Let $v_1=(1,2,2)$, $v_2=(1,0,1)$ in $R^3$
- Find an orthonormal basis $\{u_1,u_2\}$ to $Sp\{v_1,v_2\}$
- Find vector $w \in Sp\{v_1,v_2\}$ so $w \bullet u_1=3$ and $w \bullet u_2=9$
My solution
I'm new to orthogonal/orthonormal.
So orthonormal is when the normalization of vector equals to 1.
Therefore I'll check for $v_1$ and $v_2$
$||v_1||=\sqrt{1^2+2^2+2^2}=3$
$||v_2||=\sqrt{1^2+0^2+1^2}=\sqrt{2}$
Both $v_1$ and $v_2$ are not orthonormal.
Lets use gram-schmidt in order to change it to orthonormal:
$$v_1^*=v_1=(1,2,2)$$
$$v_1^*=v_2-\frac{v_2 \bullet v_1^*}{||v_1^*||^2}*v_2^*$$
$$v_1^*=(1,0,1)-\frac{(1,0,1) \bullet (1,2,2)}{9}*(1,0,1)=(1,0,1)-\frac{3}{9}(1,0,1)=(\frac{2}{3},-\frac{2}{3},\frac{1}{3})$$
Now $||v_1^*||=1$
Lets do the same thing for $v_2$
$$v_2^*=v_2=(1,0,1)$$
$$v_2^*=v_1-\frac{v_1 \bullet v_2^*}{||v_2^*||^2}*v_1^*$$
$$v_2^*=(1,2,2)-\frac{(1,0,1) \bullet (1,2,2)}{2}*(1,2,2)=(1,2,2)-\frac{3}{2}(1,2,2)=(1,2,2)+(-\frac{3}{2},-3,-3)=(\frac{1}{2},-1,-1)$$
But $||v_2^*|| \neq 1$
What did I do wrong?
You should have something like $u_1 = {1 \over \|v_1\|} v_1 = { 1\over 3} (1,2,2)$.
Then $u_2^* = v_2 - \langle u_1, v_2 \rangle u_1 = {1 \over 3} (2, -2, 1) $, $u_2 = { 1\over \|u_2^*\|} u_2^* = {1 \over 3} (2, -2, 1)$.
Then $\operatorname{sp} \{ v_1, v_2 \} = \operatorname{sp} \{ u_1, u_2 \}$ and $u_1,u_2$ are orthonormal.
If $w \in \operatorname{sp} \{ v_1, v_2 \} = \operatorname{sp} \{ u_1, u_2 \}$, then $w = \sum_k \alpha_k u_k$, and we have $\langle w, u_k \rangle = \alpha_k$. Since the $u_k$ form a basis, the $\alpha_k$ are unique. This should answer the second part of your question.