Vector Spaces and Linear Subspaces: Examples and Counterexamples

Let $\mathbb{R}$ be the real numbers (The real numbers), a field (The reals form a field) and an ordered field (The reals form a totally ordered field, Ordered field), and let $\mathbb{Q}$ be the rationals, a field (The rationals form a field).

1. $\mathbb{R}$ is a vector space over itself (Vector space over a field): the vectors are the reals, the vector addition is the field addition, the zero vector is $0$, and the scalar multiplication is the field multiplication.
2. Let $\iota :\mathbb{Q}\to \mathbb{R}$ be the unique field homomorphism (The unique embedding of ℚ into an ordered field, Field homomorphism and embedding), which is injective and order-preserving. Its image ${\mathbb{Q}}_{\mathbb{R}}:=\iota [\mathbb{Q}]=\{\iota (q):q\in \mathbb{Q}\}$ is a subfield of $\mathbb{R}$ (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations), and by restriction of scalars (A field is a vector space over itself, and over any subfield $K\subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars) $\mathbb{R}$ is a vector space over ${\mathbb{Q}}_{\mathbb{R}}$, the scalar multiplication being the field multiplication restricted to ${\mathbb{Q}}_{\mathbb{R}}\times \mathbb{R}$.
3. Setting $q\cdot x:=\iota (q)x$ for $q\in \mathbb{Q}$ and $x\in \mathbb{R}$ makes $\mathbb{R}$ a vector space over $\mathbb{Q}$ itself.

The subfield is the image, not $\mathbb{Q}$. $\mathbb{Q}$ is not a subset of $\mathbb{R}$ in this library: a rational is a class of pairs of integers and a real is a class of Cauchy sequences of rationals (The real numbers). What sits inside $\mathbb{R}$ as a subfield is the image ${\mathbb{Q}}_{\mathbb{R}}$ of the embedding, and claim 2 is a statement about that image. Claim 3 is the statement about $\mathbb{Q}$ itself, and it is proved directly rather than by restricting scalars, because restriction of scalars requires a subfield.

• Three structures on one set. The vectors are the same reals throughout and the addition is the same in all three cases; what changes is which scalars are allowed to act. Claims 2 and 3 differ only in bookkeeping: the scalars are the elements of ${\mathbb{Q}}_{\mathbb{R}}$ in one and the elements of $\mathbb{Q}$ in the other, and $\iota$ matches them up bijectively, being injective onto its image.

• Nothing here is said about size. How big $\mathbb{R}$ is as a vector space over $\mathbb{Q}$ is a question about bases and dimension, which are developed on a later page; no claim about either is made above, and the verification uses neither.

• Why the ordered field hypothesis appears at all. The embedding $\iota$ is supplied by The unique embedding of ℚ into an ordered field, which is stated for an ordered field, and $\mathbb{R}$ is one. The order plays no further role: once $\iota$ is in hand, every step above uses only that it is a field homomorphism.

Let $F$ be a field (Field) and consider the vector space $F^3$ of functions $3\to F$ with the pointwise operations (The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$). Since $3=\{0,1,2\}$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$), an element is written $x=(x_0,x_1,x_2)$ with $x_i:=x(i)$, indexed from $0$. For $j<3$ let $e_j\in F^3$ be given by $e_j(j)=1_F$ and $e_j(i)=0_F$ for $i\mathrm{\ne }j$, and put

$L_j:=\{x\in F^3:x_i=0_F\text{ for every }i<3\text{ with }i\mathrm{\ne }j\}\mathrm{.}$

Then:

1. $L_j=\{\lambda e_j:\lambda \in F\}=\mathrm{span}\{e_j\}$, so each $L_j$ is a linear subspace of $F^3$ (Linear subspace of a vector space, Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$);
2. $F^3={⨁}_{j<3}{L}_j$ (Internal direct sum $V={⨁}_{i<n}{U}_i$: the sum is everything and each summand meets the sum of the others only in $0_V$), the unique decomposition of $x\in F^3$ being $x=x_0{e}_0+x_1{e}_1+x_2{e}_2$;
3. $F^0$ is the zero space: it has exactly one element, the empty function.

The sets $L_j$ are called the coordinate lines of $F^3$; the word "line" is used informally, since dimension is not available here and nothing below uses it.

• Every index here starts at $0$. The coordinates are $x_0,x_1,x_2$ and the summands are $L_0,L_1,L_2$, because $3=\{0,1,2\}$ as a von Neumann natural. Writing the same example with coordinates $x_1,x_2,x_3$ would not match the definition of $F^n$ used here (The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$).

• $F^0$ is not empty. There is exactly one function $\mathrm{\varnothing }\to F$, so $F^0$ has exactly one element and is the zero space. It is also the internal direct sum of the empty family of its linear subspaces (Internal direct sum $V={⨁}_{i<n}{U}_i$: the sum is everything and each summand meets the sum of the others only in $0_V$), which is the only space that is.

• Nothing above is special to $3$. The same computation works for any $n\in \mathbb{N}$ and gives $F^n={⨁}_{j<n}{L}_j$, at $n=0$ degenerating to the statement that the zero space is the direct sum of the empty family. It is written out at $n=3$ so that the finite sums are the explicit $(u_0+u_1)+u_2$ rather than an induction.

Let $F$ be a field (Field) and let $F^{\mathbb{N}}$ be the function space of all families $x:\mathbb{N}\to F$ with the pointwise operations (The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$), written $x=(x_n{)}_{n\in \mathbb{N}}$ with $x_n:=x(n)$; the index runs over $\mathbb{N}$, which contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)). Put

$E:=\{x\in F^{\mathbb{N}}:\text{there is }N\in \mathbb{N}\text{ with }{x}_n=0_F\text{ for every }n\ge N\},$

the set of eventually zero families, and for $k\in \mathbb{N}$ let $e_k\in F^{\mathbb{N}}$ be the standard unit family given by $e_k(k)=1_F$ and $e_k(n)=0_F$ for $n\mathrm{\ne }k$. Then:

1. $E$ is a linear subspace of $F^{\mathbb{N}}$ (Linear subspace of a vector space);
2. $E=\mathrm{span}\{e_k:k\in \mathbb{N}\}$ (Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$);
3. $E\mathrm{\ne }{F}^{\mathbb{N}}$, the constant family at $1_F$ lying outside $E$.

• The spanning set is infinite and every combination is finite. The span of an infinite set consists of the vectors built from finitely many of its elements ($\mathrm{span}(S)$ is exactly the set of linear combinations of finite lists of elements of $S$, and $\mathrm{span}(\mathrm{\varnothing })=\{0_V\}$), which is exactly why the span of all the $e_k$ is the eventually zero families and not all of $F^{\mathbb{N}}$. Claim 3 is the concrete form of that distinction.

• No topology and no convergence are involved. "Eventually zero" is a purely algebraic condition on a family indexed by $\mathbb{N}$: some tail is identically $0_F$. Nothing here needs an order or a metric on $F$, and $F$ is an arbitrary field.

• The witness $N$ is not unique and nothing above assumes it is. If $N$ witnesses that $x$ is eventually zero then so does every larger natural number, which is what makes the argument in step 1.3 work: two families are handled by taking the larger of their witnesses.

False claim: if $V$ is a vector space over a field $F$ and $W\subseteq V$ contains $0_V$ and is closed under the vector addition, then $W$ is a linear subspace of $V$ (Linear subspace of a vector space).

The first quadrant of ${\mathbb{R}}^2$ refutes it. Take $F=\mathbb{R}$ (The reals form a field) and $V={\mathbb{R}}^2$ (The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$), and put

$Q:=\{x\in {\mathbb{R}}^2:0\le x_0\text{ and }0\le x_1\}\mathrm{.}$

Then $0_V\in Q$ and $Q$ is closed under addition, but the vector $e_0$ with coordinates $(1,0)$ lies in $Q$ while $(-1)e_0$, with coordinates $(-1,0)$, does not. So $Q$ is not closed under scalar multiplication and is not a linear subspace.

• Exactly one of the three conditions fails. $Q$ satisfies (W1) and (W2) and fails (W3), and it fails it at a single scalar, $-1$. The reverse failure, a subset closed under scalar multiplication but not under addition, is recorded in The union of the two coordinate axes of $F^2$ is closed under scalar multiplication and is not closed under addition, so neither closure condition implies the other, so neither closure condition implies the other.

• What the order is doing here. The example needs a field in which some element is not the negative of a nonnegative one, so it needs an order; over an arbitrary field there is no "first quadrant" to speak of. That is why this witness is stated over $\mathbb{R}$ while its companion is stated over an arbitrary field.

• $Q$ is closed under multiplication by nonnegative scalars. If $0\le \lambda$ and $x\in Q$ then $\lambda x\in Q$, by the closure of $P$ under multiplication together with the zero cases. So the failure is confined to the negative scalars; a subset with this weaker closure property is a cone, not a linear subspace, and the difference is exactly what the example isolates.

False claim: if $V$ is a vector space over a field $F$ and $W\subseteq V$ contains $0_V$ and is closed under scalar multiplication, then $W$ is a linear subspace of $V$ (Linear subspace of a vector space).

The union of the two coordinate axes of $F^2$ refutes it, over any field $F$. Let $e_0,e_1\in F^2$ be the vectors with coordinates $(1_F,0_F)$ and $(0_F,1_F)$ (The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$), let $L_j:=\mathrm{span}\{e_j\}$ (Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$), and put $A:=L_0\cup L_1$. Then $0_V\in A$ and $A$ is closed under scalar multiplication, while $e_0,e_1\in A$ and $e_0+e_1\mathrm{\notin }A$.

Together with The first quadrant of ${\mathbb{R}}^2$ contains $0$ and is closed under addition and is not a linear subspace, since it is not closed under multiplication by $-1$, which exhibits a subset closed under addition and not under scalar multiplication, this shows that neither of the two closure conditions in Linear subspace of a vector space implies the other.