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?



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 $\langle x, \beta_2' \rangle = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.