Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider two unit-length vectors $x=[x_1,\dots,x_n]^T\in\mathbb{R}^n$ and $w=[w_1,\dots,w_n]^T\in\mathbb{R}^n$. The vector $w$ is given and $w_i>0$ for all $i$. The vector $x$ is variable but non-zero elements in $x$ are not with the same sign.

Clearly the angle subtended by $x$ and $w$ cannot be zero. Can we determine the minimum angle subtended by the two vectors?

share|cite|improve this question
"non-zero elements in $x$ always have different signs". So there are at most $2$ non-zero elements in $x$? Also you do not show a minimum exists (indeed it doesn't), so you should say infimum. – Marc van Leeuwen Sep 7 '12 at 8:38
do you want to minimize the angle between $x$ and $w$? Please clarify 'angle subtended by the two vectors'. – ajay Sep 7 '12 at 8:38
Also does "constant" mean "given"? Any vector is constant. – Marc van Leeuwen Sep 7 '12 at 8:41
At most two components of $x$ can be nonzero, but to minimize the angle it does not help to have a component be negative, so it suffices to only consider $x$ as a standard basis vector, as in Marc's answer. For fixed $w$, the minimum angle is then attained by the basis vector along which $w$ has the largest projection. – Rahul Sep 7 '12 at 10:06
@MarcvanLeeuwen: My original question is inaccurate. The non-zero elements in $w$ are not with the same sign. And there may exist more than two nonzero elements. – Shiyu Sep 7 '12 at 10:40
up vote 3 down vote accepted

The infimum is zero; take $x$ a standard basis vector, and $w$ as close to that as you like.

[Added] Now that it appears that $w$ is given, and that $x$ is required to have at least one positive and one negative entry (I'll skip over the difficulty at $n=1$), it seems one can always make the angle smaller by (assuming wlog the initial angle acute) moving a negative entry of $x$ closer to $0$, and rescaling to a unit vector. So for the infimum we may suppose all entries of $x$ nonnegative with at least one equal to $0$. But then, fixing the set of positions where $x$ has coordinate $0$ by projecting $w$ onto the subspace with coordinates $0$ in those positions, one sees that one can do no better for $x$ than to choose its remaining coordinates equal (up tu a scalar) to those of $w$. I think it is not hard to see that the smallest angle is obtained by selecting the coordinate of minimal value in $w$ and making that zero (or infinitesimally negative in the original problem) in $x$, and rescaling.

share|cite|improve this answer
I suspect that you are onto something in one of your comments, and that $w$ is meant to be given. – Gerry Myerson Sep 7 '12 at 9:47
The vector $w$ is given, and the infimum then is not zero. – Shiyu Sep 7 '12 at 10:48
Thanks. I got it now. By making the minimal element in $w$ to zero, we actually obtain the largest projection of $w$ onto a plane that is perpendicular to a coordinate axis. The angle between $w$ and the projection is the infimum of the angle between $x$ and $w$. – Shiyu Sep 7 '12 at 13:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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