Can you find a vector not orthogonal to N others? | lisyarus blog

Can you find a vector not orthogonal to N others?

2026 Jun 15

I was reading something about quiver representations and some proof started with take any vector not orthogonal to any of these N vectors. This seems trivial, but...cut can you actually always do that?

Contents

Intuition

I mean, obviosly you can! Say, you have a bunch of (non-zero) vectors:

In 2D, the vectors orthogonal to a single vector form a line through the origin:

If we do this for all our vectors, we get a union of lines:

And the question now is: can we find a vector not on any of the red lines (i.e. in the white area)? It's visually obvious that yes indeed we can — pretty much a random vector will do:

The same works for any dimension \(\geq 2\). E.g. in 3D you'd get a union of a finite set of planes through the origin, and you just need to find a vector not in any of these planes.

In 1D things are a bit stupid: any two non-zero vectors are collinear and cannot be orthogonal, so you can just pick any non-zero vector — it won't be orthogonal to any of your input vectors.

However, it is often the case in math that a visually obvious fact can be notoriously hard to prove, or even turn out to be false in higher dimensions.

So how can we prove this?

Measure-theoretic proof

Let's try to formalize the intuition. The idea is that lines occupy a small portion of the space, and there's still a ton of space left if we exclude a finite set of lines. Let's take the (K-dimensional!) volume of the space and subtract the volume of the subspaces orthogonal to each of the input vector. The problem is that, last time I checked, the whole space was infinite, and subtracting something from infinity is usually a shady idea at best (you can get to things like \(\infty - \infty = \infty\)).

So, instead let's bound our space. Let's say we're only interested in the resulting vectors with length no larger than \(1\). In other words, we're interested in what happens inside a ball with radius \(1\). The exact measure (aka K-dimensional volume) of this ball isn't relevant to us, it's OK to just know that it's non-zero.

A subspace (line/plane/etc) orthogonal to some input vector cuts away a certain smaller-dimensional ball, — e.g. in 3D we'd have a 2D plane cutting through a 3D ball and resulting in a 2D circle (with interior) of forbidden vectors. What we want is some vector inside the big ball but not inside one of those balls of smaller dimension.

But those smaller balls have measure (volume) zero! A simple way to prove this is to wrap them into a (K-dimensional) box with one axis parallel to an input vector and all other axes orthogonal to it (and thus parallel to the forbidden ball). E.g. in 3D we can take a circle of forbidden vectors corresponding to one input vector and take a bounding square. However, we have 3 dimensions, so we extend this square a bit along the normal to the circle, which is parallel to the input vector (which in turn is orthogonal to the circle, rememeber?), and we can extend it by any non-zero amount to get a proper 3D box. In particular, we can extend it by smaller and smaller amounts, — leading to smaller and smaller volumes of this box, — while still containing the forbidden circle.

Now, if a set (the forbidden circle) is contained in each of a sequence sets (the boxes) with volume converging to zero (because their height converges to zero), it itself has volume zero (this follows from the basic properties of a measure). This is all to say that those smaller-dimensional balls have measure (aka volume) zero, simply because they live in a subspace of smaller dimension.

A union of (a finite collection of) sets of measure zero again has measure zero, and the measure of the set of allowed vectors — that is, the unit ball excluding the forbidden vectors — is thus some positive number (volume of the K-dimensional ball) minus zero, which is still that positive number. A set with positive measure cannot be empty, thus there exists a vector not in one of the forbidden sets. Hurray!

The proof is a bit involved if you've never worked with measure theory; on the other hand, it probably sounds trivial if you've taken a proper analysis class. In any case, it captures the essence of our intuition: there does exist a vector not orthogonal to a set of input vectors simply because those vectors orthogonal to any of the input vector form a tiny (lower-dimensional) subset of the whole space.

However, I'm not really satisfied with this proof. Firstly, it feels like a huge overkill to use measure theory to something that talks about simple geometry. Secondly, it doesn't work for other number fields! E.g. it doesn't work if I'm only interested in vectors with rational coefficients, or even coefficients from a finite field!

And if you thing that vectors with coefficients in a finite field is some useless toy for algebraists to play with, I regret to inform you that modern...