TITLE: How to find Dual Basis
QUESTION [1 upvotes]: I think that I have a gap in my understanding on how to find a dual basis to a given basis. So gven a 3-dimensional vector space $V$ with base $B :=\{b_1,b_2,b_3\}$. We have to find $B^* := {b^*_1,b^*_2,b^*_3}$ such that: $b^*_i(b_j) = \delta_{ij}$. Can't we just construct the basis functions of $B^*$ without explicity given them, since we know that for each vector $v$ of $V$ we can write them as a linear combination using $\{b_1,b_2,b_3\}$. Now define the function: $b^{*}_{j}(p) := \sum_{1\leq i\leq 3}\lambda_{i}^{(v)} \delta_{ij}$ for $j=1,2,3$, where $\lambda_i^v$ represent the coordintes of $v$. Then follows $b^*_i(b_j) = \delta_{ij}$. So what's the difference between this and explicitly giving the basis and how would you use this to calculate it. Thanks for any help in advance

REPLY [0 votes]: I don't know if I completely understand this question .
You say

We have to find $B^∗:=b^*_1,b^*_2,b^*_3$ such that: $b^*_i(b_j)=\delta_{ij}$

This means you have to define a functions $b^*_i : V \rightarrow \mathbb{R}$. This means you need to define a process by which you would you calculate the output of each of these functions. You've already described such a process. As you say, we define $b^*_i(v)$ as the coefficient of $b_i$ when $v$ is represented as a linear combination of $b_1,b_2$ and $b_3$. This function you describe will satisfy the necessary conditions that $b_i^*(b_j) =\delta_{ij}$.

So what's the difference between this and explicitly giving the basis and how would you use this to calculate it

I'm not sure what you mean by "explicitly" giving the basis. You already pretty much explicitly described the basis. You've basically given an algorithm for computing the output for a given input. You can't get more explicit than that .
{"set_name": "stack_exchange", "score": 1, "question_id": 4104759}