find the dual basis of $\mathbb R_{\leq2}[x]$ given a basis $B=(x^2-1,x^2-x,x^2+x)$
I know what a dual basis means, however I seem to miss something. here's my attempt:
I need to find a basis $B^\ast=(\phi_1,\phi_2,\phi_3)\, $ such that $\phi_1(x^2-1)=1 \text{ and 0 else}$ and same goes for $\phi_2 \text{ and } \phi_3$
so I wrote $\phi_i(a+bx+c^2) = x_1a+x_2b+x_3c\,$ and applied it to the coordinate vectors of $B$ and I got the following matrix:
$$ \begin{pmatrix} -1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix} $$
and found it's inverse :
$$ \begin{pmatrix} -1 & 1/2 & 1/2 \\ 0 & -1/2 & 1/2 \\ 0 & 1/2 & 1/2 \\ \end{pmatrix} $$
So I concluded that each column of this matrix is the coefficients of $\phi_i$ but then what I actually get is a transformation that doesn't return a scalar but a polynomial. What am I doing wrong?
I'm not just interested in a solution, I guess I have a bad understanding of what's a dual basis , so a clarification would be nice, especially when it comes to polynomial spaces. Thanks ahead.
There’s one more step that you have to perform. When you constructed the original matrix, you computed the coordinate vectors of elements of $B$ relative to the standard ordered basis $(1,x,x^2)$. The columns of the inverse matrix are indeed the coordinates of the elements of $B^*$—they are coefficients of linear combinations of some other set of basis vectors of $V^*$—so you still have to compute the functionals that they represent.
Since you used the standard basis of the dual space to construct the original matrix, you must interpret the elements of the inverse matrix relative to the dual of this basis. This dual basis consists of the functionals $$f_1:p\mapsto p[0] \\ f_2:p\mapsto p'[0] \\ f_3:p\mapsto \frac12p''[0].$$ Putting this more succinctly, $f_i$ returns the coefficient of $x^{i-1}$ in the polynomial. (Verify that this is indeed the dual to the standard basis of $V$.) Applying the coefficients from the inverse matrix, we therefore have $$\phi_1= -f_1\\\phi_2=\frac12(f_1-f_2+f_3)\\\phi_3=\frac12(f_1+f_2+f_3).$$ I’ll leave it to you to verify that these produce the correct values when applied to the elements of $B$.