Question

Find the dual basis of {(1,0,1),(1,1,0),(0,1,1)} in R^3

Answer 1

We need to find vectors:

$e_1'=(x_1,y_1,z_1) \newline e_2'=(x_2,y_2,z_2) \newline e_3'=(x_3,y_3,z_3) \newline$

The condition equals three systems of three equations. Each system will give us one vector for the dual base.

First vector:

$$x_1+z_1=1 \newline x_1+y_1=0\newline y_1+z_1=0 \newline$$

Solving these equations we get:

$$x_1=\frac{1}{2}; y_1=\frac{-1}{2};z_1=\frac{1}{2} \newline e'_1=(\frac{1}{2},\frac{-1}{2},\frac{1}{2})$$

Second vector:

$$x_2+z_2=0 \newline x_2+y_2=1\newline y_2+z_2=0 \newline$$

Solving these equations we get:

$$x_2=\frac{1}{2}; y_2=\frac{1}{2};z_2=\frac{-1}{2} \newline e'_2=(\frac{1}{2},\frac{1}{2},\frac{-1}{2})$$

Third vector:

$$x_3+z_3=0 \newline x_3+y_3=0\newline y_3+z_3=1 \newline$$

Solving these equations we get:

$$x_3=\frac{-1}{2}; y_3=\frac{1}{2};z_3=\frac{1}{2} \newline e'_3=(\frac{-1}{2},\frac{1}{2},\frac{1}{2})$$

Hence, dual basis are:

$$e'_1=(\frac{1}{2},\frac{-1}{2},\frac{1}{2}) \newline e'_2=(\frac{1}{2},\frac{1}{2},\frac{-1}{2}) \newline e'_3=(\frac{-1}{2},\frac{1}{2},\frac{1}{2})$$