Linear Algebra/Subspaces and Spanning sets

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One of the examples that led us to introduce the idea of a vector space was the solution set of a homogeneous system. For instance, we've seen in Example 1.4 such a space that is a planar subset of $\mathbb{R}^3$ . There, the vector space $\mathbb{R}^3$ contains inside it another vector space, the plane.

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 $\mathbb{R}^2$ is

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

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.^[1]

$S$ 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

"The following are equivalent" means that each pair of statements are equivalent.

$(1)\!\iff\!(2) \qquad (2)\!\iff\!(3) \qquad (3)\!\iff\!(1)$

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 $\vec{s}_1+\vec{s}_2$ in $S$ equals the sum $\vec{s}_1+\vec{s}_2$ in $V$ , and that equals the sum $\vec{s}_2+\vec{s}_1$ in $V$ (because $V$ is a vector space, its addition is commutative), and that in turn equals the sum $\vec{s}_2+\vec{s}_1$ in $S$ . 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 and are saved for Problem 14.

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 in Example 2.2). To parametrize, we can take $x - 2 y + z = 0$ to be a one-equation linear system and expressing the leading variable in terms of the free variables $x = 2 y - z$ .

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

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).

Parametrization is an easy technique, but it is important. We shall use it often.

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 " $span(S)$ " and " $sp(S)$ ".

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 ${3 x - x 2,2 x}$ . 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 $a 2 x 2 + a 1 x + a 0$ of $\mathcal{P}_2$ are there $c 1$ and $c 2$ such that $a 2 x 2 + a 1 x + a 0 = c 1 (3 x - x 2) + c 2 (2 x)$ ? 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 $a 0 + a 1 x + a 2 x 2$ as in the span. For instance, for the polynomial $0 - 4 x + 3 x 2$ , 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 ${x, x 2, - x + 2 x 2}$ 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.

So far in this chapter we have seen that to study the properties of linear combinations, the right setting is a collection that is closed under these combinations. In the first subsection we introduced such collections, vector spaces, and we saw a great variety of examples. In this subsection we saw still more spaces, ones that happen to be subspaces of others. In all of the variety we've seen a commonality. Example 2.19 above brings it out: vector spaces and subspaces are best understood as a span, and especially as a span of a small number of vectors. The next section studies spanning sets that are minimal.

[edit] Exercises

This exercise is recommended for all readers.

Problem 1

Which of these subsets of the vector space of $2 \! \times \! 2$ matrices are subspaces under the inherited operations? For each one that is a subspace, parametrize its description. For each that is not, give a condition that fails.

$\{\begin{pmatrix} a &0 \\ 0 &b \end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}$
$\{\begin{pmatrix} a &0 \\ 0 &b \end{pmatrix} \,\big|\, a+b=0\}$
$\{\begin{pmatrix} a &0 \\ 0 &b \end{pmatrix} \,\big|\, a+b=5\}$
$\{\begin{pmatrix} a &c \\ 0 &b \end{pmatrix} \,\big|\, a+b=0, c\in\mathbb{R}\}$

Answer

By Lemma 2.9, to see if each subset of $\mathcal{M}_{2 \! \times \! 2}$ is a subspace, we need only check if it is nonempty and closed.

Yes, it is easily checked to be nonempty and closed. This is a parametrization.

$\{a\begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix} +b\begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}$

By the way, the parametrization also shows that it is a subspace, it is given as the span of the two-matrix set, and any span is a subspace.
Yes; it is easily checked to be nonempty and closed. Alternatively, as mentioned in the prior answer, the existence of a parametrization shows that it is a subspace. For the parametrization, the condition $a + b = 0$ can be rewritten as $a = - b$ . Then we have this.

$\{\begin{pmatrix} -b &0 \\ 0 &b \end{pmatrix} \,\big|\, b\in\mathbb{R}\} =\{b\begin{pmatrix} -1 &0 \\ 0 &1 \end{pmatrix} \,\big|\, b\in\mathbb{R}\}$
No. It is not closed under addition. For instance,

$\begin{pmatrix} 5 &0 \\ 0 &0 \end{pmatrix} +\begin{pmatrix} 5 &0 \\ 0 &0 \end{pmatrix} =\begin{pmatrix} 10 &0 \\ 0 &0 \end{pmatrix}$

is not in the set. (This set is also not closed under scalar multiplication, for instance, it does not contain the zero matrix.)
Yes.

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

This exercise is recommended for all readers.

Problem 2

Is this a subspace of $\mathcal{P}_2$ : $\{a_0+a_1x+a_2x^2\,\big|\, a_0+2a_1+a_2=4\}$ ? If it is then parametrize its description.

Answer

No, it is not closed. In particular, it is not closed under scalar multiplication because it does not contain the zero polynomial.

This exercise is recommended for all readers.

Problem 3

Decide if the vector lies in the span of the set, inside of the space.

$\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ , $\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \}$ , in $\mathbb{R}^3$
$x - x 3$ , ${x 2,2 x + x 2, x + x 3}$ , in $\mathcal{P}_3$
$\begin{pmatrix} 0 &1 \\ 4 &2 \end{pmatrix}$ , $\{\begin{pmatrix} 1 &0 \\ 1 &1 \end{pmatrix}, \begin{pmatrix} 2 &0 \\ 2 &3 \end{pmatrix} \}$ , in $\mathcal{M}_{2 \! \times \! 2}$

Answer

Yes, solving the linear system arising from

$r_1\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+r_2\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} =\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$

$r 1 = 2$ and $r 2 = 1$ .
Yes; the linear system arising from $r 1 (x 2) + r 2 (2 x + x 2) + r 3 (x + x 3) = x - x 3$

$\begin{array}{*{3}{rc}r} & &2r_2 &+ &r_3 &= &1 \\ r_1 &+ &r_2 & & &= &0 \\ & & & &r_3 &= &-1 \end{array}$

gives that $- 1(x 2) + 1(2 x + x 2) - 1(x + x 3) = x - x 3$ .
No; any combination of the two given matrices has a zero in the upper right.

Problem 4

Which of these are members of the span $[{cos 2 x,sin 2 x}]$ in the vector space of real-valued functions of one real variable?

$f (x) = 1$
$f (x) = 3 + x 2$
$f (x) = sin x$
$f (x) = cos(2 x)$

Answer

Yes; it is in that span since $1\cdot\cos^2x+1\cdot\sin^2x=f(x)$ .
No, since $r 1 cos 2 x + r 2 sin 2 x = 3 + x 2$ has no scalar solutions that work for all $x$ . For instance, setting $x$ to be $0$ and $π$ gives the two equations $r_1\cdot 1+r_2\cdot 0=3$ and $r_1\cdot 1+r_2\cdot 0=3+\pi^2$ , which are not consistent with each other.
No; consider what happens on setting $x$ to be $π / 2$ and $3π / 2$ .
Yes, $\cos (2x)=1\cdot\cos^2(x)-1\cdot\sin^2(x)$ .

This exercise is recommended for all readers.

Problem 5

Which of these sets spans $\mathbb{R}^3$ ? That is, which of these sets has the property that any three-tall vector can be expressed as a suitable linear combination of the set's elements?

$\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \}$
$\{ \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \}$
$\{ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} \}$
$\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \}$
$\{ \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 6 \\ 0 \\ 2 \end{pmatrix} \}$

Answer

Yes, for any $x,y,z\in\mathbb{R}$ this equation

$r_1\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +r_2\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} +r_3\begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

has the solution $r 1 = x$ , $r 2 = y / 2$ , and $r 3 = z / 3$ .
Yes, the equation

$r_1\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} +r_2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} +r_3\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

gives rise to this

$\begin{array}{*{3}{rc}r} 2r_1 &+ &r_2 & & &= &x \\ & &r_2 & & &= &y \\ r_1 & & &+ &r_3 &= &z \\ \end{array} \;\xrightarrow[]{-(1/2)\rho_1+\rho_3}\;\;\xrightarrow[]{(1/2)\rho_2+\rho_3}\; \begin{array}{*{3}{rc}r} 2r_1 &+ &r_2 & & &= &x \\ & &r_2 & & &= &y \\ & & & &r_3 &= &-(1/2)x+(1/2)y+z \\ \end{array}$

so that, given any $x$ , $y$ , and $z$ , we can compute that $r 3 = ( - 1 / 2) x + (1 / 2) y + z$ , $r 2 = y$ , and $r 1 = (1 / 2) x - (1 / 2) y$ .
No. In particular, the vector

$\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

cannot be gotten as a linear combination since the two given vectors both have a third component of zero.
Yes. The equation

$r_1\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} +r_2\begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} +r_3\begin{pmatrix} -1\\ 0 \\ 0 \end{pmatrix} +r_4\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

leads to this reduction.

$\left(\begin{array}{*{4}{c}|c} 1 &3 &-1 &2 &x \\ 0 &1 &0 &1 &y \\ 1 &0 &0 &5 &z \end{array}\right) \xrightarrow[]{-\rho_1+\rho_3}\;\xrightarrow[]{3\rho_2+\rho_3} \left(\begin{array}{*{4}{c}|c} 1 &3 &-1 &2 &x \\ 0 &1 &0 &1 &y \\ 0 &0 &1 &6 &-x+3y+z \end{array}\right)$

We have infinitely many solutions. We can, for example, set $r 4$ to be zero and solve for $r 3$ , $r 2$ , and $r 1$ in terms of $x$ , $y$ , and $z$ by the usual methods of back-substitution.
No. The equation

$r_1\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} +r_2\begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} +r_3\begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix} +r_4\begin{pmatrix} 6 \\ 0 \\ 2 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

leads to this reduction.

$\left(\begin{array}{*{4}{c}|c} 2 &3 &5 &6 &x \\ 1 &0 &1 &0 &y \\ 1 &1 &2 &2 &z \end{array}\right) \xrightarrow[-(1/2)\rho_1+\rho_3]{-(1/2)\rho_1+\rho_2} \;\xrightarrow[]{-(1/3)\rho_2+\rho_3} \left(\begin{array}{*{4}{c}|c} 2 &3 &5 &6 &x \\ 0 &-3/2 &-3/2 &-3 &-(1/2)x+y \\ 0 &0 &0 &0 &-(1/3)x-(1/3)y+z \end{array}\right)$

This shows that not every three-tall vector can be so expressed. Only the vectors satisfying the restriction that $- (1 / 3) x - (1 / 3) y + z = 0$ are in the span. (To see that any such vector is indeed expressible, take $r 3$ and $r 4$ to be zero and solve for $r 1$ and $r 2$ in terms of $x$ , $y$ , and $z$ by back-substitution.)

This exercise is recommended for all readers.

Problem 6

Parametrize each subspace's description. Then express each subspace as a span.

The subset $\{\begin{pmatrix} a &b &c \end{pmatrix}\,\big|\, a-c=0\}$ of the three-wide row vectors
This subset of $\mathcal{M}_{2 \! \times \! 2}$

$\{\begin{pmatrix} a &b \\ c &d \end{pmatrix} \,\big|\, a+d=0\}$
This subset of $\mathcal{M}_{2 \! \times \! 2}$

$\{\begin{pmatrix} a &b \\ c &d \end{pmatrix} \,\big|\, 2a-c-d=0 \text{ and } a+3b=0 \}$
The subset $\{a+bx+cx^3\,\big|\, a-2b+c=0\}$ of $\mathcal{P}_3$
The subset of $\mathcal{P}_2$ of quadratic polynomials $p$ such that $p (7) = 0$

Answer

$\{\begin{pmatrix} c &b &c \end{pmatrix}\,\big|\, b,c\in\mathbb{R}\} =\{b\begin{pmatrix} 0 &1 &0 \end{pmatrix}+c\begin{pmatrix} 1 &0 &1 \end{pmatrix} \,\big|\, b,c\in\mathbb{R}\}$ The obvious choice for the set that spans is $\{\begin{pmatrix} 0 &1 &0 \end{pmatrix},\begin{pmatrix} 1 &0 &1 \end{pmatrix}\}$ .
$\{\begin{pmatrix} -d &b \\ c &d \end{pmatrix} \,\big|\, b,c,d\in\mathbb{R}\} =\{b\begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix} +c\begin{pmatrix} 0 &0 \\ 1 &0 \end{pmatrix} +d\begin{pmatrix} -1 &0 \\ 0 &1 \end{pmatrix} \,\big|\, b,c,d\in\mathbb{R}\}$ One set that spans this space consists of those three matrices.
The system

$\begin{array}{*{4}{rc}r} a &+ &3b & & & & &= &0 \\ 2a & & & &-c &- &d &= &0 \end{array}$

gives $b = - (c + d) / 6$ and $a = (c + d) / 2$ . So one description is this.

$\{c\begin{pmatrix} 1/2 &-1/6 \\ 1 &0 \end{pmatrix} +d\begin{pmatrix} 1/2 &-1/6 \\ 0 &1 \end{pmatrix} \,\big|\, c,d\in\mathbb{R}\}$

That shows that a set spanning this subspace consists of those two matrices.
The $a = 2 b - c$ gives $\{(2b-c)+bx+cx^3 \,\big|\, b,c\in\mathbb{R}\} =\{b(2+x)+c(-1+x^3) \,\big|\, b,c\in\mathbb{R}\}$ . So the subspace is the span of the set ${2 + x, - 1 + x 3}$ .
The set $\{a+bx+cx^2\,\big|\, a+7b+49c=0\}$ parametrized as $\{b(-7+x)+c(-49+x^2)\,\big|\, b,c\in\mathbb{R}\}$ has the spanning set ${ - 7 + x, - 49 + x 2}$ .

This exercise is recommended for all readers.

Problem 7

Find a set to span the given subspace of the given space. (Hint. Parametrize each.)

the $x z$ -plane in $\mathbb{R}^3$
$\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, 3x+2y+z=0\}$ in $\mathbb{R}^3$
$\{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\,\big|\, 2x+y+w=0 \text{ and } y+2z=0\}$ in $\mathbb{R}^4$
$\{a_0+a_1x+a_2x^2+a_3x^3\,\big|\, a_0+a_1=0 \text{ and } a_2-a_3=0\}$ in $\mathcal{P}_3$
The set $\mathcal{P}_4$ in the space $\mathcal{P}_4$
$\mathcal{M}_{2 \! \times \! 2}$ in $\mathcal{M}_{2 \! \times \! 2}$

Answer

Each answer given is only one out of many possible.

We can parametrize in this way

$\{\begin{pmatrix} x \\ 0 \\ z \end{pmatrix}\,\big|\, x,z\in\mathbb{R}\} =\{x\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +z\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\,\big|\, x,z\in\mathbb{R}\}$

giving this for a spanning set.

$\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\}$

Problem 8: Parametrize it with

$\{y\begin{pmatrix} -2/3 \\ 1 \\ 0 \end{pmatrix}+z\begin{pmatrix} -1/3 \\ 0 \\ 1 \end{pmatrix} \,\big|\, y,z\in\mathbb{R} \}$ to get $\{\begin{pmatrix} -2/3 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1/3 \\ 0 \\ 1 \end{pmatrix} \}$ .

$\{\begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2 \\ 0 \\ 0 \\ 1 \end{pmatrix} \}$
Parametrize the description as $\{-a_1+a_1x+a_3x^2+a_3x^3\,\big|\, a_1,a_3\in\mathbb{R} \}$ to get ${ - 1 + x, x 2 + x 3}.$
${1, x, x 2, x 3, x 4}$
$\{ \begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &0 \\ 1 &0 \end{pmatrix}, \begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix} \}$

Problem 9

Is $\mathbb{R}^2$ a subspace of $\mathbb{R}^3$ ?

Answer

Technically, no. Subspaces of $\mathbb{R}^3$ are sets of three-tall vectors, while $\mathbb{R}^2$ is a set of two-tall vectors. Clearly though, $\mathbb{R}^2$ is "just like" this subspace of $\mathbb{R}^3$ .

$\{\begin{pmatrix} x \\ y \\ 0 \end{pmatrix}\,\big|\, x,y\in\mathbb{R}\}$

This exercise is recommended for all readers.

Problem 10

Decide if each is a subspace of the vector space of real-valued functions of one real variable.

The even functions $\{f:\mathbb{R}\to \mathbb{R} \,\big|\, f(-x)=f(x) \text{ for all } x\}$ . For example, two members of this set are $f 1 (x) = x 2$ and $f 2 (x) = cos(x)$ .
The odd functions $\{f:\mathbb{R}\to \mathbb{R} \,\big|\, f(-x)=-f(x) \text{ for all } x\}$ . Two members are $f 3 (x) = x 3$ and $f 4 (x) = sin(x)$ .

Answer

Of course, the addition and scalar multiplication operations are the ones inherited from the enclosing space.

This is a subspace. It is not empty as it contains at least the two example functions given. It is closed because if $f 1, f 2$ are even and $c 1, c 2$ are scalars then we have this.

$(c_1f_1+c_2f_2)\,(-x) =c_1\,f_1(-x)+c_2\,f_2(-x) =c_1\,f_1(x)+c_2\,f_2(x) =(c_1f_1+c_2f_2)\,(x)$
This is also a subspace; the check is similar to the prior one.

Problem 11

Example 2.16 says that for any vector $\vec{v}$ that is an element of a vector space $V$ , the set $\{r\cdot\vec{v}\,\big|\, r\in\mathbb{R}\}$ is a subspace of $V$ . (This is of course, simply the span of the singleton set $\{\vec{v}\}$ .) Must any such subspace be a proper subspace, or can it be improper?

Answer

It can be improper. If $\vec{v}=\vec{0}$ then this is a trivial subspace. At the opposite extreme, if the vector space is $\mathbb{R}^1$ and $\vec{v}\neq\vec{0}\,$ then the subspace is all of $\mathbb{R}^1$ .

Problem 12

An example following the definition of a vector space shows that the solution set of a homogeneous linear system is a vector space. In the terminology of this subsection, it is a subspace of $\mathbb{R}^n$ where the system has $n$ variables. What about a non-homogeneous linear system; do its solutions form a subspace (under the inherited operations)?

Answer

No, such a set is not closed. For one thing, it does not contain the zero vector.

Problem 13

Example 2.19 shows that $\mathbb{R}^3$ has infinitely many subspaces. Does every nontrivial space have infinitely many subspaces?

Answer

No. The only subspaces of $\mathbb{R}^1$ are the space itself and its trivial subspace. Any subspace $S$ of $\mathbb{R}$ that contains a nonzero member $\vec{v}$ must contain the set of all of its scalar multiples $\{r\cdot\vec{v}\,\big|\, r\in\mathbb{R}\}$ . But this set is all of $\mathbb{R}$ .

Problem 14

Finish the proof of Lemma 2.9.

Answer

Item (1) is checked in the text.

Item (2) has five conditions. First, for closure, if $c\in\mathbb{R}$ and $\vec{s}\in S$ then $c\cdot\vec{s}\in S$ as $c\cdot\vec{s}=c\cdot\vec{s}+0\cdot\vec{0}$ . Second, because the operations in $S$ are inherited from $V$ , for $c,d\in\mathbb{R}$ and $\vec{s}\in S$ , the scalar product $(c+d)\cdot\vec{s}\,$ in $S$ equals the product $(c+d)\cdot\vec{s}\,$ in $V$ , and that equals $c\cdot\vec{s}+d\cdot\vec{s}\,$ in $V$ , which equals $c\cdot\vec{s}+d\cdot\vec{s}\,$ in $S$ .

The check for the third, fourth, and fifth conditions are similar to the second conditions's check just given.

Problem 15

Show that each vector space has only one trivial subspace.

Answer

An exercise in the prior subsection shows that every vector space has only one zero vector (that is, there is only one vector that is the additive identity element of the space). But a trivial space has only one element and that element must be this (unique) zero vector.

This exercise is recommended for all readers.

Problem 16

Show that for any subset $S$ of a vector space, the span of the span equals the span $[[S]] = [S]$ . (Hint. Members of $[S]$ are linear combinations of members of $S$ . Members of $[[S]]$ are linear combinations of linear combinations of members of $S$ .)

Answer

As the hint suggests, the basic reason is the Linear Combination Lemma from the first chapter. For the full proof, we will show mutual containment between the two sets.

The first containment $[ [S] ]\supseteq[S]$ is an instance of the more general, and obvious, fact that for any subset $T$ of a vector space, $[T]\supseteq T$ .

For the other containment, that $[ [S] ]\subseteq[S]$ , take $m$ vectors from $[S]$ , namely $c_{1,1}\vec{s}_{1,1}+\cdots+c_{1,n_1}\vec{s}_{1,n_1}$ , ..., $c_{1,m}\vec{s}_{1,m}+\cdots+c_{1,n_m}\vec{s}_{1,n_m}$ , and note that any linear combination of those

$r_1(c_{1,1}\vec{s}_{1,1}+\cdots+c_{1,n_1}\vec{s}_{1,n_1})+\cdots +r_m(c_{1,m}\vec{s}_{1,m}+\cdots+c_{1,n_m}\vec{s}_{1,n_m})$

is a linear combination of elements of $S$

$= (r_1c_{1,1})\vec{s}_{1,1}+\cdots+(r_1c_{1,n_1})\vec{s}_{1,n_1}+\cdots +(r_mc_{1,m})\vec{s}_{1,m}+\cdots+(r_mc_{1,n_m})\vec{s}_{1,n_m}$

and so is in $[S]$ . That is, simply recall that a linear combination of linear combinations (of members of $S$ ) is a linear combination (again of members of $S$ ).

Problem 17

All of the subspaces that we've seen use zero in their description in some way. For example, the subspace in Example 2.3 consists of all the vectors from $\mathbb{R}^2$ with a second component of zero. In contrast, the collection of vectors from $\mathbb{R}^2$ with a second component of one does not form a subspace (it is not closed under scalar multiplication). Another example is Example 2.2, where the condition on the vectors is that the three components add to zero. If the condition were that the three components add to one then it would not be a subspace (again, it would fail to be closed). This exercise shows that a reliance on zero is not strictly necessary. Consider the set

$\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, x+y+z=1\}$

under these operations.

$\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} \qquad r\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} rx-r+1 \\ ry \\ rz \end{pmatrix}$

Show that it is not a subspace of $\mathbb{R}^3$ . (Hint. See Example 2.5).
Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply.
Show that any subspace of $\mathbb{R}^3$ must pass through the origin, and so any subspace of $\mathbb{R}^3$ must involve zero in its description. Does the converse hold? Does any subset of $\mathbb{R}^3$ that contains the origin become a subspace when given the inherited operations?

Answer

It is not a subspace because these are not the inherited operations. For one thing, in this space,

$0\cdot\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

while this does not, of course, hold in $\mathbb{R}^3$ .
We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If $r 1, r 2, x 1, x 2, y 1, y 2, z 1, z 2$ are in $\mathbb{R}$ then

$r_1\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} +r_2\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} r_1x_1-r_1+1 \\ r_1y_1 \\ r_1z_1 \end{pmatrix} +\begin{pmatrix} r_2x_2-r_2+1 \\ r_2y_2 \\ r_2z_2 \end{pmatrix} =\begin{pmatrix} r_1x_1-r_1+r_2x_2-r_2+1 \\ r_1y_1+r_2y_2 \\ r_1z_1+r_2z_2 \end{pmatrix}$

(note that the definition of addition in this space is that the first components combine as $(r 1 x 1 - r 1 + 1) + (r 2 x 2 - r 2 + 1) - 1$ , so the first component of the last vector does not say " $+ 2$ "). Adding the three components of the last vector gives $r_1(x_1-1+y_1+z_1)+r_2(x_2-1+y_2+z_2)+1=r_1\cdot0+r_2\cdot0+1=1$ . Most of the other checks of the conditions are easy (although the oddness of the operations keeps them from being routine). Commutativity of addition goes like this.

$\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} =\begin{pmatrix} x_2+x_1-1 \\ y_2+y_1 \\ z_2+z_1 \end{pmatrix} =\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}+\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}$

Associativity of addition has

$(\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}) +\begin{pmatrix} x_3 \\ y_3 \\ z_3 \end{pmatrix} =\begin{pmatrix} (x_1+x_2-1)+x_3-1 \\ (y_1+y_2)+y_3 \\ (z_1+z_2)+z_3 \end{pmatrix}$

while

$\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} +(\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}+\begin{pmatrix} x_3 \\ y_3 \\ z_3 \end{pmatrix}) =\begin{pmatrix} x_1+(x_2+x_3-1)-1 \\ y_1+(y_2+y_3) \\ z_1+(z_2+z_3) \end{pmatrix}$

and they are equal. The identity element with respect to this addition operation works this way

$\begin{pmatrix} x \\ y \\ z \end{pmatrix}+\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} =\begin{pmatrix} x+1-1 \\ y+0 \\ z+0 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

and the additive inverse is similar.

$\begin{pmatrix} x \\ y \\ z \end{pmatrix}+\begin{pmatrix} -x+2 \\ -y \\ -z \end{pmatrix} =\begin{pmatrix} x+(-x+2)-1 \\ y-y \\ z-z \end{pmatrix} =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

The conditions on scalar multiplication are also easy. For the first condition,

$(r+s)\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} (r+s)x-(r+s)+1 \\ (r+s)y \\ (r+s)z \end{pmatrix}$

while

$r\begin{pmatrix} x \\ y \\ z \end{pmatrix}+s\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} rx-r+1 \\ ry \\ rz \end{pmatrix}+\begin{pmatrix} sx-s+1 \\ sy \\ sz \end{pmatrix} =\begin{pmatrix} (rx-r+1)+(sx-s+1)-1 \\ ry+sy \\ rz+sz \end{pmatrix}$

and the two are equal. The second condition compares

$r\cdot(\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}) =r\cdot\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} =\begin{pmatrix} r(x_1+x_2-1)-r+1 \\ r(y_1+y_2) \\ r(z_1+z_2) \end{pmatrix}$

with

$r\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}+r\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} rx_1-r+1 \\ ry_1 \\ rz_1 \end{pmatrix} +\begin{pmatrix} rx_2-r+1 \\ ry_2 \\ rz_2 \end{pmatrix} =\begin{pmatrix} (rx_1-r+1)+(rx_2-r+1)-1 \\ ry_1+ry_2 \\ rz_1+rz_2 \end{pmatrix}$

and they are equal. For the third condition,

$(rs)\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} rsx-rs+1 \\ rsy \\ rsz \end{pmatrix}$

while

$r(s\begin{pmatrix} x \\ y \\ z \end{pmatrix}) =r(\begin{pmatrix} sx-s+1 \\ sy \\ sz \end{pmatrix}) =\begin{pmatrix} r(sx-s+1)-r+1 \\ rsy \\ rsz \end{pmatrix}$

and the two are equal. For scalar multiplication by $1$ we have this.

$1\cdot\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} 1x-1+1 \\ 1y \\ 1z \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

Thus all the conditions on a vector space are met by these two operations. Remark. A way to understand this vector space is to think of it as the plane in $\mathbb{R}^3$

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

displaced away from the origin by $1$ along the $x$ -axis. Then addition becomes: to add two members of this space,

$\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix},\;\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}$

(such that $x 1 + y 1 + z 1 = 1$ and $x 2 + y 2 + z 2 = 1$ ) move them back by $1$ to place them in $P$ and add as usual,

$\begin{pmatrix} x_1-1 \\ y_1 \\ z_1 \end{pmatrix}+\begin{pmatrix} x_2-1 \\ y_2 \\ z_2 \end{pmatrix} =\begin{pmatrix} x_1+x_2-2 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix} \qquad\text{(in }P\text{)}$

and then move the result back out by $1$ along the $x$ -axis.

$\begin{pmatrix} x_1+x_2-1 \\ y_1+y_2 \\ z_1+z_2 \end{pmatrix}.$

Scalar multiplication is similar.
For the subspace to be closed under the inherited scalar multiplication, where $\vec{v}$ is a member of that subspace,

$0\cdot\vec{v}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

must also be a member. The converse does not hold. Here is a subset of $\mathbb{R}^3$ that contains the origin

$\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\}$

(this subset has only two elements) but is not a subspace.

Problem 18

We can give a justification for the convention that the sum of zero-many vectors equals the zero vector. Consider this sum of three vectors $\vec{v}_1+\vec{v}_2+\vec{v}_3$ .

What is the difference between this sum of three vectors and the sum of the first two of these three?
What is the difference between the prior sum and the sum of just the first one vector?
What should be the difference between the prior sum of one vector and the sum of no vectors?
So what should be the definition of the sum of no vectors?

Answer

$(\vec{v}_1+\vec{v}_2+\vec{v}_3)-(\vec{v}_1+\vec{v}_2) =\vec{v}_3$
$(\vec{v}_1+\vec{v}_2)-(\vec{v}_1)=\vec{v}_2$
Surely, $\vec{v}_1$ .
Taking the one-long sum and subtracting gives ( $\vec{v}_1)-\vec{v}_1=\vec{0}$ .

Problem 19

Is a space determined by its subspaces? That is, if two vector spaces have the same subspaces, must the two be equal?

Answer

Yes; any space is a subspace of itself, so each space contains the other.

Problem 20

Give a set that is closed under scalar multiplication but not addition.
Give a set closed under addition but not scalar multiplication.
Give a set closed under neither.

Answer

The union of the $x$ -axis and the $y$ -axis in $\mathbb{R}^2$ is one.
The set of integers, as a subset of $\mathbb{R}^1$ , is one.
The subset $\{\vec{v}\}$ of $\mathbb{R}^2$ is one, where $\vec{v}$ is any nonzero vector.

Problem 21

Show that the span of a set of vectors does not depend on the order in which the vectors are listed in that set.

Answer

Because vector space addition is commutative, a reordering of summands leaves a linear combination unchanged.

Problem 22

Which trivial subspace is the span of the empty set? Is it

$\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\}\subseteq \mathbb{R}^3, \quad\text{or}\quad \{0+0x\}\subseteq \mathcal{P}_1,$

or some other subspace?

Answer

We always consider that span in the context of an enclosing space.

Problem 23

Show that if a vector is in the span of a set then adding that vector to the set won't make the span any bigger. Is that also "only if"?

Answer

It is both "if" and "only if".

For "if", let $S$ be a subset of a vector space $V$ and assume $\vec{v}\in S$ satisfies $\vec{v}=c_1\vec{s}_1+\dots+c_n\vec{s}_n$ where $c_1,\ldots,c_n$ are scalars and $\vec{s}_1,\ldots,\vec{s}_n\in S$ . We must show that $[S\cup\{\vec{v}\} ]=[S]$ .

Containment one way, $[S]\subseteq[S\cup\{\vec{v}\} ]$ is obvious. For the other direction, $[S\cup\{\vec{v}\} ]\subseteq[S]$ , note that if a vector is in the set on the left then it has the form $d_0\vec{v}+d_1\vec{t}_1+\dots+d_m\vec{t}_m$ where the $d$ 's are scalars and the $\vec{t}\,$ 's are in $S$ . Rewrite that as $d_0(c_1\vec{s}_1+\dots+c_n\vec{s}_n) +d_1\vec{t}_1+\cdots+d_m\vec{t}_m$ and note that the result is a member of the span of $S$ .

The "only if" is clearly true— adding $\vec{v}$ enlarges the span to include at least $\vec{v}$ .

This exercise is recommended for all readers.

Problem 24

Subspaces are subsets and so we naturally consider how "is a subspace of" interacts with the usual set operations.

If $A, B$ are subspaces of a vector space, must $A\cap B$ be a subspace? Always? Sometimes? Never?
Must $A\cup B$ be a subspace?
If $A$ is a subspace, must its complement be a subspace?

(Hint. Try some test subspaces from Example 2.19.)

Answer

Always. Assume that $A, B$ are subspaces of $V$ . Note that their intersection is not empty as both contain the zero vector. If $\vec{w},\vec{v}\in A\cap B$ and $r, s$ are scalars then $r\vec{v}+s\vec{w}\in A$ because each vector is in $A$ and so a linear combination is in $A$ , and $r\vec{v}+s\vec{w}\in B$ for the same reason. Thus the intersection is closed. Now Lemma 2.19 applies.
Sometimes (more precisely, only if $A\subseteq B$ or $B\subseteq A$ ). To see the answer is not "always", take $V$ to be $\mathbb{R}^3$ , take $A$ to be the $x$ -axis, and $B$ to be the $y$ -axis. Note that

$\begin{pmatrix} 1 \\ 0 \end{pmatrix}\in A \text{ and }\begin{pmatrix} 0 \\ 1 \end{pmatrix}\in B \quad\text{but}\quad \begin{pmatrix} 1 \\ 0 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \end{pmatrix}\not\in A\cup B$

as the sum is in neither $A$ nor $B$ . The answer is not "never" because if $A\subseteq B$ or $B\subseteq A$ then clearly $A\cup B$ is a subspace. To show that $A\cup B$ is a subspace only if one subspace contains the other, we assume that $A\not\subseteq B$ and $B\not\subseteq A$ and prove that the union is not a subspace. The assumption that $A$ is not a subset of $B$ means that there is an $\vec{a}\in A$ with $\vec{a}\not\in B$ . The other assumption gives a $\vec{b}\in B$ with $\vec{b}\not\in A$ . Consider $\vec{a}+\vec{b}$ . Note that sum is not an element of $A$ or else $(\vec{a}+\vec{b})-\vec{a}$ would be in $A$ , which it is not. Similarly the sum is not an element of $B$ . Hence the sum is not an element of $A\cup B$ , and so the union is not a subspace.
Never. As $A$ is a subspace, it contains the zero vector, and therefore the set that is $A$ 's complement does not. Without the zero vector, the complement cannot be a vector space.

This exercise is recommended for all readers.

Problem 25

Does the span of a set depend on the enclosing space? That is, if $W$ is a subspace of $V$ and $S$ is a subset of $W$ (and so also a subset of $V$ ), might the span of $S$ in $W$ differ from the span of $S$ in $V$ ?

Answer

The span of a set does not depend on the enclosing space. A linear combination of vectors from $S$ gives the same sum whether we regard the operations as those of $W$ or as those of $V$ , because the operations of $W$ are inherited from $V$ .

Problem 26

Is the relation "is a subspace of" transitive? That is, if $V$ is a subspace of $W$ and $W$ is a subspace of $X$ , must $V$ be a subspace of $X$ ?

Answer

It is; apply Lemma 2.19. (You must consider the following. Suppose $B$ is a subspace of a vector space $V$ and suppose $A\subseteq B\subseteq V$ is a subspace. From which space does $A$ inherit its operations? The answer is that it doesn't matter— $A$ will inherit the same operations in either case.)

This exercise is recommended for all readers.

Problem 27

Because "span of" is an operation on sets we naturally consider how it interacts with the usual set operations.

If $S\subseteq T$ are subsets of a vector space, is $[S]\subseteq[T]$ ? Always? Sometimes? Never?
If $S, T$ are subsets of a vector space, is $[S\cup T]=[S]\cup[T]$ ?
If $S, T$ are subsets of a vector space, is $[S\cap T]=[S]\cap[T]$ ?
Is the span of the complement equal to the complement of the span?

Answer

Always; if $S\subseteq T$ then a linear combination of elements of $S$ is also a linear combination of elements of $T$ .
Sometimes (more precisely, if and only if $S\subseteq T$ or $T\subseteq S$ ). The answer is not "always" as is shown by this example from $\mathbb{R}^3$

$S=\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\},\quad T=\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\}$

because of this.

$\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\in[S\cup T] \qquad \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\not\in[S]\cup [T]$

The answer is not "never" because if either set contains the other then equality is clear. We can characterize equality as happening only when either set contains the other by assuming $S\not\subseteq T$ (implying the existence of a vector $\vec{s}\in S$ with $\vec{s}\not\in T$ ) and $T\not\subseteq S$ (giving a $\vec{t}\in T$ with $\vec{t}\not\in S$ ), noting $\vec{s}+\vec{t}\in[S\cup T]$ , and showing that $\vec{s}+\vec{t}\not\in[S]\cup[T]$ .
Sometimes. Clearly $[S\cap T] \subseteq[S]\cap[T]$ because any linear combination of vectors from $S\cap T$ is a combination of vectors from $S$ and also a combination of vectors from $T$ . Containment the other way does not always hold. For instance, in $\mathbb{R}^2$ , take

$S=\{\begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \end{pmatrix}\},\quad T=\{\begin{pmatrix} 2 \\ 0 \end{pmatrix}\}$

so that $[S]\cap[T]$ is the $x$ -axis but $[S\cap T]$ is the trivial subspace. Characterizing exactly when equality holds is tough. Clearly equality holds if either set contains the other, but that is not "only if" by this example in $\mathbb{R}^3$ .

$S=\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\}, \quad T=\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\}$
Never, as the span of the complement is a subspace, while the complement of the span is not (it does not contain the zero vector).

Problem 28

Reprove Lemma 2.15 without doing the empty set separately.

Answer

Call the subset $S$ . By Lemma 2.9, we need to check that $[S]$ is closed under linear combinations. If $c_1\vec{s}_1+\dots+c_n\vec{s}_n, c_{n+1}\vec{s}_{n+1}+\dots+c_m\vec{s}_m\in[S]$ then for any $p,r\in\mathbb{R}$ we have

$p\cdot(c_1\vec{s}_1+\cdots+c_n\vec{s}_n)+ r\cdot(c_{n+1}\vec{s}_{n+1}+\cdots+c_m\vec{s}_m) = pc_1\vec{s}_1+\cdots+pc_n\vec{s}_n +rc_{n+1}\vec{s}_{n+1}+\cdots+rc_m\vec{s}_m$

which is an element of $[S]$ . (Remark. If the set $S$ is empty, then that "if ... then ..." statement is vacuously true.)

Problem 29

Find a structure that is closed under linear combinations, and yet is not a vector space. (Remark. This is a bit of a trick question.)

Answer

For this to happen, one of the conditions giving the sensibleness of the addition and scalar multiplication operations must be violated. Consider $\mathbb{R}^2$ with these operations.

$\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} +\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix} \qquad r\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

The set $\mathbb{R}^2$ is closed under these operations. But it is not a vector space.

$1\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix} \neq\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

[1]

Linear Algebra/Subspaces and Spanning sets

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