Relationship between angle of vectors and orthogonalization

Question

I'm trying to understand the following question:

Let $u_1,...,u_d$ be a set of orthonormal vectors in $\mathbb{R}^n$. Let $a,b$ be unit vectors in $\mathbb{R}^n$ such that $\angle(a,b) < \epsilon$. If we orthogonalize $a$ and $b$ with respect to the $u_i's$ (with some Holder Transformation / Gram-Schmidt for both vectors) to $v_a,v_b$, is there a bound to $\angle(v_a,v_b)$ related to $\epsilon$? It seems like the angles should basically be the same, but I can't prove it nor can I find a good reference about what this bound should be, even if not $\epsilon$.

EDIT: To give some context: The main question I'm looking to show that there are orthonormal bases $<u_1,...,u_d,v_a>$ and $<u_1,...,u_d,v_b>$ for the spaces spanned by $<u_1,...,u_d,a>, <u_1,...,u_d,b>$ respectively, such that $\angle(v_a,v_b) < \epsilon$ or some function of $\epsilon$, given that $u_1,...,u_d$ are orthonormal and $a,b$ are unit vectors with $\angle(a,b)<\epsilon$. I see why Gram-Schmidt may not work given Eric's example in the comments, but is there an answer to this question?

ADDITIONAL EDIT: We assume the the subspaces mentioned in the above edit are not the same.

Any help would be great!

Answer 1

Just to answer the question and add to @AlDante 's response above, while it is true that $\|v_a-v_b\|$ is small, this doesn't show that the angle between $v_a$ and $v_b$ is small. In fact, it does not have to be. if $a$ and $b$ are close together and very very close to any of the vectors $u_i$, then the perpendicular components $v_a,v_b$ have very small entries and the norm of their difference is small, But the could even be orthogonal (It's actually a great geometric exercise to convince yourself of why). I appreciate everyone's help!

Answer 2

Let $U$ be the subspace of $\mathbb{R}^n$ spanned by $u_1,...,u_d$. Both $a$ and $b$ can be uniquely decomposed into their projections $u_a$, $u_b$ on $U$ and a vector $v_a$, $v_b$ orthogonal to $U$. As $a$ and $b$ are similar (the angle between them is small) their projections are similar.

$v_a$, $v_b$ in their turn are similar.

$\left\| a - b \right\| = \left\| (u_a + v_a) - (u_b + v_b) \right\| $ As $\left\| a - b \right\|$ is small (similar vectors) and $\left\| u_a - u_b \right\|$ is small by projection, $\left\| v_a - v_b \right\|$ is small.

As the starting points ($u_a$ resp. $u_b)$ and endpoints ($a$ resp. $b$) are close to one another the angle between $v_a$ and $v_b$ is therefore also small.