Finding dual basis to some given basis Can someone briefly explain me how to find dual basis to some given basis?
Let's say we've got $n$-dimensional linear space with given basis $(e_1,e_2,...,e_n)$. How to find $(e_1^*, e_2^*, ...., e_n^*)$?
 A: Put the vectors $e_i$ in the columns of some matrix $A$. If you have a matrix $B$ having as columns the vectors $e_i^*$, then $B^*A=I$, where $B^*$ is the transpose of $B$. Hence, $B=(A^{-1})^*$.
So, the elements of the dual basis are the columns of the matrix $(A^{-1})^*$.
A: Let $V$ be a vector space over a field $k$ with basis $e_1, ...  e_n$.  The dual $V^{\ast}$ is defined to be the vector space consisting of all linear transformations from $V$ to $k$.  The definitions of addition and scalar multiplication in $V^{\ast}$ are obvious, as is the fact that $V^{\ast}$ is actually a vector space.
Every element in $V$ can be uniquely written as a sum $c_1e_1 + \cdots + c_ne_n$, where $c_i$ are in $k$.  For each $j$, define a function $e_j^{\ast}: V \rightarrow k$ by the formula $$e_j^{\ast}(c_1e_1 + \cdots + e_nv_n) = c_j$$  You can check that for any $x, y \in V$, and any $c \in k$, that $e_j^{\ast}(x+y) = e_j^{\ast}(x) + e_j^{\ast}(y)$, and $e_j^{\ast}(cx) = c e_j^{\ast}(x)$.  In other words, $e_j^{\ast}$ is a linear transformation from $V$ to $k$, so $e_j^{\ast} \in V^{\ast}$.
The basic result is that the elements $e_i^{\ast}$ form a basis for the vector space $V^{\ast}$.  Is this what you're having trouble proving?
A: Okej, so let's suppose we've got $([1,1,1],[1,1,-1],[1,-1,-1])$, some basis of $\mathbb{R}^3$. That's how I can generate dual basis:
Let
\begin{bmatrix}
1 &1  &1 \\ 
1 &1  &-1 \\ 
1 &-1  &-1 
\end{bmatrix}
be the $A$ matrix. I have to find $A^{-1}$, which is
\begin{bmatrix}
0,5 &0  &0,5 \\ 
0 &0,5  &-0,5 \\ 
0,5 &-0,5  &0 
\end{bmatrix}
And that's all. Dual basis is $((\frac{1}{2}x^2+\frac{1}{2}), (\frac{1}{2}x -\frac{1}{2}), (\frac{1}{2}x^2-\frac{1}{2}x))$
Right? And I cannot avoid calculation $A^{-1}$, right?
