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I have seen many articles on forums that did not conclusively write answers to which conditions enable moving from vector subspaces to vector spaces.

I will first provide definitions and a theorem. From my research in mathematics textbooks, the definitions of vector spaces and vector subspaces and a theorem that links them are as follows:

Definition(Vector Space): A vector space is a set $V$ endowed with two operators:

Addition: $$ +: V \times V \rightarrow V \quad(x, y) \mapsto x+y \quad \text { and } $$ Scalar Multiplication:

$$\therefore \mathbb{R} \times V \rightarrow V \quad(\alpha, x) \mapsto \alpha \cdot x $$ such that

$$ \begin{array}{lc} \forall x, y, z \in V: & x+(y+z)=(x+y)+z \\ \forall x, y \in V: & x+y=y+x \\ \exists \underline{0} \in V: \forall x \in V: & x+\underline{0}=x \\ \forall x \in V: \exists(-x) \in V: & x+(-x)=\underline{0} \\ \forall \alpha \in \mathbb{R}: \forall x, y \in V: & \alpha(x+y)=\alpha x+\alpha y \\ \forall \alpha, \beta \in \mathbb{R}: \forall x \in V: & (\alpha+\beta) x=\alpha x+\beta x \\ \forall \alpha, \beta \in \mathbb{R}: \forall x \in V: & \alpha(\beta x)=(\alpha \beta) x \\ \forall x \in V: & 1 x=x \end{array} $$

Notation: We write $x-y$ for $x+(-y)$.

Definition(Vector Subspace): Let $X$ be a vector space. $Y \subseteq X$ is a vector subspace of $X$ if $Y$ is a vector space under the operations defined on $X$.

Theorem(Vector Subspace): Let $X$ be a vector space and $Y \subseteq X$. Then $Y$ is a vector subspace of $X$ if and only if $x+\lambda y \in Y$ for all $x, y \in Y$ and $\lambda \in \mathbb{R}$.

By the provided theorem, which is the most common and fastest way to establish the vector subspaces, many properties (e.g. Commutativity) are directly inherited from the vector (super)space of the vector subspace. However, from these definitions and theorem, one can deduce that from vector spaces one can go to vector subspaces and not the other way due to the definition being only one way(without actually proving the vector subspace is a vector space itself).

My definitive question is: If we establish that $Y$ is a vector subspace of vector space $X$ with the Theorem above, under what additional conditions can $Y$ be a vector space itself? Can we immediately call $Y$ a vector space, or if not, what can be the counterexamples? Can you provide me with a source for your answer to have a reliable source?

Thanks.

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  • $\begingroup$ $Y$ is a vector space in its own right. That’s essentially what motivates the definition and theorem. I’m still not sure why you’re not convinced by the definition and theorem and why you call it “one-way”. $\endgroup$
    – peek-a-boo
    Oct 28, 2023 at 0:37
  • $\begingroup$ The source of confusion here is difficult to follow. Let me just say something, and see if it helps you. A vector subspace has an alternative definition: a subset $Y$ of a vector space $(X, F, +, \cdot)$ is a subspace if we have operations $\oplus : Y \times Y \to Y$ and $\odot : F \times Y \to Y$ such that $(Y, \oplus, \odot, F)$ is a vector space (in its own right) and $y_1 + y_2 = y_1 \oplus y_2$, and $\lambda \cdot y = \lambda \odot y$ for all $y_1, y_2, y \in Y$ and $\lambda \in F$. In other words, we can restrict $+, \cdot$ to $Y$ (to get $\oplus, \odot$) and get a vector space. $\endgroup$ Oct 28, 2023 at 1:27
  • $\begingroup$ ... is that something you'd want us to prove: that this definition is equivalent to the one given (once you remember to restrict $Y$ to be non-empty)? $\endgroup$ Oct 28, 2023 at 1:27

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The whole point of $Y$ being a subspace is that you get, very cheaply, that $Y$ is itself a vector space. All the axioms for $Y$ follow from the fact that $X$ satisfies them, together with the simple condition in the Vector Subspace Theorem.

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