# Finding coefficients using dual basis

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 $$\{1,x,x^2,x^3\}$$?

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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.

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thank you.