Is every vector space a proper subspace of a larger vector space?

Question

Is every vector space a proper subspace of a larger vector space?

I would think this is true, but was unable to come up with a construction.

Answer 1

Here is the sketch of a concrete construction using the existence of free vector spaces over a given set and cantors theorem.

Let K be a field an M be a set.

Definition

A free K vector space over M is a K vector space that has M as a basis.

Theorem

There is a free K vector space over M

Sketch of proof:

Consider the set $K^{(M)}:=\{f:M\to K~|~|f^{-1}(K\setminus\{0\})|\in\mathbb N\}$ of functions that take only finitely many values $\neq 0$.

This is a vector space of dimension $|M|$ since the characteristic functions $\chi_{\{m\}}:M\to K$ for $m\in M$ form a basis.

The only problem is, that M is not a subset of that space. By "exchanging" $\chi_{\{m\}}$ with m for every m we get a new vector space with M as a basis (make sure you understand why this works!).

Now we know that for any set M there is a K vector space V that has M as a basis. This reduces your question to the following:

For a given set M, ist there a superset N with strictly bigger cardinality?

If so, take a free K vector space over that set and you are done.

The answer to that question is "yes". First of all note that cantor's theorem states $|\mathfrak{P}(M)|>|M|$ (the proof is easy and can be found on wikipedia). Swap every $\{m\}\in \mathfrak P(M)$ with m to obtain a set with the cardinality of $\mathfrak{P}(M)$ with M as a subset (again, make sure to make sense of my sloppiness!).

I hope this answers your question to your satisfaction, feel free to ask if anything is unclear.

Greetings, Tobi

PS: I'm not a native English speaker and I'm not used to writing mathematics in English. Also, this is my first answer on math.stackexchange. I hope my text is intelligible and up to standards and would gladly receive any feedback concerning that.

Answer 2

Yes. Let $V$ be any vector space your heart desires. Let $\beta$ be a basis for $V.$ Let $\vec{s}$ be a formal symbol, which we will regard as linearly independent of the vectors in $\beta.$ Then $\beta\cup \{\vec{s}\}$ is a basis for some "new" vector space, and $$V\subset \text{span}(\beta\cup \{\vec{s}\})$$ and $V$ is proper, nontrivial.

Answer 3

First the usual bit: if $V$ is a vector space over the field $F$, then (knowing that $F$ is a vector space over itself) you can construct the vector space $V\times F$, in which elements are pairs from the Cartesian product $V\times F$ and the operations are defined coordinate-wise, i.e. $(v_1, \alpha_1)+(v_2, \alpha_2)=(v_1+v_2, \alpha_1+\alpha_2)$, $\lambda(v,\alpha)=(\lambda v, \lambda\alpha)$.

(Note: Instead of $F$, I could use any other nontrivial vector space over $F$, but let's be specific for the purpose of this answer.)

Now, is $V$ a proper subspace of $V\times F$?

What is obvious is that there is an injective homomorphism $V\to V\times F$, mapping $v\in V$ to $(v,0_F)\in V\times F$. Thus we can say that we can "identify" $V$ with its image $V\times\{0_F\}$, and we can "see" $V$ as a proper subspace of $V\times F$. Namely, $V$ is then isomorphic to $V\times\{0_F\}$ and the latter is a proper subspace, knowing that it does not contain $(0_V, 1_F)$.

Okay, but isn't this a bit hand-wavy?

After all, $V$ is $V$ and $V\times\{0_F\}$ is $V\times\{0_F\}$, we didn't really extend $V$ but have made something else, that looks pretty much as $V$, contained in a bigger vector space.

The answer is obvious when you look at it, but is rarely given up to detail. We will actually take the set $V\times F$, cut out the image of the injective homomorphism ($V\times\{0_F\}$), and add the actual $V$. In other words, we will define a new set:

$$\overline{V}=((V\times F)\setminus(V\times\{0_F\}))\cup V$$

On that set we need to define the operations to "stitch together" $V$ with the rest of the set difference $(V\times F)\setminus(V\times\{0_F\})$: for all $x,y\in\overline{V}$:

$$x+y:=\begin{cases}x+y&x,y\in V\\(x+w,\beta)&x\in V,y=(w,\beta)\in V\times F,\beta\ne 0_F\\(v+y,\alpha)&x=(v,\alpha)\in V\times F, \alpha\ne 0_F, y\in V\\(v+w, \alpha+\beta)&x=(v,\alpha)\in V\times F, y=(w,\beta)\in V\times F, \alpha,\beta,\alpha+\beta\ne 0_F\\v+w\in V&x=(v,\alpha)\in V\times F, y=(w,\beta)\in V\times F, \alpha,\beta\ne 0_F,\alpha+\beta=0_F\end{cases}$$

and:

$$\lambda x:=\begin{cases} \lambda x\in V&x\in V\\(\lambda v,\lambda\alpha) & x=(v,\alpha)\in V\times F, \alpha\ne 0_F\end{cases}$$

Now these operations need to be proven to be well defined, and $\overline{V}$ would need to be proven to be a vector space over $F$, i.e. that all the vector space axioms are satisfied. That is why you'll rarely see it done - it is very tedious, it will need to distinguish a lot of cases, and at the end of the day, all you learn is what you intuitively already knew, in the "hand-wavy" way, that it will extend $V$ to a properly containing vector space $\overline{V}$. Hope you will forgive me for also omitting this tedious proof.

Answer 4

The answer to this question requires only one line, and it does not require using a basis, or considering cardinality.

Let $V$ be denote your vector space over a field $F$. Let $X$ be any non-zero vector space over $F$, for example $X = F$ if you like. Then $V$ imbeds in $W = V \oplus X$ via $v \mapsto (v, 0)$, with image $V \oplus (0) \subsetneq W$.