what does it mean to be a vector space over a field

Vector Spaces

Vector spaces and linear transformations are the principal objects of study in linear algebra. A vector space (which I'll define below) consists of ii sets: A set of objects called vectors and a field (the scalars).

Definition. A vector space V over a field F is a set V equipped with an performance called (vector) addition, which takes vectors u and v and produces some other vector .

There is besides an operation called scalar multiplication, which takes an element $a \in F$ and a vector $u \in V$ and produces a vector $au \in V$ .

These operations satisfy the following axioms:

Vector addition is associative: If $u, v, w \in V$ , then

Vector addition is commutative: If $u, v \in V$ , then

There is a zilch vector 0 which satisfies

$0 + u = u = u + 0 \quad\hbox{for all}\quad u \in V.$

For every vector $u \in V$ , there is a vector $-u \in V$ which satisfies

If $a, b \in F$ and $x \in V$ , then

If $a, b \in F$ and $x \in V$ , then

If $a \in F$ and $x, y \in V$ , and then

If $x \in V$ , then

$1\cdot x = x.$

The elements of V are called vectors; the elements of F are called scalars. Every bit usual, the employ of words like "multiplication" does not imply that the operations involved await like ordinary "multiplication".

Example. If F is a field, then denotes the set

$F^n = \{(a_1, \ldots, a_n) \mid a_1, \ldots, a_n \in F\}.$

is called {the vector space of northward-dimensional vectors over F}. The elements , ..., are chosen the vector'southward components.

becomes a vector space over F with the following operations:

$(a_1, \ldots, a_n) + (b_1, \ldots, b_n) = (a_1 + b_1, \ldots, a_n + b_n).$

$p\cdot (a_1, \ldots, a_n) = (pa_1, \ldots, pa_n), \quad\hbox{where}\quad p \in F.$

It's easy to bank check that the axioms concord. For instance, I'll check Axiom 6. Let $p, q \in F$ , and let $(a_1, \ldots, a_n) \in F^n$ . Then

$\matrix{(p + q)(a_1, \ldots, a_n) & = & \left((p + q)a_1, \ldots, (p + q)a_n\right) & \hbox{Definition of scalar multiplication} \cr & = & \left(pa_1 + qa_1, \ldots, pa_n + qa_n\right) & \hbox{Field axiom: Distributivity} \cr & = & (pa_1, \ldots, pa_n) + (qa_1, \ldots, qa_n) & \hbox{Definition of vector additon} \cr & = & p(a_1, \ldots, a_n) + q(a_1, \ldots, a_n) & \hbox{Definition of scalar multiplication} \cr}.$

Every bit a specific example, $\real^3$ consists of three-dimensional vectors with real components, like

$(3, -2, \pi) \quad\hbox{or}\quad \left(\dfrac{1}{2}, 0, -1.234\right).$

Y'all're probably familiar with addition and scalar multiplication for these vectors:

$7\cdot (-2, 0, 3) = (7\cdot (-2), 7\cdot 0, 7\cdot 3) = (-14, 0, 21).$

(Sometimes people write $\langle 3, -2, \pi\rangle$ , using angle brackets to distinguish vectors from points. I'll use angle brackets when at that place'southward a danger of confusion.)

$\integer_3^2$ consists of 2-dimensional vectors with components in $\integer_3$ . Since each of the two components tin can be any element in $\{0, 1, 2\}$ , there are $3\cdot 3 = 9$ such vectors:

Here are examples of vector improver and multiplication in $\integer_3^2$ :

$2\cdot (2,1) = (2\cdot 2, 2\cdot 1) = (1,2).\quad\halmos$

Example. The prepare $\real[x]$ of polynomials with real coefficients is a vector infinite over $\real$ , using the standard operations on polynomials. For example,

$4\cdot (-3x^2 + 10) = -12x^2 + 40.$

Dissimilar $\real^n$ , $\real[x]$ is space dimensional (in a sense to be made more precise presently). Intuitively, y'all need an space set of polynomials, like

$1, x, x^2, x^3, \ldots$

to "construct" all the elements of $\real[x]$ .

Example. Allow denote the continuous real-valued functions defined on the interval $0 \le x \le 1$ . Add functions pointwise:

$(f + g)(x) = f(x) + g(x) \quad\hbox{for}\quad f, g \in C[0,1].$

From calculus, you know that the sum of continuous functions is a continuous function.

If $a \in \real$ and $f \in C[0,1]$ , define scalar multiplication in pointwise fashion:

$(af)(x) = a\cdot f(x).$

For example, if and , and then

These operations make into an $\real$ -vector infinite.

Similar $\real[x]$ , is infinite dimensional. Notwithstanding, its dimension is uncountably infinite, while $\real[x]$ has countably infinite dimension over $\real$ .

Yous tin can as well define a "dot product" for two vectors $f, g \in C[0,1]$ :

$f\cdot g = \int_0^1 f(x)g(x)\,dx.$

The product of continuous functions is continuous, and so the integral of is defined. This example shows that abstract vectors do non have to look like trivial arrows!.

Proposition. Let V be a vector space over a field F.

$0\cdot x = 0$ for all $x \in V$ .
$(-1)\cdot x = -x$ for all $x \in V$ .
for all $x \in V$ .

Proof. (a) Note that the "0" on the left is the nada {\information technology scalar} in F, whereas the "0" on the right is the zero {\information technology vector} in Five.

$0\cdot x = (0 + 0)\cdot x = 0\cdot x + 0\cdot x.$

Subtracting $0\cdot x$ from both sides, I become $0 = 0\cdot x$ .

(b) (The "-1" on the left is the scalar -1; the " " on the correct is the "negative" of $x \in V$ .)

$(-1)\cdot x + x = (-1)\cdot x + 1\cdot x = \left((-1) + 1\right)\cdot x = 0\cdot x = 0.$

(c)

$-(-x) = (-1)\cdot [(-1)\cdot x] = [(-1)\cdot (-1)]x = 1\cdot x = x.\quad\halmos$

Definition. Let V exist a vector space over a field F, and allow $W \subset V$ , $W \ne \emptyset$ . W is a subspace of V if:

If $u, v \in W$ , then $u + v \in W$ .
If $k \in F$ and $u \in W$ , and then $ku \in W$ .

In other words, West is airtight under addition of vectors and under scalar multiplication.

Lemma. Let Due west be a subspace of a vector space V.

The null vector is in West.
If $w \in W$ , then $-w \in W$ .

Proof. (a) Accept any vector $w \in W$ (which you can do because W is nonempty), and take $0 \in F$ . Since W is closed nether scalar multiplication, $0\cdot w \in W$ . Simply $0\cdot w = 0$ , and then $0 \in W$ .

(b) Since $w \in W$ and $-1 \in F$ , $(-1)\cdot w = -w$ is in W.

Example. If 5 is a vector space over a field F, $\{0\}$ and V are subspaces of Five.

Example. Consider the real vector space $\real^2$ , the usual x-y plane. Then

$W_1 = \{(x,0) \mid x \in \real\} \quad\hbox{and}\quad W_2 = \{(0,y) \mid y \in \real\}$

are subspaces of $\real^2$ . (These are just the x and y-axes, of course.)

I'll check that is a subspace. Commencement, I accept to show that two elements of add together to an element of . An element of is a pair with the second component 0. So hither are 2 elements of : , . Add them:

is in , because its second component is 0. Thus, is closed nether sums.

Next, I accept to show that is airtight under scalar multiplication. Have a scalar $k \in \real$ and a vector $(x,0) \in W_1$ . Take their product:

$k\cdot (x,0) = (kx,0).$

The product is in because its 2nd component is 0. Therefore, is closed nether scalar multiplication.

Thus, is a subspace.

Notice that in doing the proof, I did not use specific vectors in like or . I'm trying to prove statements almost arbitrary elements of , so I use "variable" elements.

In general, the subspaces of $\real^2$ are $\{0\}$ , $\real^2$ , and lines passing through the origin. (Why tin't a line which doesn't pass through the origin be a subspace?)

In $\real^3$ , the subspaces are the $\{0\}$ , $\real^3$ , and lines or planes passing through the origin.

Then on.

Case. Prove or disprove: The post-obit subset of $\real^3$ is a subspace:

$W = \{(x, y, 1) \mid x, y \in \real\}.$

If yous're trying to decide whether a ready is a subspace, it's always expert to check whether it contains the zero vector before yous start checking the axioms. In this case, the set consists of iii-dimensional vectors whose third components are equal to 1. Plain, the null vector doesn't satisfy this condition.

Since West doesn't contain the nix vector, information technology'south non a subspace of $\real^3$ .

Case. Permit

$W = \left\{(x, \sin x) \mid x \in \real\right\}.$

Prove or disprove: West is a subspace of $\real^2$ .

Annotation that $(0,0) = (0, \sin 0) \in W$ . This is not one of the axioms for a subspace, just it's a good thing to check first because you can usually exercise it quickly. If the nothing vector is not in a prepare, then the lemma above shows that the set is non a subspace. In this example, the zero vector is in Westward, so the issue isn't settled, and I'll endeavour to check the subspace axioms.

Commencement, I might try to check that the gear up is closed under sums. I take two vectors in W --- say $(x, \sin x)$ and $(y, \sin y)$ . I add them:

$(x, \sin x) + (y, \sin y) = (x + y, \sin x + \sin y).$

The concluding vector isn't in the right form --- information technology would exist if $\sin x + \sin y$ was equal to $\sin (x + y)$ . That doesn't sound right, and so I doubtable that Due west is not a subspace. I attempt to get a specific counterexample to contradict closure under addition.

First,

$\left(\dfrac{\pi}{2}, \sin \dfrac{\pi}{2}\right) = \left(\dfrac{\pi}{2}, 1\right) \in W \quad\hbox{and}\quad (\pi, \sin \pi) = (\pi, 0) \in W.$

On the other hand,

$\left(\dfrac{\pi}{2}, \sin \dfrac{\pi}{2}\right) + (\pi, \sin \pi) = \left(\dfrac{\pi}{2}, 1\right) + (\pi, 0) = \left(\dfrac{3\pi}{2}, 1\right) \not\in W.$

For I have $\sin \dfrac{3\pi}{2} = -1 \ne 1$ .

Since Westward is not airtight nether vector addition, it is not a subspace.

Instance. Permit F be a field, and let $A, B \in M(n, F)$ . Consider the following subset of :

$W = \{v \in F^n \mid Av = Bv\}.$

Prove or disprove: W is a subspace.

This fix is defined by a property rather than by appearance, and axiom checks for this kind of set oftentimes requite people trouble. The problem is that elements of W don't "wait like" anything --- if y'all need to refer to a couple of arbitrary elements of W, you might call them u and v (for case). At that place'south nothing most the symbols u and v which tells y'all that they belong to W. Merely u and v are like people who belong to a society: You can't tell from their advent that they're club members, but they're carrying membership cards in their pockets.

With this in listen, I'll check closure under addition. Let $u, v \in W$ . I must show that $u + v \in W$ .

Since u and v are in W,

$Au = Bu \quad\hbox{and}\quad Av = Bv.$

Adding the equations and factoring out, I get

$\eqalign{Au + Av &= Bu + Bv \cr A(u + v) &= B(u + v) \cr}$

The terminal equation shows that $u + v \in W$ .

Warning: Don't say " $A(u + v) + B(u + v) \in W$ " --- it doesn't make sense! " " is an equation that satisfies; it can't be an element of W, because elements of W are vectors.

Side by side, I'll check closure under scalar multiplication. Permit $k \in F$ and let $v \in W$ . Since $v \in W$ , I have

Multiply both sides by k, then commute the matrices and the scalar:

$\eqalign{k(Av) &= k(Bv) \cr A(kv) &= B(kv) \cr}$

The final equation says that $kv \in W$ .

Since Westward is closed under addition and scalar multiplication, it'due south a subspace.

Example. Consider the following subsets of the polynomial ring $\real[x]$ :

$V_1 = \{f(x) \in \real[x] \mid f(2) = 0\}, \qquad V_2 = \{f(x) \in \real[x] \mid f(2) = 1\}.$

is a subspace; information technology consists of all polynomials having as a root.

is non a subspace. 1 way to see this is to notice that the nada polynomial (i.e. the zero vector) is non in , because the goose egg polynomial does not requite 1 when you plug in .

Alternatively, the abiding polynomial is an chemical element of --- it gives 1 when you plug in 2 --- but $2\cdot f(x)$ is not. So is not airtight under scalar multiplication.

Lemma. If A is an $m \times n$ matrix over the field F, the set of n-dimensional vectors 10 which satisfy

is a subspace of (the solution space of the organisation).

Proof. If and , then

Therefore, if x and y are in the set, so is .

If and yard is a scalar, then

$A(kx) = k(Ax) = k\cdot 0 = 0.$

Therefore, if ten is in the set, and so so is .

Therefore, the solution space is a subspace.

Example. Consider the following system of linear equations over $\real$ :

$\left[\matrix{1 & 1 & 0 & 1 \cr 0 & 0 & 1 & 3 \cr}\right] \left[\matrix{w \cr x \cr y \cr z \cr}\right] = \left[\matrix{0 \cr 0 \cr 0 \cr 0 \cr}\right].$

The solution can be written equally

$w = -s - t, \quad x = s, \quad y = -3t, \quad z = t.$

Thus,

$\left[\matrix{w \cr x \cr y \cr z \cr}\right] = \left[\matrix{-s - t \cr s \cr -3t \cr t \cr}\right].$

The Lemma says that the set of all vectors of this form constitute a subspace of $\real^4$ .

For case, if you add together two vectors of this form, you lot get another vector of this grade:

$\left[\matrix{-s - t \cr s \cr -3t \cr t \cr}\right] + \left[\matrix{-s' - t' \cr s' \cr -3t' \cr t' \cr}\right] = \left[\matrix{-(s + s') - (t + t') \cr s + s' \cr -3(t + t') \cr t + t' \cr}\right].$

You tin check that the set up is also closed under scalar multiplication.

Definition. If , , ..., are vectors in a vectors space Five, a linear combination of the 5'due south is a vector

$k_1v_1 + k_2v_2 + \cdots + k_nv_n,$

where the k's are scalars.

Instance. Take $u = \langle 1,2\rangle$ and $v = \langle -3,7\rangle$ in $\real^2$ . Hither is a linear combination of u and v:

$2u - 5v = 2\cdot \langle 1,2\rangle - 5\langle -3,7\rangle = \langle 17,-31\rangle.$

$(\sqrt{2} - 17)u + \dfrac{\pi^2}{4}v$ is also a linear combination of u and five. u and v are themselves linear combinations of u and v, as is the zero vector (why?).

In fact, information technology turns out that any vector in $\real^2$ is a linear combination of u and v.

On the other hand, there are vectors in $\real^2$ which are not linear combinations of $p = \langle 1,-2\rangle$ and $q = \langle -2,4\rangle$ . Do yous run across how this pair is different from the first?

Definition. If S is a subset of a vector space 5, the bridge $\langle S\rangle$ of Due south is the fix of all linear combinations of vectors in South.

Theorem. If Southward is a subset of a vector space V, the span $\langle S\rangle$ of South is a subspace of 5.

Proof. Here are typical elements of the span of Due south:

$j_1u_1 + j_2u_2 + \cdots + j_nu_n, \quad k_1v_1 + k_2v_2 + \cdots + k_nv_n,$

where the j's and k's are scalars and the u's and v's are elements of S.

Take 2 elements of the span and add together them:

$(j_1u_1 + j_2u_2 + \cdots + j_nu_n) + (k_1v_1 + k_2v_2 + \cdots + k_mv_m) = j_1u_1 + j_2u_2 + \cdots + j_nu_n + k_1v_1 + k_2v_2 + \cdots + k_mv_m.$

This humongous sum is an element of the span, because it's a sum of vectors in S, each multiplied past a scalar. Thus, the span is closed under taking sums.

Take an element of the span and multiply it by a scalar:

$k\cdot(k_1v_1 + k_2v_2 + \cdots k_nv_n) = kk_1v_1 + kk_2v_2 + \cdots + kk_nv_n.$

This is an element of the span, because it's a sum of vectors in Due south, each multiplied by a scalar. Thus, the bridge is airtight under scalar multiplication.

Therefore, the bridge is a subspace.

Example. Prove that the span of $\langle 3,1,0\rangle$ and $\langle 2,1,0\rangle$ in $\real^3$ is

$V = \left\{\langle a, b, 0\rangle \mid a, b \in \real\right\}.$

To bear witness that ii sets are equal, you need to show that each is independent in the other. To do this, take a typical chemical element of the first set and testify that it's in the second prepare. Then take a typical chemical element of the 2nd set up and show that it'due south in the kickoff set.

Let Westward exist the span of $\langle 3,1,0\rangle$ and $\langle 2,1,0\rangle$ in $\real^3$ . A typical chemical element of W is a linear combination of the two vectors:

$x\cdot \langle 3,1,0\rangle + y\cdot \langle 2,1,0\rangle = \langle 3x + 2y, x + y, 0\rangle.$

Since the sum is a vector of the course $\langle a, b, 0\rangle$ for $a, b \in \real$ , information technology is in Five. This proves that $W \subset V$ .

At present let $\langle a, b, 0\rangle \in V$ . I accept to prove that this vector is a linear combination of $\langle 3,1,0\rangle$ and $\langle 2,1,0\rangle$ . This means that I have to find real numbers x and y such that

$x\cdot \langle 3,1,0\rangle + y\cdot \langle 2,1,0\rangle = \langle a, b, 0\rangle.$

If I expand the left side, I get

$\langle 3x + 2y, x + y, 0\rangle = \langle a, b, 0\rangle.$

Equating corresponding components, I get

$3x + 2y = a, \quad x + y = b.$

This is a system of linear equations which y'all can solve by row reduction or matrix inversion (for instance). The solution is

$x = a - 2b, \quad y = -a + 3b.$

In other words,

$(a - 2b)\cdot \langle 3,1,0\rangle + (-a + 3b)\cdot \langle 2,1,0\rangle = \langle a, b, 0\rangle.$

Since $\langle a, b, 0\rangle$ is a linear combination of $\langle 3,1,0\rangle$ and $\langle 2,1,0\rangle$ , it follows that $\langle a, b, 0\rangle \in W$ . This proves that $V \subset W$ .