Answer to Question #86510 in Linear Algebra for RAKESH DEY

Question

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

Answer 1

SOLUTION

We need to find vectors:

$$e^1=(x_1,y_1,z_1),\quad e^2=(x_2,y_2,z_2),\quad e^3=(x_3,y_3,z_3)$$

The dual base vectors should satisfy this:

$$e_i\cdot e_j=\delta_i^j$$

The condition equals to 3 systems of 3 equations in 3 unknowns. Solving each system will give you one vector for the dual base. Here goes the first one which would be for the first dual base vector:

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

Then,

$$\boxed{x_1=\frac{1}{2}\quad y_1=\frac{1}{2}\quad z_1=0}$$

for second vector:

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

Then,

$$\boxed{x_2=0\quad y_2=\frac{1}{2}\quad z_2=\frac{1}{2}}$$

for third vector:

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

Then,

$$\boxed{x_3=\frac{1}{2}\quad y_3=0\quad z_3=\frac{1}{2}}$$

ANSWER

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