Question

Find the basis (a_1, a_2, a₃) that is dual to the following basis of R³

{u₁ = (1, -1, 3), u₂ = (0, 1, -1), u₃ = (0, 3, -2)}

Answer 1

The linear functional may be expressed in the form

$a_1(x,y,z)=m_1x+m_2y+m_3z \ , \ a_2(x,y,z)=n_1x+n_2y+n_3z$ ,

$a_3(x,y,z)=p_1x+p_2y+p_3z$ .

By definition of dual basis ,

$a_i(u_i)=0 \ for \ i \neq j \ but \ a_i(u_i)=1 \ for \ i=j$ .

We define $a_1$ by setting $a_1((u_1)=1,a_1(u_2)=0,a_1(u_3)=0$ , This yields

$a_1(1,-1,3)=m_1-m_2+3m_3=1 \ , \ a_1(0,1,-1)=m_2-m_3=0$ ,

$a_1(0,3,-2)=3m_2-2m_3=0$

solving the system of equations yields ,

$m_1=1,m_2=0,m_3=0$

Thus $a_1(x,y,z)=x$

Now , we find $a_2$ by setting , $a_2(u_1)=0,a_2(u_2)=1, a_2(u_3)=0$ . This yields

$a_2(1,-1,3)=n_1-n_2+3n_3=0 \ , \ a_2(0,1,-1)=n_2-n_3=1$ ,

$a_2(0,3,-2)=3n_2-2n_3=0$

Solving the system of equation yields ,

$n_1=7,n_2=-2,n_3=-3$

Thus $a_2(x,y,z)=7x-2y-3z$ .

we find $a_3$ by setting ,

$a_3(u_1)=0,a_3(u_2)=0,a_3(u_3)=1$ , This yields

$a_3(1,-1,3)=p_1-p_2+3p_3=0 \ , \ a_3(0,1,-1)=p_2-p_3=0$ ,

$a_3(0,3,-2)=3p_2-2p_3=1$ ,

Solving the system of equation yields

$p_1=-2,p_2=1,p_3=1$

Thus $a_3(x,y,z)=-2x+y+z$