Question

Consider the basis e1 = (−2,4,−1), e2 = (−1,3,−1) and e3 = (1,−2,1) of R
3
over R. Find the dual basis of {e1, e2, e3}.

Answer 1

Answer on Question #79868 – Math – Linear Algebra

Question

Consider the basis $e_1 = (-2, 4, -1), e_2 = (-1, 3, -1)$ and $e_3 = (1, -2, 1)$ of $\mathbb{R}^3$ over $\mathbb{R}$. Find the dual basis of $\{e_1, e_2, e_3\}$.

Solution

For every basis $\{e_1, e_2, e_3\}$ there is a basis $\{v_1, v_2, v_3\}$ such that

$$v_i(e_j) = \begin{cases} 1 & \text{if } i \ne j \\ 0 & \text{if } i = j \end{cases}$$

The basis $\{v_1, v_2, v_3\}$ is known as the dual basis of $\{e_1, e_2, e_3\}$. Clearly, it is unique.

\begin{aligned} &\langle v_i, e_j \rangle = \delta_{ij} \\ AM = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad M = \begin{pmatrix} -2 & 4 & -1 \\ -1 & 3 & -1 \\ 1 & -2 & 1 \end{pmatrix} \\ A = M^{-1} \\ \begin{pmatrix} -2 & 4 & -1 & 1 & 0 & 0 \\ -1 & 3 & -1 & 0 & 1 & 0 \\ 1 & -2 & 1 & 0 & 0 & 1 \end{pmatrix} \xrightarrow{R_1 / (-2)} \begin{pmatrix} 1 & -2 & 1/2 & -1/2 & 0 & 0 \\ -1 & 3 & -1 & 0 & 1 & 0 \\ 1 & -2 & 1 & 0 & 0 & 1 \end{pmatrix} \\ \begin{pmatrix} 1 & -2 & 1/2 & -1/2 & 0 & 0 \\ -1 & 3 & -1 & 0 & 1 & 0 \\ 1 & -2 & 1 & 0 & 0 & 1 \end{pmatrix} \xrightarrow{R_2 + R_1} \begin{pmatrix} 1 & -2 & 1/2 & -1/2 & 0 & 0 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 1 & -2 & 1 & 0 & 0 & 1 \end{pmatrix} \\ \begin{pmatrix} 1 & -2 & 1/2 & -1/2 & 0 & 0 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 1 & -2 & 1 & 0 & 0 & 1 \end{pmatrix} \xrightarrow{R_3 - R_1} \begin{pmatrix} 1 & -2 & 1/2 & -1/2 & 0 & 0 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \\ \begin{pmatrix} 1 & -2 & 1/2 & -1/2 & 0 & 0 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \xrightarrow{R_1 + (2)R_2} \begin{pmatrix} 1 & 0 & -1/2 & -3/2 & 2 & 0 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \\ \begin{pmatrix} 1 & 0 & -1/2 & -3/2 & 2 & 0 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \xrightarrow{R_1 + R_3} \begin{pmatrix} 1 & 0 & 0 & -1 & 2 & 1 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \\ \begin{pmatrix} 1 & 0 & 0 & -1 & 2 & 1 \\ 0 & 1 & -1/2 & -1/2 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \xrightarrow{R_2 + R_3} \begin{pmatrix} 1 & 0 & 0 & -1 & 2 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \\ \begin{pmatrix} 1 & 0 & 0 & -1 & 2 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1/2 & 1/2 & 0 & 1 \end{pmatrix} \xrightarrow{(2)R_3} \begin{pmatrix} 1 & 0 & 0 & -1 & 2 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 2 \end{pmatrix} \end{array} $$A = M ^ {- 1} = \left( \begin{array}{c c c} - 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{array} \right)$$ $$v _ {1} = (- 1, 0, 1),$$ $$v _ {2} = (2, 1, 0),$$ $$v _ {3} = (1, 1, 2).$$

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