Answer to Question #160249 in Linear Algebra for Nikhil Singh

Question

Using the Gram-Schmidt prodecure find an orthornornal set of vectors corresponding to the ordered basis B {(1,1,1),(1,1,0),(1,0,0)} of R^3. Also find a basis dual to B.

Answer 1

Answer

a)

If { v₁, v₂, v₃) is any ordered basis, then according to Gram-Schmidt procedure,

u_k=v_k- $\displaystyle\sum_{i=1}^{k-1}$ proj _uj (v_k), where proj_u (v) = $u.v \over u.u$u.

and the normalized vector is e_k= $u_k \over \sqrt{u_x.u_x }$

Here v₁= $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, v₂= $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ v₃= $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$

Now u₁=v₁= $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and e₁= $u_1 \over \sqrt{u_1.u_1 }$= $1 \over \sqrt{3 } ​ ​$$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

u₂= v₂- $u_1.v_2 \over u_1.u_1$u. = $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ - $2 \over 3$    $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ = $\begin{bmatrix} 2 \over 3 \\ 2 \over 3 \\ 2 \over 3 \end{bmatrix}$  

u₂.u₂= $1 \over 9$ + $1 \over 9$ + $4 \over 9$ = $6 \over 9$ = $2 \over 3$

e₁= $\sqrt{3} \over \sqrt{2}$$\begin{bmatrix} 1 \over 3 \\ 1 \over 3 \\ -2 \over 3 \end{bmatrix}$

u₃= v₃- $u_1.v_3 \over u_1.u_1$u₁ - $u_2.v_3 \over u_2.u_2$u₂

u₃= $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ - $1 \over 3$ $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ - $1 \over 2$$\begin{bmatrix} 1 \over 3 \\ 1 \over 3 \\ -2 \over 3 \end{bmatrix}$

u₁.v_{3 = 1}

u₂.v_{3 =} $1 \over 3$

u₁.u_{1 = 3}

u₂.u_{2 =} $2 \over 3$

b)

By definition, if B = { u_1,u_{2, ---,}u_k,} is basis of a vector space, then

B^* = { f_1,f_{2, ---,}f_k,} is dual basis of dual space,

where f_i (u_j)=d_{kij ,} d_kij= $\begin{cases} 1 &\text{if } i=j \\ 0 &\text{if } i\not =j \end{cases}$

Now given basis is B= {(1,1,1), (1,1,0), (1,0,0)}, then {  u_1,u_2,u₃}

We have (u,y,z) = z (1,1,1)+ (y-z)(1,1,0) + (u-y)(1,0,0)

Now f₁(u,y,z) =  z f₁ (1,1,1)+ (y-z) f₁(1,1,0) + (u-y) f₁(1,0,0)

=z+ (y-z).0+ (u-y).0

So  f₁(u,y,z) =  z

 f₂(u,y,z) =  z  f₂ (1,1,1)+ (y-z) f₂(1,1,0) + (u-y) f₂(1,0,0)

= z.0+ (y-z).1+ (u-y).0

so  f₂(u,y,z) = y-z

 f₃(u,y,z) =  z   f₃ (1,1,1)+ (y-z) f₃(1,1,0) + (u-y)  f₃(1,0,0)

               = z.0+ (y-z).0+ (u-y).1

so  f₃(u,y,z) = u-y

Thus  B^* = { f_1,f_{2, ,}f₃), where  f₁(u,y,z) = z, f₂(u,y,z) = y-z, f₃(u,y,z) = u-y.