# Question about Gram-Schmidt & orthonormal basis

Suppose $\alpha_1, \dots, \alpha_n$ are vectors of norm 1 in some $\mathbf R^d$. Let $\beta_1, \dots, \beta_n$ be the orthogonal basis vectors from the Gram-Schmidt process, i.e. $\beta_1 = \alpha_1, \beta_2 = \alpha_2 - (\alpha_2 . \beta_1) \beta_1$ etc.

Now suppose I have some vector $x$ such that $(x.\beta_i)^2 \geq \lambda$, for $i = 1, \dots, n$.

(1) Does this depend on the ordering of the vectors $\alpha_1, \dots, \alpha_n$? That is, if I reorder the vectors $\alpha$, and get a new set of vectors $\beta'$ (which span the same space, but in a different order), do I also have $(x. \beta'_i)^2 \geq \lambda$? (THIS IS ANSWERED IN THE NEGATIVE, SEE BELOW)

(2) If not, is there any condition I can derive on $x$, that does not depend on the order of the vectors $\alpha_1, \dots, \alpha_n$? (I.e. I get the same condition for any ordering of those vectors)? For example, must it be the case that $(x.\alpha_i)^2 \geq \lambda$ for $i = 1, \dots, n$? Or something similar to this?

Thanks!

The order matters. Take $\alpha_1=(1,0), \alpha_2 = (1,1)$, $x=\alpha_2$.
Then $\beta_1 = (1,0), \beta_2 = (0,1)$, $\langle x, \beta_i \rangle = 1$.
Now switch the $\alpha_i$, then $\beta_1' = {1 \over \sqrt{2}} (1,1), \beta_2' = {1 \over \sqrt{2}} (1,-1)$, but $\langle x, \beta_2' \rangle = 0$.