Education: Subspace and Spanning sets

Subspace:


Definition 2.1

For any vector space, a subspace is a subset that is itself a vector space, under the inherited operations.

Example 2.2

The plane from the prior subsection,

P=\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, x+y+z=0\}

is a subspace of

\mathbb{R}^3

. As specified in the definition, the operations are the ones that are inherited from the larger space, that is, vectors add in

P

as they add in

\mathbb{R}^3

\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix}

and scalar multiplication is also the same as it is in

\mathbb{R}^3

. To show that

P

is a subspace, we need only note that it is a subset and then verify that it is a space. Checking that

P

satisfies the conditions in the definition of a vector space is routine. For instance, for closure under addition, just note that if the summands satisfy that

x_1+y_1+z_1=0

and

x_2+y_2+z_2=0

then the sum satisfies that

(x_1+x_2)+(y_1+y_2)+(z_1+z_2)=(x_1+y_1+z_1)+(x_2+y_2+z_2)=0

Example 2.3

The

x

-axis in

\mathbb{R}^2

is a subspace where the addition and scalar multiplication operations are the inherited ones.

\begin{pmatrix} x_1 \\ 0 \end{pmatrix} + \begin{pmatrix} x_2 \\ 0 \end{pmatrix} = \begin{pmatrix} x_1+x_2 \\ 0 \end{pmatrix} \qquad r\cdot\begin{pmatrix} x \\ 0 \end{pmatrix} =\begin{pmatrix} rx \\ 0 \end{pmatrix}

As above, to verify that this is a subspace, we simply note that it is a subset and then check that it satisfies the conditions in definition of a vector space. For instance, the two closure conditions are satisfied: (1) adding two vectors with a second component of zero results in a vector with a second component of zero, and (2) multiplying a scalar times a vector with a second component of zero results in a vector with a second component of zero.

Example 2.4

Another subspace of

\mathbb{R}^2

\{\begin{pmatrix} 0 \\ 0 \end{pmatrix}\}

its trivial subspace.

Any vector space has a trivial subspace

\{\vec{0}\,\}

. At the opposite extreme, any vector space has itself for a subspace. These two are the improper subspaces. Other subspaces are proper.

Example 2.5

The condition in the definition requiring that the addition and scalar multiplication operations must be the ones inherited from the larger space is important. Consider the subset

\{1\}

of the vector space

\mathbb{R}^1

. Under the operations

1+1=1

and

r\cdot 1=1

that set is a vector space, specifically, a trivial space. But it is not a subspace of

\mathbb{R}^1

because those aren't the inherited operations, since of course

\mathbb{R}^1

has

1+1=2

Example 2.6

All kinds of vector spaces, not just

\mathbb{R}^n

's, have subspaces. The vector space of cubic polynomials

\{a+bx+cx^2+dx^3\,\big|\, a,b,c,d\in\mathbb{R}\}

has a subspace comprised of all linear polynomials

\{m+nx\,\big|\, m,n\in\mathbb{R}\}

Example 2.7

Another example of a subspace not taken from an

\mathbb{R}^n

is one from the examples following the definition of a vector space. The space of all real-valued functions of one real variable

f:\mathbb{R}\to \mathbb{R}

has a subspace of functions satisfying the restriction

(d^2\,f/dx^2)+f=0

Example 2.8

Being vector spaces themselves, subspaces must satisfy the closure conditions. The set

\mathbb{R}^+

is not a subspace of the vector space

\mathbb{R}^1

because with the inherited operations it is not closed under scalar multiplication: if

\vec{v}=1

then

-1\cdot\vec{v}\not\in\mathbb{R}^+

The next result says that Example 2.8 is prototypical. The only way that a subset can fail to be a subspace (if it is nonempty and the inherited operations are used) is if it isn't closed.

Lemma 2.9

For a nonempty subset

S

of a vector space, under the inherited operations, the following are equivalent statements.

$S$ is a subspace of that vector space
$S$ is closed under linear combinations of pairs of vectors: for any vectors $\vec{s}_1,\vec{s}_2\in S$ and scalars $r_1,r_2$ the vector $r_1\vec{s}_1+r_2\vec{s}_2$ is in $S$
$S$ is closed under linear combinations of any number of vectors: for any vectors $\vec{s}_1,\ldots,\vec{s}_n\in S$ and scalars $r_1, \ldots,r_n$ the vector $r_1\vec{s}_1+\cdots+r_n\vec{s}_n$ is in $S$ .

Briefly, the way that a subset gets to be a subspace is by being closed under linear combinations.

Proof

We will show this equivalence by establishing that

(1)\implies (3)\implies (2)\implies (1)

. This strategy is suggested by noticing that

(1)\implies (3)

and

(3)\implies (2)

are easy and so we need only argue the single implication

(2)\implies (1)

For that argument, assume that

S

is a nonempty subset of a vector space

V

and that

S

is closed under combinations of pairs of vectors. We will show that

S

is a vector space by checking the conditions.

The first item in the vector space definition has five conditions. First, for closure under addition, if

\vec{s}_1,\vec{s}_2\in S

then

\vec{s}_1+\vec{s}_2\in S

, as

\vec{s}_1+\vec{s}_2=1\cdot\vec{s}_1+1\cdot\vec{s}_2

. Second, for any

\vec{s}_1,\vec{s}_2\in S

, because addition is inherited from

V

, the sum

\vec{s}_1+\vec{s}_2

S

equals the sum

\vec{s}_1+\vec{s}_2

V

, and that equals the sum

\vec{s}_2+\vec{s}_1

V

(because

V

is a vector space, its addition is commutative), and that in turn equals the sum

\vec{s}_2+\vec{s}_1

S

. The argument for the third condition is similar to that for the second. For the fourth, consider the zero vector of

V

and note that closure of

S

under linear combinations of pairs of vectors gives that (where

\vec{s}

is any member of the nonempty set

S

)

0\cdot\vec{s}+0\cdot\vec{s}=\vec{0}

is in

S

; showing that

\vec{0}

acts under the inherited operations as the additive identity of

S

is easy. The fifth condition is satisfied because for any

\vec{s}\in S

, closure under linear combinations shows that the vector

0\cdot\vec{0}+(-1)\cdot\vec{s}

is in

S

; showing that it is the additive inverse of

\vec{s}

under the inherited operations is routine.

The checks for item 2 are similar.

We usually show that a subset is a subspace with

(2)\implies (1)

Remark 2.10

At the start of this chapter we introduced vector spaces as collections in which linear combinations are "sensible". The above result speaks to this.

The vector space definition has ten conditions but eight of them— the conditions not about closure— simply ensure that referring to the operations as an "addition" and a "scalar multiplication" is sensible. The proof above checks that these eight are inherited from the surrounding vector space provided that the nonempty set

S

satisfies Lemma 2.9's statement (2) (e.g., commutativity of addition in

S

follows right from commutativity of addition in

V

). So, in this context, this meaning of "sensible" is automatically satisfied.

In assuring us that this first meaning of the word is met, the result draws our attention to the second meaning of "sensible". It has to do with the two remaining conditions, the closure conditions. Above, the two separate closure conditions inherent in statement (1) are combined in statement (2) into the single condition of closure under all linear combinations of two vectors, which is then extended in statement (3) to closure under combinations of any number of vectors. The latter two statements say that we can always make sense of an expression like

r_1\vec{s}_1+r_2\vec{s}_2

, without restrictions on the

r

's— such expressions are "sensible" in that the vector described is defined and is in the set

S

This second meaning suggests that a good way to think of a vector space is as a collection of unrestricted linear combinations. The next two examples take some spaces and describe them in this way. That is, in these examples we parametrize, just as we did in Chapter One to describe the solution set of a homogeneous linear system.

Example 2.11

This subset of

\mathbb{R}^3

S=\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, x-2y+z=0\}

is a subspace under the usual addition and scalar multiplication operations of column vectors (the check that it is nonempty and closed under linear combinations of two vectors is just like the one inExample 2.2). To parametrize, we can take

x-2y+z=0

to be a one-equation linear system and expressing the leading variable in terms of the free variables

x=2y-z

S =\{\begin{pmatrix} 2y-z \\ y \\ z \end{pmatrix}\,\big|\, y,z\in\mathbb{R}\} =\{y\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}+z\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\,\big|\, y,z\in\mathbb{R}\}

Now the subspace is described as the collection of unrestricted linear combinations of those two vectors. Of course, in either description, this is a plane through the origin.

Example 2.12

This is a subspace of the

2 \! \times \! 2

matrices

L=\{\begin{pmatrix} a &0 \\ b &c \end{pmatrix} \,\big|\, a+b+c=0\}

(checking that it is nonempty and closed under linear combinations is easy). To parametrize, express the condition as

a=-b-c

L =\{\begin{pmatrix} -b-c &0 \\ b &c \end{pmatrix} \,\big|\, b,c\in\mathbb{R}\} =\{b\begin{pmatrix} -1 &0 \\ 1 &0 \end{pmatrix} +c\begin{pmatrix} -1 &0 \\ 0 &1 \end{pmatrix} \,\big|\, b,c\in\mathbb{R}\}

As above, we've described the subspace as a collection of unrestricted linear combinations (by coincidence, also of two elements).

Spanning:

Definition 2.13
The span(or linear closure) of a nonempty subset $S$ of a vector space is the set of all linear combinations of vectors from $S$ .

[S] =\{ c_1\vec{s}_1+\cdots+c_n\vec{s}_n \,\big|\, c_1,\ldots, c_n\in\mathbb{R} \text{ and } \vec{s}_1,\ldots,\vec{s}_n\in S \}

The span of the empty subset of a vector space is the trivial subspace.

No notation for the span is completely standard. The square brackets used here are common, but so are "

\mbox{span}(S)

" and "

\mbox{sp}(S)

Remark 2.14

In Chapter One, after we showed that the solution set of a homogeneous linear system can be written as

\{c_1\vec{\beta}_1+\cdots+c_k\vec{\beta}_k\,\big|\, c_1,\ldots,c_k\in\mathbb{R}\}

, we described that as the set "generated" by the

{\vec{\beta}}

's. We now have the technical term; we call that the "span" of the set

\{\vec{\beta}_1,\ldots,\vec{\beta}_k\}

Recall also the discussion of the "tricky point" in that proof. The span of the empty set is defined to be the set

\{\vec{0}\}

because we follow the convention that a linear combination of no vectors sums to

\vec{0}

. Besides, defining the empty set's span to be the trivial subspace is a convienence in that it keeps results like the next one from having annoying exceptional cases.

Lemma 2.15

In a vector space, the span of any subset is a subspace.

Proof

Call the subset

S

. If

S

is empty then by definition its span is the trivial subspace. If

S

is not empty then by Lemma 2.9 we need only check that the span

[S]

is closed under linear combinations. For a pair of vectors from that span,

\vec{v}=c_1\vec{s}_1+\cdots+c_n\vec{s}_n

and

\vec{w}=c_{n+1}\vec{s}_{n+1}+\cdots+c_m\vec{s}_m

, a linear combination

p\cdot(c_1\vec{s}_1+\cdots+c_n\vec{s}_n)+ r\cdot(c_{n+1}\vec{s}_{n+1}+\cdots+c_m\vec{s}_m)

= pc_1\vec{s}_1+\cdots+pc_n\vec{s}_n +rc_{n+1}\vec{s}_{n+1}+\cdots+rc_m\vec{s}_m

(

p

r

scalars) is a linear combination of elements of

S

and so is in

[S]

(possibly some of the

\vec{s}_i

's forming

\vec{v}

equal some of the

\vec{s}_j

's from

\vec{w}

, but it does not matter).

The converse of the lemma holds: any subspace is the span of some set, because a subspace is obviously the span of the set of its members. Thus a subset of a vector space is a subspace if and only if it is a span. This fits the intuition that a good way to think of a vector space is as a collection in which linear combinations are sensible.

Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset

S

of a vector space is the smallest subspace containing all the members of

S

Example 2.16

In any vector space

V

, for any vector

\vec{v}

, the set

\{r\cdot\vec{v} \,\big|\, r\in\mathbb{R}\}

is a subspace of

V

. For instance, for any vector

\vec{v}\in\mathbb{R}^3

, the line through the origin containing that vector,

\{k\vec{v}\,\big|\, k\in\mathbb{R} \}

is a subspace of

\mathbb{R}^3

. This is true even when

\vec{v}

is the zero vector, in which case the subspace is the degenerate line, the trivial subspace.

Example 2.17

The span of this set is all of

\mathbb{R}^2

\{\begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ -1 \end{pmatrix}\}

To check this we must show that any member of

\mathbb{R}^2

is a linear combination of these two vectors. So we ask: for which vectors (with real components

x

and

y

) are there scalars

c_1

and

c_2

such that this holds?

c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}

Gauss' method

\begin{array}{rcl} \begin{array}{*{2}{rc}r} c_1 &+ &c_2 &= &x \\ c_1 &- &c_2 &= &y \end{array} &\xrightarrow[]{-\rho_1+\rho_2} &\begin{array}{*{2}{rc}r} c_1 &+ &c_2 &= &x \\ & &-2c_2 &= &-x+y \end{array} \end{array}

with back substitution gives

c_2=(x-y)/2

and

c_1=(x+y)/2

. These two equations show that for any

x

and

y

that we start with, there are appropriate coefficients

c_1

and

c_2

making the above vector equation true. For instance, for

x=1

and

y=2

the coefficients

c_2=-1/2

and

c_1=3/2

will do. That is, any vector in

\mathbb{R}^2

can be written as a linear combination of the two given vectors.

Since spans are subspaces, and we know that a good way to understand a subspace is to parametrize its description, we can try to understand a set's span in that way.

Example 2.18

Consider, in

\mathcal{P}_2

, the span of the set

\{3x-x^2, 2x\}

. By the definition of span, it is the set of unrestricted linear combinations of the two

\{c_1(3x-x^2)+c_2(2x)\,\big|\, c_1,c_2\in\mathbb{R}\}

. Clearly polynomials in this span must have a constant term of zero. Is that necessary condition also sufficient?

We are asking: for which members

a_2x^2+a_1x+a_0

\mathcal{P}_2

are there

c_1

and

c_2

such that

a_2x^2+a_1x+a_0=c_1(3x-x^2)+c_2(2x)

? Since polynomials are equal if and only if their coefficients are equal, we are looking for conditions on

a_2

a_1

, and

a_0

satisfying these.

\begin{array}{*{2}{rc}r} -c_1 & & &= &a_2 \\ 3c_1 &+ &2c_2 &= &a_1 \\ & &0 &= &a_0 \end{array}

Gauss' method gives that

c_1=-a_2

c_2=(3/2)a_2+(1/2)a_1

, and

0=a_0

. Thus the only condition on polynomials in the span is the condition that we knew of— as long as

a_0=0

, we can give appropriate coefficients

c_1

and

c_2

to describe the polynomial

a_0+a_1x+a_2x^2

as in the span. For instance, for the polynomial

0-4x+3x^2

, the coefficients

c_1=-3

and

c_2=5/2

will do. So the span of the given set is

\{a_1x+a_2x^2\,\big|\, a_1,a_2\in\mathbb{R}\}

This shows, incidentally, that the set

\{x,x^2\}

also spans this subspace. A space can have more than one spanning set. Two other sets spanning this subspace are

\{x,x^2,-x+2x^2\}

and

\{x,x+x^2,x+2x^2,\ldots\,\}

. (Naturally, we usually prefer to work with spanning sets that have only a few members.)

Example 2.19

These are the subspaces of

\mathbb{R}^3

that we now know of, the trivial subspace, the lines through the origin, the planes through the origin, and the whole space (of course, the picture shows only a few of the infinitely many subspaces). In the next section we will prove that

\mathbb{R}^3

has no other type of subspaces, so in fact this picture shows them all.

The subsets are described as spans of sets, using a minimal number of members, and are shown connected to their supersets. Note that these subspaces fall naturally into levels— planes on one level, lines on another, etc.— according to how many vectors are in a minimal-sized spanning set.

18/09/2014

Subspace and Spanning sets

Subspace:

Spanning: