Suppose we take a set $S$ which contains just one vector space $V$ over any field $\mathbb{F}$. If we define scalar multiplication in such way, that we multiply every vector in $V$ with a scalar from $\mathbb{F}$, and if we take the regular sum of vector subspaces (set of sums of all pairs) for addition, then $S$ becomes a trivial vector space over $\mathbb{F}$ (because $V$ is then a neutral element for addition and hence it is a zero vector).

Does anyone know if we could define addition and scalar multiplication over a larger set $S$ (with more than one element) of vector subspaces of some vector space $V$ and choose some field $\mathbb{F}$ so that $S$ would become a non-trivial vector space over $\mathbb{F}$? In the example I have given, there is only one subspace in $S$, that is entire $V$. $V$ and $S$ can be over the same field, but they do not have to be over the same field.

  • $\begingroup$ If you want to sum subspaces of $V$, then $V$ is not a neutral element; it’s an absorbing element. Similar to adding $\infty$ to a positive number. We get $$V + W = V$$ not $$V + W = W$$ for $W$ a subspace of $V$. $\endgroup$
    – shoteyes
    Jul 12, 2021 at 12:29
  • $\begingroup$ Not necessery for some abstract summation. @shoteyes $\endgroup$
    – nonuser
    Jul 12, 2021 at 12:33
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    $\begingroup$ There are at least two approaches to your question: 1. Pick an uncountable family $S$ of subspaces of $(V, +, \cdot)$, pick a bijection of $S$ with some vector space, and transfer the operations from that space. This is possible, but probably uninteresting. 2. Define the vector space operations on $S$ in a way that uses the vector space structure of $(V, +, \cdot)$. It's doubtful this can be accomplished non-trivially, both for the reasons shoteyes mentions, and because scalar multiplication maps a subspace to itself. <> Can you clarify your intent? $\endgroup$ Jul 12, 2021 at 15:22
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    $\begingroup$ With such open-ended requirements you should probably start by looking for a way to make the set of subspaces into an abelian group, and worry about scalar multiplications later. For a construction that makes honest sense you'll want that an isomorphism between vector spaces should induce an isomorphism of your additive structure of subspaces; this will severely restrict what the zero element of your abelian group can be. Then somehow concoct an "addition" operation such that there's always a "negative" of each subspace ... $\endgroup$ Jul 12, 2021 at 18:11
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    $\begingroup$ Ultimately I think you'd end up basically creating something that is extremely contrived and perhaps only technically a vector space but so wonky that it doesn't mean much (you'd probably lose all relationship to the parent vector space). There are a lot of hurdles to overcome. I specifically have in mind the sum of two subspaces where one is a subspace of the other. $\endgroup$ Jul 12, 2021 at 18:41

1 Answer 1


$\newcommand{\Reals}{\mathbf{R}}$This answer has two parts: 1. Remarks about bijections; and 2. Transfer of operations.

Part I. For definiteness, let's work with finite-dimensional vector spaces with real scalars. Any two vector spaces $(V, +, \cdot)$ of positive dimension contain the same number of elements, and any two vector spaces of dimension at least $2$ contain the same number of vector subspaces.

There are numerous ways to pick families $S$ of subspaces in a way that $|S| = |\Reals|$. Many of these amount to picking "nearly structureless" collections of subspaces. On the other hand, some ways are not entirely uninteresting. Here is a small selection:

  1. In the Cartesian plane $(\Reals^{2}, +, \cdot)$, for each real $t$ let $V_{t}$ denote the line spanned by $(1, t)$, namely the line of slope $t$ through the origin. (The union of these lines is the complement of the vertical axis.) More generally, by viewing $\Reals^{n+1} = \Reals \times \Reals^{n}$, we obtain an "$\Reals^{n}$'s-worth" of lines spanned by $(1, v)$ as $v$ runs over $\Reals^{n}$. A similar idea gives, in $\Reals^{n+1}$, families of $k$-dimensional subspaces parametrized by $\Reals^{n-k+1}$.

  2. In the plane $(\Reals^{2}, +, \cdot)$, for each real $t$ with $-1 < t < 1$, let $V_{t}$ denote the line spanned by $(t, 1)$. (The union is the set of $(x, y)$ with $|x| < |y|$ together with the origin.) One natural generalization is to write $\Reals^{n+1} = \Reals^{n} \times \Reals$, where we obtain an open ball's worth of lines spanned by $(v, 1)$ with $|v| < 1$.

  3. In space $(\Reals^{3}, +, \cdot)$, let $S$ be the set of lines on the right circular cone $x^{2} + y^{2} - z^{2} = 0$. Or, let $S$ be the set of planes tangent to this cone. Or, the union of these sets of lines and planes....

Each of these sets is defined by finitely many algebraic equations and inequalities, and a bijection with the set of reals can be given more-or-less explicitly (in fact, is given explicitly in the first two examples, where for the second we use the inverse hyperbolic function $\tanh^{-1}:(-1, 1) \to \Reals$).

More typically, $S$ is an arbitrary collection of subspaces of varying dimension. For instance, let $M$ be a dense subset of the reals such that $M \cap (a, b)$ is Lebesgue non-measurable for every interval $(a, b)$, and let $S$ be the set lines in example 1. above whose slopes lie in $M$. Or, perform some analogous construction in higher dimension. Still, no matter how pathologically we choose, each subspace in $S$ is spanned by some finite collection from our original vector space, so the cardinality of $S$ is no greater than $|\Reals|$, even if a bijection with some $\Reals^{n}$ is inconvenient to write down.

Part II. Suppose $(V, +, \cdot)$ is a (finite-dimensional, real) vector space, $S$ is an arbitrary set of cardinality $|\Reals|$, and $\phi:S \to V$ is an arbitrary bijection. We may define the induced vector space structure $(S, \oplus, \odot)$ by defining \begin{align*} U \oplus V &= \phi^{-1}(\phi(U) + \phi(V)), \\ c \odot V &= \phi^{-1}(c \cdot \phi(V)),\quad \text{$c$ real.} \end{align*} These equations tell us how to add elements of $S$, and how to multiple elements of $S$ by scalars. Conceptually, the bijection attaches an "avatar" $\phi(U) = U'$ to each element of $S$. We know how to perform vector space operations on avatars, and each algebraic result is itself the avatar of a unique element of $S$. Applying $\phi$ to each side and using primes to denote avatars, the preceding equations may be written $$ (U \oplus V)' = U' + V',\qquad (c \odot V)' = c \cdot V'. $$ It should be clear that $(V, +, \cdot)$ and $(S, \oplus, \odot)$ are abstractly identical (here, naturally isomorphic as vector spaces).

The same idea holds any time we have a set with operations or distinguished collections of subsets, and a bijection with some other arbitrary set $S$.

In example 1. above, the transfer of operations under the bijection $\phi(V_{t} = t$ is particularly simple: The real number indices acts like vectors: $$ V_{s} \oplus V_{t} = V_{s+t},\qquad c \odot V_{t} = V_{ct}. $$ In example 2., the bijection $\tanh^{-1}:(-1, 1) \to \Reals$ allows the induced operation on subspaces to expressed almost as nicely: $$ V_{s} \oplus V_{t} = V_{\tanh(\tanh^{-1}s + \tanh^{-1}t)},\qquad c \odot V_{t} = V_{\tanh(c\tanh^{-1}t)}. $$


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