Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question

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 $f_2(3,1) = 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.