# Consider the basis e_1 = (−2,4,−1), e_2 = (−1,3,−1) and e_3 = (1,−2,1) of R^3 over R. Find the dual basis of {e_1, e_2, e_3}.?

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Feb 11, 2018

See below.

#### Explanation:

Calling ${e}^{i}$ the dual basis of ${e}_{j}$ we have

$\left\langle{e}^{i} , {e}_{j}\right\rangle = {\delta}_{j}^{i}$ or

$\left({e}^{1} , {e}^{2} , {e}^{3}\right) \left(\begin{matrix}{e}_{1} \\ {e}_{2} \\ {e}_{3}\end{matrix}\right) = {I}_{3}$ then if

$M = \left(\begin{matrix}{e}_{1} \\ {e}_{2} \\ {e}_{3}\end{matrix}\right) = \left(\begin{matrix}- 2 & 4 & - 1 \\ - 1 & 3 & - 1 \\ 1 & - 2 & 1\end{matrix}\right)$ then

${M}^{-} 1 = \left({e}^{1} , {e}^{2} , {e}^{3}\right) = \left(\begin{matrix}- 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 2\end{matrix}\right)$

or

${e}^{1} = {\left(- 1 , 0 , 1\right)}^{\top}$
${e}^{2} = {\left(2 , 1 , 0\right)}^{\top}$
${e}^{3} = {\left(1 , 1 , 2\right)}^{\top}$

form the dual basis.

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