# Trying to understand vector subspaces

I am studying a chapter on vector subspaces and projections. I came across this definition:

We can split the set of all vectors $$\mathbb{R}^3$$ into two disjoint sets: vectors entirely contained in $$S$$ and vectors perpendicular to $$S$$. We say $$\mathbb{R}^3$$ decomposes into the direct sum of the subspaces $$S$$ and $$S^\bot$$:

$$\mathbb{R}^3 = S \oplus S^\bot$$

And then about projection onto a line:

Let $$l$$ be a line passing through the origin and is in $$\mathbb{R}^3$$ with direction vector $$\vec{v}$$:

$$l: \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z) = t\vec{v}, t \in \mathbb{R}^3 \}$$

The orthogonal space to the line $$l$$ consists of all vectors perpendicular to the direction vector $$\vec{v}$$:

$$l^\bot : \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z).\vec{v} = 0 \}$$ ...

The orthogonal space for a line $$l$$ with direction vector $$\vec{v}$$ is a plane with normal vector $$\vec{v}$$.

All makes sense except the last sentence which says the orthogonal space is a plane with normal vector $$\vec{v}$$. What my intuition says is that the orthogonal space should be the set of all planes (not a single plane) in $$\mathbb{R}^3$$ with normal vector $$\vec{v}$$ because:

$$\mathbb{R}^3 = S \oplus S^\bot$$

I hope someone can tell me what is it that I am getting wrong here.

• The orthogonal subspace contains the origin though, which fixes one plane.
– Paul
Dec 26, 2020 at 19:47

"We can split the set of all vector in $$\mathbb{R}^{3}$$ into two disjoint sets: vectors entirely contained in $$S$$ and vectors perpendicular to $$S$$."

I don't like this line for several reasons, the two main reasons being that

1. These two sets $$S$$ and $$S^{\perp}$$ are not disjoint, they share the $$0$$ vector;
2. it seems to suggest that every vector in $$\mathbb{R}^{3}$$ is either in $$S$$ or $$S^{\perp}$$, which is not the case. There are lots of vectors that are neither contained in $$S$$, nor are orthogonal to $$S$$. What is true is that every vector in $$\mathbb{R}^{3}$$ is a sum of a vector in $$S$$ and a vector in $$S^{\perp}$$.

The orthogonal space for a line $$\ell$$ with direction $$v$$ is a plane with normal vector $$v$$.

This is absolutely correct. The orthogonal space is a collection of vectors, not a collection of planes. And all vectors are rooted at the origin. So the orthogonal space consists of all vectors in the plane containing the origin which has normal vector $$v$$ (this is essentially the definition of the normal vector of a plane).

• Thank you very much. I think the key point I was missing here was that the vectors in orthogonal space are rooted at the origin. This way, there is only one plane that its vectors are perpendicular to the line. Dec 27, 2020 at 6:32

Actually you're confusing with vectorial spaces and affines spaces. With vectorial spaces there exists only one subspace that has a given normal vector, because a plane is define by two non colinear vectors.

But then if you work with affines spaces, you also need a point : that's what we are used to.

• what is wrong with my answer?
– math
Dec 26, 2020 at 19:48
• Thanks mate. I don't know about affine space yet. btw I didn't down vote :) Dec 27, 2020 at 6:30