Let $V$ be a Inner product space, $(\epsilon_1....\epsilon_k)$ orthonormal sequence, let $\vec{v_1},\vec{v_2}\in V$. $\quad$ $GS(u_1...u_k)$ is the set of the vectors that recieved after doing GS procces on the set $(u_1...u_k)$
(A) Is $GS(\vec{v_1},\vec{v_2})= (\vec{v_1},\vec{v_2})$ ?
(B) Assume that $(\vec{v_1},\vec{v_2})$ are linearly independent and $\vec{u}\in SPAN(\vec{v_1},\vec{v_2})$. What is $GS(\vec{v_1},\vec{v_2},\vec{u})$?
$\quad$
My idea on (A) was to say that if I have orthonormal sequence so I can complete it to orthonormal basis for $V$ than I can say that $\vec{v_1}= 5\epsilon_1$ (for example) so if doing the GS procces on the $\vec{v_1}$ I get that $\vec{v'_1}=\epsilon_1$, so I am not sure if I can argue that they are not equal because they are linearly dependent, but on the other hand we not talking about SPAN.
on (B) I have no clue how approach, I think I need to use the given $GS(u_1...u_k)$ but no idea how.
thank you kindly.
(A) You are right. Unless $\left\{\vec{v_1},\vec{v_2}\right\}$ is an orthonormal set, $\operatorname{GS}\left(\left\{\vec{v_1},\vec{v_2}\right\}\right)\ne\left\{\vec{v_1},\vec{v_2}\right\}$, for the reasons that you have mentioned.
(B) $\operatorname{GS}\left(\left\{\vec{v_1},\vec{v_2},\vec u\right\}\right)=\operatorname{GS}\left(\left\{\vec{v_1},\vec{v_2}\right\}\right)$, since $\vec u$ is a linear combination of $\vec{v_1}$ and $\vec{v_2}$. Because of this, when you apply the Gram-Schimdt process to $\left\{\vec{v_1},\vec{v_2},\vec u\right\}$, the third vector will be the the null vector.