Let $V$ be the space of polynomials of degree at most $2$, with basis $\{1,x,x^2 \}$.Find the dual basis to $\{1,x,x^2\}$.

I know that the dual basis of $\{f_1,f_2,f_3\}$ is the basis $\{f_1^*,f_2^*,f_3^*\}$ where $f_i^*(a_1f_1,\ldots,a_3f_3)$ but I'm not sure how to actually compute the dual basis.

  • $\begingroup$ Wouldn't that mean that $f_1(x)=1$ $f_2(x)=x$ $f_3(x)=x$ ? $\endgroup$
    – lupus nox
    Mar 18 at 12:16

A dual basis functional acting on a polynomial would extract the coefficient at the corresponding monomial. The natural candidate for the dual basis therefore is the linear differential operators $D^k=\tfrac1{k!}\tfrac{d^k} {dx^k}\big|_{x=0}$, $k=0,1,2,..$ -- the familiar formula for the coefficients of the Taylor (Maclaurin) series.


We want the dual basis $\{f_1, f_2, f_3\}$ to satisfy $f_i{x^j}=1$ if $i=j$ and $0$ otherwise. Note that $f_i$ is a linear map taking polynomials of degree less than or equal to two to $\Bbb R$. Since if this is so then:

$f_i(a_0x^0 + a_1x +a_2x^2) = f_i(a_0 x^0) + f_i(a_1x^1) + f_i(a_2x^2) = a_if_i(x^i) = a_i$

The $f_i$ are the unique linear maps satisfying this.


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