Let $P[t]{_2}$ = $V$ a vector space. A basis $B$ = $(1,t,t^2)$ of V and $B$* = ($e_1$,$e_2$,$e_3$) the dual basis of $B$. $f_a$: $V$ $->$ $R$ , $p(t)$ $->$ $p(a)$ (evaluation). Show that f_a, f_b, f_c form a basis of V*. So i found that $f_a$ = $f_a$$(1)$$e_1$ + $f_a$$(t)$$e_2$ + $f_a$$(t^2)$$e_3$ = $e_1$ + $ae_2$ + $a^2$$e_3$
Then the same goes for the evaluation in b and c. In the end i obtain
1)$e_1$ + $ae_2$ + $a^2$$e_3$
2)$e_1$ + $be_2$ + $b^2$$e_3$
3)$e_1$ + $ce_2$ + $c^2$$e_3$
I'm supposed to find constants x,y,z such that x(1) + y(2) + z(3) = 0 -> x = y = z = 0?? So i thought id put it in a matrix and set up a homogenous equation and multiply the 3 equations which would put into a big matrix A by (x y z) column matrix so give (0 0 0) then find that x y z = 0 ? But apparently it just says that if these 3 f_a f_b f_c form a basis then the matrix
$1$ $a$ $a^2$
$1$ $b$ $b^2$
$1$ $c$ $c^2$
would have a non zero determinant. How can we suppose that would mean it forms a basis? maybe im missing an important property or something that is so obvious that i cant see at the moment :P thanks!
It is non zero for two reasons: (i) It is the determinant of a basis : (ii) it is a Vandermonde determinant, and its value is $$\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}=(a-b)(b-c)(c-a)$$ (make the relevant column operations to see that).