Question #169300

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


1
Expert's answer
2021-03-08T19:15:16-0500

We need to find vectors:


e1=(x1,y1,z1)e2=(x2,y2,z2)e3=(x3,y3,z3)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:

{x1+z1=1x1+y1=0y1+z1=0\begin{cases} x_1+z_1=1 \\ x_1+y_1=0 \\ y_1+z_1=0 \\ \end{cases}

x1=12;y1=12;z1=12x_1=\frac{1}{2}; y_1=-\frac{1}{2}; z_1=\frac{1}{2}

e1=(12,12,12)e'_1 = (\frac{1}{2},-\frac{1}{2},\frac{1}{2}) \\

Second vector:

{x2+z2=0x2+y2=1y2+z2=0\begin{cases} x_2+z_2=0 \\ x_2+y_2=1 \\ y_2+z_2=0 \\ \end{cases}

x2=12;y2=12;z2=12x_2=\frac{1}{2}; y_2=\frac{1}{2}; z_2=-\frac{1}{2}

e2=(12,12,12)e'_2 = (\frac{1}{2},\frac{1}{2},-\frac{1}{2}) \\

Third vector:

{x3+z3=0x3+y3=0y3+z3=1\begin{cases} x_3+z_3=0 \\ x_3+y_3=0 \\ y_3+z_3=1 \\ \end{cases}

x3=12;y3=12;z3=12x_3=-\frac{1}{2}; y_3=\frac{1}{2}; z_3=\frac{1}{2}

e3=(12,12,12)e'_3 = (-\frac{1}{2},\frac{1}{2},\frac{1}{2}) \\

Answer: e1=(12,12,12)e'_1 = (\frac{1}{2},-\frac{1}{2},\frac{1}{2}) \\ ; e2=(12,12,12)e'_2 = (\frac{1}{2},\frac{1}{2},-\frac{1}{2}) \\ ; e3=(12,12,12)e'_3 = (-\frac{1}{2},\frac{1}{2},\frac{1}{2}) \\.


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