Answer to Question #169300 in Linear Algebra for Nikhil

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:

$$e'_1 = (x_1,y_1,z_1) \\ e'_2 = (x_2,y_2,z_2) \\ e'_3 = (x_3,y_3,z_3) \\$$

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

First vector:

$$\begin{cases} x_1+z_1=1 \\ x_1+y_1=0 \\ y_1+z_1=0 \\ \end{cases}$$

$$x_1=\frac{1}{2}; y_1=-\frac{1}{2}; z_1=\frac{1}{2}$$

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

Second vector:

$$\begin{cases} x_2+z_2=0 \\ x_2+y_2=1 \\ y_2+z_2=0 \\ \end{cases}$$

$$x_2=\frac{1}{2}; y_2=\frac{1}{2}; z_2=-\frac{1}{2}$$

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

Third vector:

$$\begin{cases} x_3+z_3=0 \\ x_3+y_3=0 \\ y_3+z_3=1 \\ \end{cases}$$

$$x_3=-\frac{1}{2}; y_3=\frac{1}{2}; z_3=\frac{1}{2}$$

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

Answer: $e'_1 = (\frac{1}{2},-\frac{1}{2},\frac{1}{2}) \\$ ; $e'_2 = (\frac{1}{2},\frac{1}{2},-\frac{1}{2}) \\$ ; $e'_3 = (-\frac{1}{2},\frac{1}{2},\frac{1}{2}) \\$.