[ALGEBRA] Subspaces

A subspace is a subset that is itself a vector space under the operations it inherits. This page reduces the ten axioms to a three-condition test and applies it to vectors, polynomials and matrices.

Inheriting the vector-space structure

[DEFINITION]

Vector space

A vector space $V$ over a field $F$ is a non-empty set of vectors equipped with an addition rule $V\times V\to V$ and a scalar-multiplication rule $F\times V\to V$ satisfying:

Closure under addition: if $u,v\in V$ , then $u+v\in V$ .
Associativity of addition: $(u+v)+w=u+(v+w)$ .
Commutativity of addition: $u+v=v+u$ .
Zero vector: there is a $0\in V$ such that $0+u=u+0=u$ .
Additive inverses: for every $u\in V$ , there is a $-u\in V$ such that $u+(-u)=(-u)+u=0$ .
Closure under scalar multiplication: if $u\in V$ and $\lambda\in F$ , then $\lambda u\in V$ .
Associativity of scalar multiplication: $\lambda(\mu u)=(\lambda\mu)u$ .
Multiplication by one: $1u=u$ .
Scalar distributivity: $(\lambda+\mu)u=\lambda u+\mu u$ .
Vector distributivity: $\lambda(u+v)=\lambda u+\lambda v$ .

[DEFINITION]

Subspace

Let $V$ be a vector space over $F$ . A subset $S\subseteq V$ is a subspace of $V$ when it is a vector space over the same field $F$ and uses the addition and scalar multiplication inherited from $V$ . A proper subspace is a subspace that is not equal to $V$ .

Two elements of $S$ are added as they are in $V$ and scalar multiplication uses the same scalars. The sets $\{0\}$ and $V$ are always subspaces of $V$ . When $m\leq n$ , the standard embedding

(x_1,\ldots,x_m)\longmapsto(x_1,\ldots,x_m,\underbrace{0,\ldots,0}_{n-m})

identifies $\mathbb R^m$ with the coordinate subspace $\{(x_1,\ldots,x_m,0,\ldots,0)\}\subseteq\mathbb R^n$ . For $n\ge2$ , a plane through the origin in $\mathbb R^n$ is another example.

Testing closure

Most axioms hold in $S$ automatically because they hold in $V$ , so there is no need to check all ten again.

[THEOREM]

Subspace test

Let $S$ be a subset of a vector space $V$ over $F$ , with the operations inherited from $V$ . Then $S$ is a subspace of $V$ if and only if:

$S$ is non-empty;
$S$ is closed under vector addition: $u,v\in S$ implies $u+v\in S$ ;
$S$ is closed under scalar multiplication: $u\in S$ and $\lambda\in F$ imply $\lambda u\in S$ .

Non-emptiness and scalar closure supply the zero vector and additive inverses, so these three conditions are enough.

[1]:Subspace
[2]:Vector space
[3]:The kernel is a subspace
[4]:The image is a subspace
[5]:a set of polynomial divisors
[6]:subspaces of $M_{2\times2}(\mathbb R)$
8 more

[PROOF]

If $S$ is a subspace, the three properties are part of the vector-space axioms. Conversely, the inherited operations already satisfy associativity, commutativity and distributivity because they do so in $V$ . Choose $u\in S$ using non-emptiness. Scalar closure gives

0u=0\in S

and, for every $u\in S$ ,

(-1)u=-u\in S.

Thus $S$ has the zero vector and additive inverses, while the three listed properties provide the remaining requirements.

[1]:Subspace test

[COROLLARY]

The zero-vector check

Every subspace contains the zero vector of its underlying vector space. Therefore, a subset that does not contain $0$ is not a subspace.

[1]:Subspace test
[2]:Sets that are not subspaces

Keep $\{0\}$ and $\varnothing$ distinct. The set $\{0\}$ has one element and is a subspace; the empty set has none and fails the non-emptiness condition.

Examples and non-examples

[EXAMPLE]

A line through the origin

For a fixed non-zero direction vector $w\in\mathbb R^n$ , consider

S=\{x\in\mathbb R^n:x=\lambda w,\ \lambda\in\mathbb R\}.

The zero vector is in $S$ by taking $\lambda=0$ . If $x_1=\lambda_1w$ and $x_2=\lambda_2w$ , then

x_1+x_2=(\lambda_1+\lambda_2)w\in S.

For $\mu\in\mathbb R$ ,

\mu x_1=(\mu\lambda_1)w\in S.

The subspace test shows that $S$ is a subspace of $\mathbb R^n$ .

[routine]

[1]:Subspace test

The same test applies when membership is a homogeneous condition on a polynomial or a matrix.

[EXAMPLE]

Polynomials satisfying a homogeneous condition

Consider

S=\{p\in P_3(\mathbb R):p(1)=0\}.

The zero polynomial belongs to $S$ . If $p_1,p_2\in S$ , then

(p_1+p_2)(1)=p_1(1)+p_2(1)=0,

and if $p\in S$ and $\lambda\in\mathbb R$ , then

(\lambda p)(1)=\lambda p(1)=0.

Thus $S$ is a subspace of $P_3(\mathbb R)$ . It is not merely $\{0\}$ ; for example, $x-1\in S$ .

[routine]

[1]:Subspace test

[EXAMPLE]

Matrices annihilating a fixed vector

For a fixed $x\in\mathbb R^n$ , let

S=\{A\in M_{m\times n}(\mathbb R):Ax=0\}.

The zero matrix belongs to $S$ . If $A_1,A_2\in S$ , then

(A_1+A_2)x=A_1x+A_2x=0.

If $A\in S$ and $\lambda\in\mathbb R$ , then

(\lambda A)x=\lambda(Ax)=0.

Therefore $S$ is a subspace of $M_{m\times n}(\mathbb R)$ .

[routine]

[1]:Subspace test

To reject a proposed subspace, one failed check is enough: a missing zero vector or a failed closure condition.

[EXAMPLE]

Sets that are not subspaces

For a fixed non-zero $u\in\mathbb R^n$ , the set

\{x\in\mathbb R^n:x=\lambda u,\ -10\leq\lambda\leq10\}

is not closed under addition: $10u$ belongs to it but $10u+10u=20u$ does not.

The affine line

\{x\in\mathbb R^2:x_1+2x_2=1\}

does not contain the zero vector. Likewise, neither

\{p\in P_3(\mathbb R):p(x)=1\text{ for every }x\in\mathbb R\} \qquad\text{nor}\qquad \{p\in P_2(\mathbb R):p(0)=1\}

contains the zero polynomial. Finally, the invertible matrices in $M_{n\times n}(\mathbb R)$ do not form a subspace because the zero matrix is not invertible.

Replacing the right-hand side $1$ by $0$ gives homogeneous conditions such as $p(x)=0$ for every $x$ or $p(0)=0$ and these do define subspaces. The same contrast between homogeneous and non-homogeneous conditions returns with kernels of linear transformations.

[routine]

[1]:Subspace test
[2]:The zero-vector check

Spans build a subspace from chosen vectors.

Subspaces

Vector space

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