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Let $\beta= \{ (2,1),(3,1) \} $ be an ordered basis for $\Bbb R^2$. Suppose that the dual basis of $\beta$ is given by $\beta^*= \{f_1,f_2 \} $ To explicitly determine a formula for $f_1$ we need to consider the equations $$1=f_1(2,1)=f_1(2e_1+e_2)=2f_1(e_1)+f_1(e_2)$$ $$0=f_1(3,1)=f_1(3e_1+e_2)=3f_1(e_1)+f_1(e_2)$$ Solving this equations, we obtain $f_1(e_1)=-1$ and $f_1(e_2)=3$, that is $f_1(x,y)=-x+3y$.

My question is why do we need to solve the above equations for 1 and 0 respectively?

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1 Answer 1

The definition of $\{f_1,f_2\}$ as the dual basis to a basis $\{v_1,v_2\}$ say, is that $f_i$ are the linear maps such that $f_i(v_j) = \delta_{ij}$, extended linearly to all of $V$. In other words, if $V$ is $n$ dimensional, then $f_i(\sum_{j=1}^n \lambda_jv_j) = \lambda_i$. In the case you're given, $v_1 = (2,1)$ and $v_2 = (3,1)$ so $f_1(2,1) =1 $and $f_2(3,1) = 0$.

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Thanks! Didn't considered the delta part. –  aortizmena Jan 2 '13 at 0:54
    
@aortizmena no worries, happy to help! –  Tom Oldfield Jan 2 '13 at 0:55

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