Finding coefficients using dual basis

  • #1

Main Question or Discussion Point

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

  • #2
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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 [itex]\mathbf{e}_\nu[/itex] be the nu'th vector of some basis for V. Let [itex]\pmb{\varepsilon}_\mu[/itex] be the mu'th vector of the dual basis for V* corresponding to the basis { [itex]\mathbf{e}_\nu[/itex] }. 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

[tex]\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,[/tex]

since, by the definition of a dual basis,

[tex]\pmb{\varepsilon}_\mu(\mathbf{e}_\nu)= \delta_{\mu\nu},[/tex]

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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  • #3
thank you.
 

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