Can one think of vector subspaces as equivalence classes?

Question

I am currently studying Linear Algebra Done Right (LADR) by Sheldon Axler and also How to prove it by Daniel Velleman. Currently I am in the chapter of Bases in LADR and I already read the chapter on equivalence relations in How to prove it. I tend to be somebody who looks for analogous things and something that I wanted to verify was whether there is a resemblance of Vector Subspaces and Equivalence Relations: Can one think of subspaces being equivalence classes? What I would suggest would be to say that vectors in a subspace form an equivalence class if they can be written as a linear combination of the basis vectors of that subspace. Moreover, there is a theorem which says: Every subspace of V is part of a direct sum equal to V. Wouldn't then the vector subspaces be analogous to a partition? Am I thinking along the right lines? Doesn't that mean that there is a sort of universal structure of mathematical topics?

Answer 1

Can one think of subspaces being equivalence classes?

Pretty much anything can form equivalence classes; the only requirement for a family of subsets to form equivalence classes is that they form a partition.

The equivalence relation corresponding to these equivalence classes will be, quite simply but also quite unenlighteningly: "two elements are equivalent iff they belong to the same equivalence class".

Unfortunately, it's impossible to partition a vector space into a partition consisting only in vector subspaces, for a very simple reason: by definition, a vector subspace must always contain vector $\bf 0$, and so two vector subspaces of the same vector space cannot be disjoint. A partition must consist in disjoint subsets, so two vector subspaces cannot be in the same partition.

Therefore a partition of a vector space into subsets can consist in at most one vector subspace; the other subsets cannot also contain $\bf 0$ and so cannot be vector subspaces.

What you could do is partition a vector space into an family of parallel affine subspaces. For instance, you can partition 3d space into the family of all horizontal planes; then, two elements are equivalent iff they are in the same horizonal plane, ie, if they are at the same altitude. But those are affine subspaces, not vector subspaces: note how the word "elements" really wants to be "points" and doesn't want to be "vectors" in my previous sentence.

What I would suggest would be to say that vectors in a subspace form an equivalence class if they can be written as a linear combination of the basis vectors of that subspace.

If a vector subspace has a basis, then by definition the basis generates the subspace, which means that any vector in that subspace can be written as a linear combination of the basis vectors. There is no "if" here, the second part of your sentence is always true.

Every subspace of V is part of a direct sum equal to V. Wouldn't then the vector subspaces be analogous to a partition?

Consider a vector space $V$ and a vector subspace $U$ of $V$. There exists a vector subspace $W$ such that $V = U \oplus W$; but $U$ and $W$ do not form a partition of $V$. There are two reasons why $(U, W)$ is not a partition of set $V$: (1) because vector $\bf 0$ is both an element of $U$ and an element of $W$, which is forbidden in a partition; (2) because it is easy to construct a vector which is neither an element of $U$ nor an element of $W$, and that's also forbidden in a partition.

If you absolutely wanted to exhibit a partition of set $V$, you could define $U' = U \setminus \{\bf 0\}$, $W' = W \setminus \{\bf 0\}$, and $T = V \setminus (U \cup W)$, and then the four subsets $(\{\bf 0\}, U', W', T)$ would form a partition of set $V$. But none of $U'$, $W'$ nor $T$ would be vector subspaces.

However, if $B_1$ is a basis of $U$ and $B_2$ is a basis of $W$, then $B_1 \cup B_2$ is a basis of $V$, and $(B_1, B_2)$ is a partition of $B_1 \cup B_2$. So you can say that the bases of subspaces form a partition of the corresponding basis of the space (if those subspaces are in direct sum).

Doesn't that mean that there is a sort of universal structure of mathematical topics?

Yes! Mathematics love structures! In fact it is sometimes said that mathematics are the art of giving different names to similar structures.

Answer 2

If I understand your thought process correctly, one of the issues (among many) that will arise is as follows: Suppose $S=\text{Span}(\{e_1,e_2\})$ and $L=\text{Span}(\{e_2,e_3\})$, where $e_i$'s are the standard basis vectors in $\mathbb{R}^3$. Now the vector $e_2=0e_1+1e_2$ and $e_2=1e_2+0e_3$, so the classes $S$ and $T$ cannot be forming a partition as equivalence classes.