Question About Gram Schmidt Procedure

Question

Let's say we have a basis A = {$\vec{a}_1, ..., \vec{a}_n$} for $\mathbb{R^n}$.

If we apply the Gram Schmidt Procedure to A and get B={$\vec{b}_1, ..., \vec{b}_n$}, an orthogonal basis for $\mathbb{R^n}$, then is bk orthogonal to $\vec{a}_1, ..., \vec{a}_{n-1}$?

Answer 1

If $A=\{(1,0),(1,1)\}$, then $B=\{(1,0),(0,1)\}$. Does this answer your question?

Answer 2

After the $k$th step of the process, we have $\operatorname{span}\left\{\vec a_1,\dots,\vec a_k\right\}=\operatorname{span}\left\{\vec b_1,\dots,\vec b_k\right\}$. The next vector that is added to the orthonormal basis is the orthogonal rejection, normalized, of $\vec a_{k+1}$ from this subspace. By construction, then, $\langle \vec a_i,\vec b_{k+1}\rangle = 0$ for all $i\lt k+1$.