subscript mu be the mu'th coefficient of alpha in this dual basis. Then

\pmb{\alpha}(\mathbf{v})= \sum_{\mu=0}^{3} \alpha_\mu \pmb{\varepsilon}_\mu(\mathbf{v}) = \sum_{\mu=0}^{3} \alpha_\mu \pmb{\varepsilon}_\mu \left ( \sum_{\nu=0}^{3}v_\nu \mathbf{e}_\nu \right ) = \sum_{\mu=0}^{1} \alpha_\mu v_\mu,

since, by the definition of a dual basis,

\pmb{\varepsilon}_\mu(\mathbf{e}_\nu)= \delta_{\mu\nu},

where the expression on the right is Kronecker's delta, equal to 1 if mu = nu, and 0 otherwise.

Notice what happens if v is one of your chosen basis vectors.

 

[Supreme_BiH]

thank you.