How the dual basis is used to transform a vector to a scalar?

Question

Suppose we have a vector space V with basis $e_1 = (1,1,2)$, $e_2 = (1,0,1)$ and $e_3 = (2,1,0)$ on $\mathbb R^3$ over $\mathbb R$. Its dual basis is found to be $e_1' = (−1/3,2/3,1/3)$, $e_2' = (2/3,−4/3,1/3)$, $e_3' = (1/3,1/3,−1/3)$.

Now, given a vector from $V$, how does this dual basis help in finding the mapping $V\rightarrow F$? That is, how is the operation that takes a vector $v$ from $V$ and turn it into a number? Or is this only a basis for linear functionals that will turn a vector from $V$ to a scalar? How do I construct the linear functionals out of this dual basis?

I'm not sure if this is a right question to ask, but I guess I have not understood some key aspect and I hope this question brings that out.

Answer 1

$e^i(e_j)=\delta _i^j$, where $\delta _i^j=\begin{cases} 1, i=j\\ 0, i\not =j\end{cases}$ is the Kronecker delta.

Use this fact and linearity to evaluate...

Example: $(-\frac13,\frac23,\frac13)(1,1,2)=-\frac13\cdot 1+\frac23\cdot 1+\frac13\cdot 2=-\frac13 +\frac23+\frac23=\frac13+\frac23=1$

Similarly, $(-\frac13,\frac23,\frac13)(1,0,1)=0$, and $(-\frac13,\frac23,\frac13)(2,1,0)=0$.

In general, $(-\frac13,\frac23,\frac13)(a,b,c)=-\frac13 a+\frac23b+\frac13 c$.

Any linear functional $\phi$ in $V^*$ will be given by a linear combination of the dual basis: $\phi= a_1e^1+\dots+a_ne^n$.

The dual basis is a basis for the dual space. Thus $V$ and $V^*$ have the same dimension (when $V$ is finite dimensional).

Answer 2

The $e'_i$ triples are actually the coordinates of the vectors of the dual basis. If I understand you right, $V=\mathbb R^3$, and so the canonical basis (that is, the dual basis of $\{(1,0,0),(0,1,0),(0,0,1)\}\subset \mathbb R^3$, is $\{\phi_1,\phi_2,\phi_3\}$, where $$\phi_1(x_1,x_2,x_3)=x_1,$$ $$\phi_2(x_1,x_2,x_3)=x_2,$$ $$\phi_3(x_1,x_2,x_3)=x_3.$$ So if your $e'_i$ are coordinates in this canonical base, you have $$e'_1=-\tfrac13\cdot \phi_1+\tfrac23\cdot\phi_2+\tfrac13\cdot\phi_3,$$ and so on for $e'_2$ and $e'3$.

So, for instance, $$e'_1(x_1,x_2,x_3)=-\tfrac13\cdot x_1+\tfrac23\cdot x_2+\tfrac13\cdot x_3,$$ and if you wanted—just to say something—$e'_1(2,-1,0)$, that would be $$e'_1(2,-1,0)=-\tfrac13\cdot 2+\tfrac23\cdot (-1)+\tfrac13\cdot 0=-\tfrac 43.$$