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Let $V$ be the vector space of $P_2[x]$ of polynomials over $\mathbb{R}$ of degree less than or equal to $2$. Let $L_1$,$L_2$, $L_3$ be the linear functions on $F$ defined by $L_1(f)=f(1)$,$L_2(f)=f(2)$, and $L_3(f)=f(3)$.

Show that the span of the $L_i$'s is a basis for $V^*$ (the dual of $V$). Also, show an explicit basis for $V$ whose dual is $\{L_1,L_2,L_3\}.$

I've been thinking about this for a while. I'm not sure how to show the first part of the second question. Also, I'm curious, is the second question simply using Lagrangean interpolation to find 3 polynomials?

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  • $\begingroup$ This doesn't make sense: $L_1(f) = 1$ (the constant function $1$) is not a linear functional and so it isn't even in $V^*$. $\endgroup$ – user296602 Mar 24 '18 at 19:18
  • $\begingroup$ You're right. I forgot that they were evaluating the polynomials at $1,2,3$. $\endgroup$ – Annalena Mar 24 '18 at 19:19
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Denote the canonical basis as $(e) = \{t^2, t, 1\}$ and let's try to find a basis $(b)$ such that $\{L_1, L_2, L_3\}$ is its dual basis.

Writing $L_1, L_2, L_3$ in terms of polynomial coefficients, we get

$$L_1(at^2 + bt + c) = a + b + c$$ $$L_2(at^2 + bt + c) = 4a + 2b + c$$ $$L_3(at^2 + bt + c) = 9a + 3b + c$$

Recall that if $I_{(e, b)}$ is a matrix representing the identity with respect to the bases $(b)$ and $(e)$, then the columns of $I_{(e,b)}$ will be the vectors of the basis $(b)$ written in the basis $(e)$.

The rows of its inverse $I_{(b,e)} = I_{(e, b)}^{-1}$ are precisely the matrix representations of the functionals $L_1, L_2, L_3$ in the canonical basis.

Hence, writing $L_i$ as rows$$I_{(b,e)} = \pmatrix{1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1}$$

and inverting the matrix gives:

$$I_{(e,b)} = \frac12\pmatrix{1 & -2 & 1 \\ -5 & 8 & -3 \\ 6 & -6 & 2}$$

Now we read the vectors of $(b)$ as columns:

$$(b) = \left\{\frac12(t^2 - 5t + 6), \frac12(-2t^2 + 8t - 6), \frac12(t^2 - 3t + 2)\right\}$$

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  • $\begingroup$ What about showing that the $L_i$'s are a basis? Clearly it could be a basis since it has three elements. But I'm not sure how to show the linear independence. $\endgroup$ – Annalena Mar 26 '18 at 3:26
  • $\begingroup$ @Annalena They are a basis since the matrix consisting of columns representing them in the standard basis is invertible. $\endgroup$ – mechanodroid Mar 27 '18 at 6:26

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