Question

Let B={u₁= (1, 1, -1) u₂= (1, 0, 1) u₃ = (3, 2, 0)} be a basis for R³. Find the dual basis of B

Answer 1

The linear function may be expressed in the form

${\phi}_1(x,y,z)=a_1x+a_2y+a_3z$ , $\phi_2(x,y,z)=b_1x+b_2y+b_3z ,$

$\phi_3(x,y,z)=c_1x+c_2y+c_3z$

By the definition of dual basis ,

$\phi_i(v_j)=0 \ for \ i\neq j$ but $\phi_i(v_j)=1$ for $i=j$ .

We find $\phi_1$ by setting $\phi_1(u_1)=1, \phi_1(u_2)=0,\phi_1(u_3)=0$ . This yields

$\phi_1(u_1)=a_1+a_2-a_3=1,\phi_1(u_2)=a_1+a_3=0 \ , \ \phi_1(u_3)=3a_1+2a_2=0$

Solving the system of equation yields

$a_1=2 \ ,\ a_2=-3 \ , \ a_3=-2$

Thus , $\phi_1(x,y,z)=2x_1-3x_2-2x_3$

We find $\phi_2$ by setting , $\phi_2(u_1)=0 \ , \ \phi_2(u_2)=1 \ , \ \phi_3(u_3)=0$ , This yields

$\phi_1(u_1)=b_1+b_2-b_3=0,\phi_1(u_2)=b_1+b_3=1 \ , \ \phi_1(u_3)=3b_1+2b_2=0$

Solving the system of equation yields

$b_1=2 \ , \ b_2=-3 \ , \ b_3=-1$

Thus , $\phi_2(x,y,z)=2x - 3x_2 - x_3$

Finally , we find $\phi_3$ by setting $\phi_1(u_1)=c_1+c_2-c_3=0,\phi_1(u_2)=c_1+c_3=0 \ , \ \phi_1(u_3)=3c_1+2c_2=1$

Solving the system of equation yields , $c_1=0 \ ,\ c_2=\frac{1}{2} \ , \ c_3=0$ .

Thus , $\phi_3(x,y,z)=\frac{1}{2}x_2$ .