Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that $\{q_1,q_2,q_3\}$ is a arthogonal basis for $\mathbb{R}^3$.
My problem is the following: I did take $v=(1,0,0)$ and I did verify that $\{q_1,q_3,v\}$ is a basis for $\mathbb{R}^3$. Then I did take $$q_2=v-\langle v|q_1\rangle q_1-\langle v|q_3\rangle q_3=(-1,-2,1)$$
And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?
Best Answer
None of $q_1$ and $q_2$ are normalized. Hence, the formula would be $$ q_2=v-{\langle v|q_1\rangle \over \langle q_1|q_1\rangle} q_1-{\langle v|q_3\rangle \over \langle q_3|q_3\rangle} q_3 = ({1 \over 2},-{1 \over 2},0)$$.