Mathematics, Statistics & Physics

Math 511: Linear Algebra

Vector Spaces

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4.2 Vector Spaces

• 4.2.1 Vector Space Examples
  • Video Lecture 1 - Vector Space Examples
  • Video Lecture 2 - Abstract Vector Spaces
• 4.2.2 The Definition of a Vector Space
  • Definition - Vector Space
  • Video Lecture 3 - Introduction to Vector Spaces
• 4.2.3 A Vector Space of Matrices
  • Theorem 1 - Real mxn Matrices Form a Vector Space
  • Definition - Vector
  • Exercise 1 - Linear Combinations Are Vectors
• 4.2.4 Polynomials
  • Definition - Polynomial
  • Definition - Degree
  • Example 1 - Degree of a Polynomial
  • Definition - Polynomial Addition
  • Definition - Polynomial Functions
  • Definition - Polynomial Multiplication
  • Definition - Polynomial Scalar Multiplication
• 4.2.5 A Vector Space of Polynomials
  • Theorem 2 - The Set of Real Polynomials is a Vector Space
  • Exercise 2 - The Set of Real Polynomials is a Vector Space
• 4.2.6 More Vector Spaces of Polynomials
  • Example 2 - Polynomial Vector Spaces $P_n$
  • Exercise 3 - Prove the Axioms
  • Example 3 - Polynomials and Vectors
  • Definition - Bijection
  • Exercise 4 - Identify the Polynomial
  • Exercise 5 - Identify the Vector
• 4.2.7 Vector Spaces of Functions
  • Exercise 6 - Show That $C[a,b]$ is a Vector SpaceVectorSpace)
• 4.2.8 Complex Numbers
  • Exercise 7 - Show that $\mathbb{C}$ is a Vector Space
• 4.2.9 Properties of Scalar Multiplication
  • Theorem 3 - Properties of Scalar Multiplication
  • Proof - Properties of Scalar Multiplication
• 4.2.10 Examples of Vector Spaces
• copyleft

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4.2.1 Vector Space Examples

In this section, we will formally define one of the central concepts to linear algebra, a vector space. Everything we worked with in this course is directly related to vector spaces. We study

1. vectors in a vector space,
2. a linear transformation from one vector space to another vector space,
3. vector spaces as mathematical objects

Vector spaces are best understood from the perspective of examples. Start at time mark 20:20 in the following video lecture:

Video Lecture 1: Vector Space Examples.

Vector Space Examples

We are going to discuss several vector space examples, formally define vector spaces, and then discuss more general examples that fit the definition of a vector space. Recall what Sanderson calls the physics student perspective, the mathematician's perspective, and the computer science student perspective of a vector. We'll be referring to these ideas, but using the mathematician's perspective in this section.

Video Lecture 2: Abstract Vector Spaces.

Abstract Vector Spaces

A Vector Space is any set of objects that has a reasonable definition for

1. adding the objects in the set to get another object in the set, and
2. multiplying objects in the set by scalars to get more objects in the set

These properties are called the closure properties of vector addition and scalar multiplication.

In a vector space we can create linear combinations of the objects in our vector space to get more of these objects in our space.

What is Reasonable?

When the definitions for adding objects and scaling object by multiplying them by scalars conforms to the same eight (8) properties of Theorem 4.1.1, they are reasonable. In vector space $\mathbb{R}^n$, these properties are inherited from the properties of real numbers. However in the wide world of vector spaces, this connection is not usually available. The definition of vector space includes these eight (8) properties as axioms in the definition.

A set of objects with a definition for adding objects and multiplying the objects by scalars is a Vector Space, when the definitions of addition and scalar multiplication adhere to the eight (8) axioms.

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4.2.2 The Definition of a Vector Space

For a set of objects to be a Vector Space the operations of vector addition and scalar multiplication must satisfy the two closure properties and the eight properties of Theorem 4.1.1. These additional properties do not need to be proved because they are now part of the definition of a vector space.

Definition

Vector Space

Let $V$ be a set of objects upon which two operations, vector addition ($\oplus$) and scalar multiplication ($\odot$) are defined. If both closure properties and all eight axioms are satisfied, then the set $V$ with these two operations is a Vector Space, $(V,\oplus,\odot)$.

For every $\mathbf{u}$ and $\mathbf{v}\in V$, and every scalar $c$, the following closure properties must be satisfied

1. $\mathbf{u} \oplus \mathbf{v}\in V$

2. $c\odot\mathbf{u}\in\ V$

In addition, for every $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}\in V$, and all scalars $c$ and $d$,

1. $\mathbf{u} \oplus \mathbf{v} = \mathbf{v} \oplus \mathbf{u}$

2. $\mathbf{u} \oplus (\mathbf{v} \oplus \mathbf{w}) = (\mathbf{u} \oplus \mathbf{v}) \oplus \mathbf{w}$

3. $V$ has a vector $\mathbf{0}$ so that $\mathbf{u} \oplus \mathbf{0} = \mathbf{u}$

4. There is a vector $-\mathbf{u}\in V$ so that $\mathbf{u} \oplus (-\mathbf{u}) = \mathbf{0}$

5. $c\odot(\mathbf{u} \oplus \mathbf{v}) = ( c\odot\mathbf{u} ) \oplus ( c\odot\mathbf{v} )$

6. $(c + d)\odot\mathbf{u} = ( c\odot\mathbf{u} ) \oplus ( d\odot\mathbf{u} )$

7. $c\odot(d\odot\mathbf{u}) = (cd)\odot\mathbf{u}$

8. $1\odot\mathbf{u} = \mathbf{u}$

Why did we use the fancy symbols for addition and scalar multiplication?

In the wide world of vector spaces, addition may look very different from adding real numbers in our abstract vector space. Likewise scaling a vector may not look like the multiplying real numbers. We need to emphasize that difference here.

The goal of the mathematician's perspective is to take advantage of abstraction and study the properties of vector spaces without worrying about the specific details of a particular vector space. If we can show that something is true for a general vector space, then it must be true for ALL Vector Spaces. Additionally, if we can show that a set of mathematical objects is closed under vector addition, scalar multiplication, and satisfies the eight axioms, then it is a vector space. The space of objects inherits ALL of the vast and powerful mathematical machinery of vector spaces.

Video Lecture 3: Introduction to Vector Spaces.

Introduction to Vector Spaces

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4.2.3 A Vector Space of Matrices

In Section 2.2 - Matrix Algebra, we discussed the operations and properties of $m\times n$ matrices. We did this because the set of $m\times n$ matrices of real values, element-wise addition, and element-wise scalar multiplication form a vector space.

We are required to prove that the set $\mathbb{R}^{m\times n}$ of all $m\times n$ matrices with element-wise addition and element-wise scalar multiplication satisfies the definition of a vector space.

Theorem 1

$\mathbb{R}^{m\times n}$ is a Vector Space

The set of all $m\times n$ matrices of real numbers, with element-wise addition and element-wise scalar multiplication is a Vector Space.

Proof of Theorem 1

This is a significant amount of work since there are two closure properties, and eight axioms.

1. The sum of two $m\times n$ matrices is an $m\times n$ matrix.

Let $A$ and $B\in\mathbb{R}^{m\times n}$. Then both $A$ and $B$ have $mn$ elements $a_{ij}$, $1\le i\le m$, and $1\le j\le n$. Matrix addition was defined in Section 2.1.4,

$$A + B = [a_{ij}] + [b_{ij}] = [ a_{ij}+b_{ij}]$$

This is also a real $m\times n$ matrix, and therefore an element of $\mathbb{R}^{m\times n}$.

2. The scalar multiple of an $m\times n$ matrix.

Let $A\in\mathbb{R}^{m\times n}$, and $c\in\mathbb{R}$. Scalar multiplication is also defined in Section 2.1.4,

$$cA = c[a_{ij}] = [ca_{ij}]$$

The product is an $m\times n$ matrix of real numbers, so $cA\in\mathbb{R}^{m\times n}$.

Now that we established the two closure properties, we must explicitly show that all eight of the axioms are also satisfied by this definition of addition and scalar multiplication.

Let $A$ and $B\in\mathbb{R}^{m\times n}$ and let $c$ and $d$ be real scalars. Recall Theorem 2.2.1.

1. Theorem 2.2.1-1 establishes axiom 1 of the definition of vector spaces.

2. Theorem 2.2.1-2 establishes axiom 2 of the definition.

3. The $m\times n$ matrix whose elements are all zeros is the zero vector $O = [\,0\,]$. Since $0_{ij}=0$ for $1\le i\le m$ and $1\le j\le n$, $$\begin{align*} A+O &= [a_{ij}] + [\,0\,] = [ a_{ij}+0 ] = [a_{ij}] = A \\ O+A &= [\,0\,] + [a_{ij}] = [ 0+a_{ij} ] = [a_{ij}] = A \end{align*}$$

4. $(-1)A$ is the additive inverse of $m\times n$ matrix $A$, and $$\begin{align*} A + (-1)A &= [a_{ij}] + (-1)[a_{ij}] = [a_{ij}] + [(-1)a_{ij}] = [a_{ij}] + [-a_{ij}] = [a_{ij}-a_{ij}] = [\,0\,] = O \\ (-1)A + A &= (-1)[a_{ij}] + [a_{ij}] = [(-1)a_{ij}] + [a_{ij}] = [-a_{ij}] + [a_{ij}] = [-a_{ij}+a_{ij}] = [\,0\,] = O \end{align*}$$ Hence we may call $(-1)A$ the additive inverse, $(-1)A = -A$.

5. Theorem 2.2.1-9 establishes axiom 5 of the definition.

6. Theorem 2.2.1-8 establishes axiom 6 of the definition.

7. Theorem 2.2.1-6 establishes axiom 7 of the definition.

8. As in axioms 3 and 4, we need only compute the product, $$1\,A = 1\,[a_{ij}] = [ 1\,a_{ij} ] = [ a_{ij} ] = A$$

Therefore the set of $m\times n$ matrices with element-wise addition and element-wise scalar multiplication, is a vector space. $\blacksquare$

This looks like a lot of work!

The work to establish that our mathematical objects with a reasonable definition of addition and scalar multiplication is a vector space allows us to use all of the techniques we will learn about vector spaces in the second half of this course. It also allows us to use all of the computational techniques we learned in the first half of this course. The happy news is that we must establish that such a set with addition and scalar multiplication is a vector space only once. We proved that $\mathbb{R}^{m\times n}$ is a vector space for all positive integers $m$ and $n$ with one Theorem.

Notice that we can now refer to $m\times n$ matrices as vectors in the vector space $\mathbb{R}^{m\times n}$. Will this have significant implications for our further study of vector spaces?

Grant Sanderson asked,

What are vectors? Are two-dimensional vectors for example, fundamentally arrows in a flat plane that we can describe with coordinates for convenience? Or, is it fundamentally that pair of real numbers which is just nicely visualized as an arrow on a flat plane? Or are both of these just manifestations of something deeper?

Definition

Vector

A vector is an object in a vector space. The elements of a vector space are called vectors.

Exercise 1 - Linear Combinations

Prove that any finite linear combination of vectors in a vector space is an element of the vector space.

View Solution

Let $(V, +, \cdot)$ be a vector space, $\mathbf{u}$ and $\mathbf{v}$ be vectors in $V$, and $a$ and $b$ be scalars. Then $a\mathbf{u}$ and $b\mathbf{v}$ are vectors in $V$ by the second closure property. Thus $a\mathbf{u}+b\mathbf{v}$ must be a vector in $V$ by the first closure property. Therefore, the linear combination $$a\mathbf{u}+b\mathbf{v}\in V.$$ Assume that any linear combination of $k$ vectors from $V$ is a vector in $V$ for positive integer $k$. Let $\mathbf{v}_i\in V$ and $t_i$ be scalars for $1\le i\le k$ so that $$\sum_{i=1}^k t_i\mathbf{v}_i\,\in V.$$ Now let $\mathbf{v}_{k+1}\in V$ and $t_{k+1}$ be a scalar. Using axiom 2 of the definition of vector spaces, $$\sum_{i=1}^{k+1} t_i\mathbf{v}_i = \left( \sum_{i=1}^k t_i\mathbf{v}_i \right) + t_{k+1}\mathbf{v}_{k+1}.$$ The product $t_{k+1}\mathbf{v}_{k+1}\in V$ by closure property 2, and $\sum_{i=1}^k t_i\mathbf{v}_i\in V$ by hypothesis. Thus their sum is in $V$ by closure property 1. Hence, a linear combination of $k+1$ vectors is a vector in $V$.
Therefore, any finite linear combination of vectors from vector space $V$ is a vector in $V$. $\blacksquare$

As a result of this exercise we say that, Vector Spaces are closed under the operation of linear combinations.

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4.2.4 Polynomials

We need to review some vocabulary to make sure everyone remembers some key terms we will use with regards to polynomials.

Definition

Polynomial

A polynomial over a field $F$ is an algebraic expression consisting of elements of the field and symbols called variables or indeterminates by means of addition, multiplication, and exponentiation to a non-negative integer power.

1. The field is usually real or complex numbers, but may be any field that does not already involve indeterminates. Encryption, computer graphics, coding theory, and computational geometry use finite fields in the construction of polynomials.

2. Field elements of a polynomial expression are called constants or coefficients.

3. Two polynomial expressions represent the same polynomial when one may be transformed into the other using the rules of commutativity, associativity, and distributivity of addition and multiplication from the field.

4. A polynomial always has a finite number of terms. Like terms are terms with the indeterminate raised to the same non-negative integer power. Each term consists of a finite number of factors.

5. The variable in a polynomial may be replaced using substitution with any element of its field. The result must be a valid expression in the field. Therefore the variable is required to follow the same rules of addition and multiplication as the underlying field.

Remarks

In mathematics one may define a polynomial ring over any ring, however algebraic constructions beyond basic calculus are not prerequisites for this course.

A univariate polynomial $p$, or a polynomial with a single variable can always be written in the form

$$p = p(x) = a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + a_nx^n = \sum_{k=0}^n a_kx^k,$$

where the symbol $x^0$ is always taken to be the unit from the field ($x^0=1$). Both labels $p$ and $p(x)$ are useful. In many cases one desires to focus on the properties of the polynomial as a mathematical object $p$. Polynomials can be evaluated by substitution at a value in the field or other mathematical object.

1. $p(0) = a_0 + a_10 + \cdots + a_{n-1}0^{n-1} + a_n0^n = a_0$
2. $p(1) = a_0 + a_11 + \cdots + a_{n-1}1^{n-1} + a_n1^n = \sum_{k=0}^n a_k$
3. $p(a) = a_0 + a_1a + \cdots + a_{n-1}a^{n-1} + a_na^n = \sum_{k=0}^n a_ka^k$

If $A\in\mathbb{R}^{n\times n}$,

4. $p(A) = a_0I + a_1A + \cdots + a_{n-1}A^{n-1} + a_nA^n = \sum_{k=0}^n a_kA^k$

For any square matrix, we consider $A^0=I$. Why would we do this? Because we want sums of exponents to work properly.

$$A\cdot I = A = A^1 = A^{1+0} = A^1A^0$$

What if $A$ is singular, or the zero matrix? We need these algebraic rules of exponents for our polynomials to be consistent. This is not always the case. In calculus we would resort to L'Hospital's rule to decide whether $O^0$ has a value or is indeterminate.

Definition

Degree

1. The degree of a nonzero term of a polynomial is the sum of the exponents of the variables that appear in the term. Hence it is the sum of non-negative integers, and there the degree of a term is a non-negative integer.

2. For a univariate polynomial, the degree of the polynomial is the highest exponent of the variable occurring in the nonzero terms of the polynomial.

Example 1 - The Degree of a Polynomial

Consider the polynomials

1. $2x^4y + 3x^5y^5 + 2xy^0z^2$

2. $x + x^3 + x^1 + x^0 $

3. $x + x^3 + x^2 + 0x^5$

What are the degrees of each term and the degrees of each polynomial?

1. $2x^4y + 3x^5y^5 + 2xy^0z^2$
   1. The degree of term $2x^4y$ is $4 + 1 = 5$.
   2. The degree of term $3x^5y^5$ is $5 + 5 = 10$.
   3. The degree of term $xy^0z^2$ is $1 + 0 + 2 = 3$.
   4. The degree of the polynomial is $\max\left\{\,5, 10, 3\,\right\} = 10$.
2. $x + x^3 + x^1 + x^0 = 1 + (x + x) + x^3 = 1 + 2x + x^3$
   1. The degree of term $1$ is $0$.
   2. The degree of term $2x$ is $1$.
   3. The degree of term $x^3$ is $3$.
   4. The degree of the polynomial is $3$.
3. $x + x^3 + x^2 + 0x^5 = x + x^2 + x^3$
   1. The degree of term $x$ is $1$.
   2. The degree of term $x^2$ is $2$.
   3. The degree of term $x^3$ is $3$.
   4. The degree of the polynomial is $3$.

Zero coefficients or constants don't count. This causes a problem. What is the degree of the zero polynomial $z = z(x) = 0$? It cannot be zero because non-zero constant terms have degree zero, $c = cx^0$. The zero polynomial either has no degree or we can make one up that works for the ways we add, subtract, and multiply polynomials.

For the remainder of this course polynomials are univariate polynomials, only one indeterminate. Everyone should consider the course Math 511, Abstract Algebra, to learn about more interesting algebraic structures.

Definition

Addition of (univariate) Polynomials

The sum of two polynomials $p$ with degree $m$,

$$p = p(x) = a_0 + \cdots + a_mx^m = \sum_{k=0}^m a_kx^k$$

and $q$ with degree $n$

$$q = q(x) = b_0 + \cdots + b_nx^n = \sum_{k=0}^n b_kx^k$$

is polynomial $p+q$ with degree $r = \max\{m,n\}$. If one of the matrices has degree strictly less than the other, pad the polynomial with zero terms $0x^k$ until both polynomials have the same number of terms. Notice that this padding does not change the degree of the polynomial. If $m < n$, then we can represent $p$ with $r=n$ terms,

$$p = a_0 + \cdots + a_mx^m + 0x^{m+1} + \cdots + 0x^r$$

Now that both polynomials have the same number of terms,

$$ \begin{align*} p+q = (p+q)(x) &= \left( a_0 + \cdots + a_mx^m + 0x^{m+1} + \cdots + 0x^r \right) + \left( b_0 + \cdots + b_rx^r \right) \\ \\ &= (a_0 + b_0) + \cdots + (a_m + b_m)x^m + \cdots + (0 + b_r)x^r \end{align*} $$

We can define polynomial functions with any polynomial.

Definition

Polynomial Functions

A function $f_p:F\rightarrow F$ can be defined using a univariate polynomial over a field $F$. For every $a\in F$ define $a\mapsto f_p(a)$ by,

$$y = f_p(a) = p(a)$$

Like matrices, we can multiply two polynomials. Two vectors in $\mathbb{R}^n$ cannot be multiplied. Will polynomial multiplication perform function composition as it does with matrices?

Definition

Polynomial Multiplication

The product of two (univariate) polynomials $p$ with degree $m$ and $q$ with degree $n$ is a polynomial with degree $m+n$ using the distributive property:

$$ \begin{align*} p\cdot q = (p\cdot q)(x) &= p(x)\cdot q(x) \\ \\ &= (a_0 + \cdots + a_mx^m)(b_0 + \cdots + b_nx^n) \\ \\ &= a_0(b_0 + \cdots + b_nx^n) + a_1x(b_0 + \cdots + b_nx^n) + \cdots + a_mx^m(b_0 + \cdots + b_nx^n) \\ \\ &= a_0b_0 + (a_0b_1 + a_1b_0)x + \cdots + a_mb_nx^{m+n} \end{align*} $$

This is not the same as function composition which would be evaluated:

$$ \begin{align*} p\circ q = (p\circ q)(x) &= p(q(x)) \\ \\ &= a_0 + a_1q(x) + \cdots + a_m(q(x))^m \\ \\ &= a_0 + a_1(b_0 + \cdots + b_nx^n) + \cdots + a_m(b_0 + \cdots + b_nx^n)^m \end{align*} $$

This polynomial has degree $mn$, however the constant term is

$$a_0 + a_1b_0 + \cdots + a_mb_0^m = \sum_{k=0}^m a_kb_0^k$$

Definition

Polynomial Scalar Multiplication

If $p$ is a univariate polynomial with degree $m$ over a field $F$, and $c\in F$, then the product

$$cp = c(a_0 + \cdots + a_mx^m) = ca_0 + \cdots + ca_mx^m$$

If $c = 1$, then

$$1p = 1(a_0 + \cdots + a_mx^m) = 1a_0 + \cdots + 1a_mx^m = a_0 + \cdots + a_mx^m = p$$

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4.2.5 A Vector Space of Polynomials

Now let us consider a vector space that does not inherit its algebraic structure from the real or complex numbers, and whose geometry differs significantly from Euclidean geometry.

Consider the set

$$P = \left\{\, \text{polynomials with real coefficients and one variable} \,\right\}$$

Define the sum of two polynomials and scalar multiplication of a polynomial using the definitions is the previous section. The $P$ with these two definitions for addition and scalar multiplication is a vector space.

Theorem 2

The Set of Real Polynomials is a Vector Space

Define the set $P$ of all univariate polynomials with real coefficients. Define the sum of two polynomials, and the scalar product of a polynomial as in the previous section. Then the set $P$ with this definition of addition and scalar multiplication is a vector space.

Exercise 2 - The Set of Real Polynomials is a Vector Space

Prove Theorem 2.

View Solution

We begin by showing that both scalar multiplication and vector addition satisfy the closure properties.

Suppose that $p$ with degree $m$ and $q$ with degree $n$ are polynomials in $P$. Then

1. Let $r=\max{m,n}$, and pad either polynomial with zero terms to that both polynomials have $r+1$ terms. Then $p+q = (a_0 + b_0) + \cdots + (a_r+b_r)x^r$ which is a polynomial with degree $r$ in $P$.

2. $cp = c(a_0 + \cdots + a_mx^m) = ca_0 + \cdots + ca_mx^m$, a polynomial with degree $\left\{ \begin{array}{rcl} m & \text{ if } & c\neq 0 \\ \text{undefined} & \text{ if } & c=0 \end{array} \right.$

Now we must establish the eight axioms of vector spaces. If, in addition, $\pi$ is a polynomial with degree $\nu$ and $d\in\mathbb{R}$, then we must define $r = \max\left\{m,n,\nu\right\}$, and pad two of the polynomials with zero terms so that all three have $r+1$ terms. Then,

1. $p + q = \sum_{k=0}^r (a_k + b_k)x^k = \sum_{k=0}^r (b_k + a_k)x^k = q + p$. (Does this look eerily similar to adding vectors in $\mathbb{R}^n$?)

2. We can add $p$, $q$ and $r$ associatively. $$ \begin{align*} (p + q) + \pi &= \sum_{k=0}^r (a_k + b_k)x^k + \sum_{k=0}^r c_kx^k \\ &= \sum_{k=0}^r \left[ (a_k + b_k)x^k + c_kx^k \right] \\ &= \sum_{k=0}^r \left[(a_k + b_k) + c_k\right]x^k \\ &= \sum_{k=0}^r \left[a_k + (b_k + c_k)\right]x^k \\ &= \sum_{k=0}^r a_kx^k + \sum_{k=0}^r (b_k + c_k)x^k \\ &= p + (q + \pi) \end{align*} $$ 3. The zero polynomial $0$ has the property that $p + 0 = 0 + p = p$.

4. The polynomial $(-1)p$ has the property that $(-1)p = -a_0 - \cdots - a_mx^m$ so that $(-1)p + p = p + (-1)p = 0$. Thus the additive inverse $-p = (-1)p$.

5. Scalar multiplication distributes over polynomial addition. Returning to $r = \max\{m,n\}$, we have $$ \begin{align*} c(p + q) &= c(a_0 + b_0) + \cdots + c(a_r + b_r)x^r \\ &= \sum_{k=0}^r c(a_k + b_k)x^k = \sum_{k=0}^r (ca_k + cb_k)x^k \\ &= \sum_{k=0}^r ca_kx^k + \sum_{k=0}^r cb_kx^k = c\sum_{k=0}^r a_kx^k + c\sum_{k=0}^r b_kx^k \\ &= cp + cq \end{align*} $$ 6. Scalar multiplication distributes over scalar addition. $$ \begin{align*} (c + d)p &= \sum_{k=0}^r (c + d)a_kx^k = \sum_{k=0}^r (ca_k + da_k)x^k = \sum_{k=0}^r ca_kx^k + \sum_{k=0}^r da_kx^k \\ &= c\sum_{k=0}^r a_kx^k + d\sum_{k=0}^r a_kx^k = cp + dp \end{align*} $$ 7. Scalar multiplication is associative. $$c(dp) = c\left( d\sum_{k=0}^r a_kx^k \right) = cd\sum_{k=0}^r a_kx^k = (cd)p$$ 8. $1p = 1\sum_{k=0}^r a_kx^k = \sum_{k=0}^r 1a_kx^k = \sum_{k=0}^r a_kx^k = p$.

Now that we established that $P$ is a vector space, the notes will denote polynomials in $P$ using vector notation. So if $p(x)$ is a polynomial in $P$, then we write $\mathbf{p}\in P$. If vector (polynomial) $\mathbf{p}\in P$ has degree $n$, then

$$\mathbf{p} = p(x) = p_0 + p_1x + \cdots + p_nx^n$$

We must agree that $\mathbf{p}$ may also be written

$$\mathbf{p} = p_nx^n + \cdots p_1x + p_0$$

without confusion. In subsequent notes, we will use ascending or descending degree terms according to which seems easiest for each application.

There are many vector spaces, which is why studying them is so important. These next examples of vector spaces are very important in modern engineering and physics.

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4.2.6 More Vector Spaces of Polynomials

If we have a polynomial $\mathbf{p}$ with degree $3$, we can write the polynomial using $4$ terms.

$$\mathbf{p} = p(x) = p_0 + p_1x + p_2x^2 + p_3x^3$$

We can write the same polynomial using $7$ terms by adding higher degree terms with zero coefficients.

$$\mathbf{p} = p(x) = p_0 + p_1x + p_2x^2 + p_3x^3 + 0x^4 + 0x^5 + 0x^6$$

The zero terms do not affect the degree of $\mathbf{p}$. The degree of $\mathbf{p}$ is still $3$. In this way polynomials with degree $k$ strictly less than positive integer $n$ can still be written using $n+1$ terms. The terms with degree greater than $k$ have zero coefficients:

$$\mathbf{p} = p(x) = p_0 + \cdots + p_kx^k + 0x^{k+1} + \dots + 0x^n$$

Define the set $P_6 = \left\{\, \mathbf{p}\in P\,:\, \textrm{deg}(\mathbf{p})\le 6\,\right\}$. Polynomials with degree $0-5$ are in $P_6$, and we can write them all using $6+1=7$ terms if we add enough terms with zero coefficients. For example, if $\mathbf{q} = 1 + 2x + x^2$, then we can write $\mathbf{q}$:

$$\mathbf{q} = q(x) = 1 + 2x + x^2 + 0x^3 + 0x^4 + 0x^5 + 0x^6$$

Example 2 - Polynomial Vector Spaces $P_n$

The set $P_n$ of polynomials with degree $\le$ positive integer $n$ is a vector space.

To prove this first show that the set $P_n$ is closed under polynomial addition and scalar multiplication. Let $\mathbf{p}$ and $\mathbf{q}$ be polynomials in $P_n$, and $c$ be any scalar.

1. $\mathbf{p}\in P_n$ implies the degree $\textrm{deg}(\mathbf{p}) \le n$, and $\mathbf{q}\in P_n$ implies that the degree $\text{deg}(\mathbf{q}) \le n$. The degree of $\textrm{deg}(\mathbf{p}+\mathbf{q}) = \max\left\{\, \textrm{deg}(\mathbf{p}),\ \textrm{deg}(\mathbf{q}) \,\right\} \le n$. Hence $(\mathbf{p}+\mathbf{q})\in P_n$.

2. Let $d = \textrm{deg}(\mathbf{p})$. Then $d\le n$ and $\mathbf{p} = p(x) = p_0 + \cdots + p_dx^d$. Hence $c\mathbf{p} = c\left(p_0 + \cdots + p_dx^d\right) = cp_0 + \cdots + cp_dx^d \in P_n$.

This establishes the closure properties of $P_n$.

Exercise 3 - Prove the Axioms

Show that $P_n$, the set of polynomials of degree at most $n$, is a vector space by proving that the eight axioms of a vector space are satisfied.

View Solution

Suppose that $\mathbf{p}$, $\mathbf{q}$, and $\mathbf{r}$ are polynomials in $P_n$, and $c$ and $d$ are scalars. Even though these three polynomials have different or equal degree less than or equal to $n$, we may write each polynomial using $n+1$ terms by padding each polynomial with enough terms with zero coefficients. $$ \begin{align*} p(x) &= p_0 + \cdots + p_nx^n = \sum_{k=0}^n p_kx^k \\ q(x) &= q_0 + \cdots + q_nx^n = \sum_{k=0}^n q_kx^k \\ r(x) &= r_0 + \cdots + r_nx^n = \sum_{k=0}^n r_kx^k \\ \end{align*} $$
1. Polynomial addition is still commutative. $$ \begin{align*} \mathbf{p} + \mathbf{q} &= \sum_{k=0}^n p_kx^k + \sum_{k=0}^n q_kx^k \\ &= \sum_{k=0}^n (p_k+q_k)x^k = \sum_{k=0}^n (q_k + p_k)x^k \\ &= \sum_{k=0}^n q_kx^k + \sum_{k=0}^n p_kx^k = \mathbf{q} + \mathbf{p} \end{align*} $$ 2. Polynomial addition is associative. $$ \begin{align*} (\mathbf{p} + \mathbf{q}) + \mathbf{r} &= \left( \sum_{k=0}^n p_kx^k + \sum_{k=0}^n q_kx^k \right) + \sum_{k=0}^n r_kx^k \\ &= \sum_{k=0}^n (p_k + q_k)x^k + \sum_{k=0}^n r_kx^k = \sum_{k=0}^n \left[ (p_k + q_k) + r_k \right]x^k \\ &= \sum_{k=0}^n \left[ p_k + (q_k + r_k) \right]x^k = \sum_{k=0}^n p_kx^k + \sum_{k=0}^n (q_k + r_k)x^k \\ &= \sum_{k=0}^np_kx^k + \left( \sum_{k=0}^n q_kx^k + \sum_{k=0}^n r_kx^k \right ) = \mathbf{p} + (\mathbf{q} + \mathbf{r}) \end{align*} $$ 3. The zero polynomial can be written using $n+1$ terms with zero coefficients. $$ \begin{align*} \mathbf{0} &= 0 + 0x + \cdots + 0x^n \\ \mathbf{p} + \mathbf{0} &= \sum_{k=0}^n p_kx^k + \sum_{k=0}^n 0x^k = \sum_{k=0}^n (p_k + 0)x^k = \sum_{k=0}^n p_kx^k \\ &= \mathbf{p} = \sum_{k=0}^n p_kx^k = \sum_{k=0}^n (0 + p_k)x^k = \sum_{k=0}^n 0x^k + \sum_{k=0}^n p_kx^k \\ &= \mathbf{0} + \mathbf{p} \end{align*} $$
The reader should finish the rest of the exercise.

Example 3 - Polynomials and Vectors

Consider the following examples of polynomials in $P_3$:

1. $-2 - 2x + 2x^2 + 0x^3$, a polynomial with degree 2
2. $-2 + 2x - x^2 + 0x^3$, a polynomial with degree 2
3. $1 + x + 0x^2 + 0x^3$, a polynomial with degree 1
4. $0 + x + 0x^2 - 2x^3$, a polynomial with degree 3
5. $1 + 0x + 0x^2 + 0x^3$, a polynomial with degree 0
6. $0 + 0x + 0x^2 + 0x^3$, a polynomial with no degree (the only one)

When we insist on writing them with four terms, we notice that there are four coefficients, even if some or all of them are zeros. This should remind everyone of $\mathbb{R}^4$ because vectors in $\mathbb{R}^4$ also require a list of exactly four numbers. In fact we can create a bijection $\Phi\,:\,P_3\rightarrow\mathbb{R}^4$ defined for every $\mathbf{p}\in P_3$ by

$$\Phi(\mathbf{p}) = \Phi( p_0 + p_1x + p_2x^2 + p_3x^3 ) = \begin{bmatrix} p_0 \\ p_1 \\ p_2 \\ p_3 \end{bmatrix}$$

Definition

Bijection

A bijection from domain $D$ to codomain $E$ is a function $f\,:\,D\rightarrow E$ that is one-to-one and onto.

A function $f\,:\,D\rightarrow E$ is one-to-one when every element in the image $f(D)\subset E$ of $f$ has an unique pre-image. If $y\in f(D)$, then $f^{-1}(y)$ is a singleton set. Recall that the definition of function requires that every element of the domain must be mapped uniquely to one element of the codomain. However several elements can be mapped to the same element of the codomain.

A constant function maps every element of the domain to one element of the codomain. $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f(x)=3$ maps every real number to the same output.

Functions that are one-to-one also map only one element of the domain to each element of the function's image set. One usually defines one-to-one as follows.

If $f(x)=f(y)$, then $x=y$. In other words, if $x$ and $y$ get mapped to the same point $f(x)=f(y)$, then $x$ and $y$ are the same point too.

A function $f\,:\,D\rightarrow E$ is onto when every element in the codomain has nonempty pre-image. That means for every $y\in E$, there is at least one $x\in D$ so that $y=f(x)$.

A function that is both one-to-one and onto maps exactly one element of the domain to every element of the codomain. This mirrors the definition of function which states, every element of the domain is mapped to exactly one element of the codomain.

Bijections are important because it means that the domain and the codomain have the same number of elements and they are matched up uniquely by the bijection.

Exercise 4 - Identify the Polynomial

Identify the polynomial in $P_3$ associated to each vector in $\mathbb{R}^4$ by our bijection $\Phi$.

1. $\langle 1, -3, 2, 2 \rangle$
2. $\langle 1, 0, 2, -1 \rangle$
3. $\langle 3, 0, 0, 2 \rangle$
4. $\langle -2, 0, 1, 0 \rangle$
5. $\langle 1, 1, 1, 3 \rangle$

View Solution

1. $\mathbf{p} = 1 - 3x + 2x^2 + 2x^3$
2. $\mathbf{p} = 1 + 2x^2 - x^3$
3. $\mathbf{p} = 3 + 2x^3$
4. $\mathbf{p} = -2 + x^2$
5. $\mathbf{p} = 1 + x + x^2 + 3x^3$

Exercise 5 - Identify the Vector

Identify the vector in $\mathbb{R}^4$ associated to each vector in $P_3$ by our bijection $\Phi$.

1. $\mathbf{p} = 1 + x - 2x^2 + x^3$
2. $\mathbf{p} = -2x + 2x^2 - x^3$
3. $\mathbf{p} = 2 - 2x^3$
4. $\mathbf{p} = 3$
5. $\mathbf{p} = x^3$

View Solution

1. $\mathbf{u} = \langle 1, 1, -2, 1 \rangle$
2. $\mathbf{v} = \langle 0, -2, 2, -1\rangle$
3. $\mathbf{w} = \langle 2, 0, 0, -2\rangle$
4. $\mathbf{x} = \langle 3, 0, 0, 0\rangle$
5. $\mathbf{y} = \langle 0, 0, 0, 1\rangle$

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4.2.7 Vector Spaces of Functions

We use the notation $C[a,b]$ to denote the set of functions that are continuous on the closed interval $[a,b]$. If we suppose, for simplicity, that the functions $f,g\in C[a,b]$ are real-valued and $\alpha$ is a real scalar, then we may talk about the operations of scalar multiplication and vector addition on this space. They are defined in the typical way.

Vector addition:

$$ (f + g)(x) := f(x) + g(x),\ \text{for every $x\in[a,b]$} $$

Scalar multiplication:

$$ (c\,f)(x) := c\,f(x),\ \text{for every $x\in[a,b]$} $$

Recall that two functions $f$ and $g$ are equal if and only if for every element of the domain of $f$ and $g$, $f(x)=g(x)$. This makes $f$ and $g$ the same function. The definitions of addition and scalar multiplication of functions clearly give us the required closure properties that we learned in our differential calculus class,

1. the sum of two continuous functions is a continuous function

2. the product of a scalar and a continuous function is a continuous function

This establishes the closure properties in $C[a,b]$.

Exercise 6 - Show that $C[a,b]$ is a Vector Space

Show that $C[a,b]$ is a vector space.

View Solution

Let $f$, $g$, and $h$ be functions continuous on the interval $[a,b]$, and $c$ and $d$ be any scalars. Then

1. Addition of functions is commutative. For every $x\in[a,b]$, $$ \begin{align*} (f + g)(x) &= f(x) + g(x) = g(x) + f(x) = (g + f)(x) \end{align*} $$ Since this equation is true for every $x\in[a,b]$, $f + g = g + f$ in $C[a,b]$.

2. Addition of functions is associative. For every $x\in[a,b]$, $$ \begin{align*} \Big(f + (g + h)\Big)(x) &= f(x) + \Big( g(x) + h(x) \Big) \\ &= \Big( f(x) + g(x) \Big) + h(x) \\ &= \Big( (f + g) + h \Big)(x) \end{align*} $$

Since this equation is true for every $x\in[a,b]$, $(f + g) + h = f + (g + h)$ in $C[a,b]$.

The reader should finish this proof.

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4.2.8 Complex Numbers

Complex numbers $\mathbb{C}$ also form a vector space. We've not discussed it explicitly, but so do the real numbers $\mathbb{R}$. The fact that real numbers are a vector space should not be surprising, since we can think of $\mathbb{R}$ as $\mathbb{R}^1$ and we've shown that all $\mathbb{R}^n$ (even $n=1$) form vector spaces.

The mechanics for showing that $\mathbb{C}$ is a vector space are pretty straightforward, with one twist. Instead of using the real numbers as our scalars, we use the complex numbers. Let $x,y,\zeta\in\mathbb{C}$ be given by

$$ x = a + bi \qquad\qquad y = c + di \qquad\qquad \zeta = \alpha + \beta i $$
where $a,b,c,d,\alpha,\beta$ are all real numbers and $i = \sqrt{-1}$ is the imaginary unit. (We'll be using Greek letter $\zeta$ (zeta) as our scalar.) Our operation of scalar multiplication

$$ \begin{align*} \zeta x &= (\alpha + \beta i)(a + bi) \\ \\ &= \alpha (a + bi) + \beta i (a + bi) \\ \\ &= \alpha a + \alpha b i + \beta a i + \beta b i^2 \\ \\ &= \alpha a + \alpha b i + \beta a i + \beta b (-1) \\ \\ &= \alpha a - \beta b + \alpha b i + \beta a i \\ \\ &= \left( \alpha a - \beta b \right) + \left( \alpha b + \beta a \right)i \in\mathbb{C} \end{align*} $$
is closed. It is much easier to show closure for vector addition.

$$ \begin{align*} x + y &= (a + bi) + (c + di) \\ \\ &= a + bi + c + di \\ \\ &= a + c + bi + di \\ \\ &= (a + c) + (b + d)i\in\mathbb{C} \end{align*} $$

Exercise 7 - Show that $\mathbb{C}$ is a Vector Space

Show that $\mathbb{C}$ is a vector space.

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4.2.9 Properties of Scalar Multiplication

Theorem 3

Properties of Scalar Multiplication

If $\mathbf{v}\in V$ is a vector in vector space $V$, and $c$ is any scalar, then

1. $0\mathbf{v} = \mathbf{0}$

2. $c\mathbf{0} = \mathbf{0}$

3. If $c\mathbf{v} = \mathbf{0}$, then $c=0$ or $\mathbf{v}=\mathbf{0}$

4. $(-1)\mathbf{v} = -\mathbf{v}$

Proof of Theorem 3

1. We can use the fact that $0 = 0 + 0$ to show that $$ \begin{align*} 0\mathbf{v} &= (0 + 0)\mathbf{v} \qquad &\text{property of real numbers} \\ 0\mathbf{v} &= 0\mathbf{v} + 0\mathbf{v} \qquad &\text{Axiom 6} \\ 0\mathbf{v} + (-0\mathbf{v}) &= \left( 0\mathbf{v} + 0\mathbf{v}\right) + (-0\mathbf{v}) \qquad &\text{Closure Property 1} \\ \mathbf{0} &= \left( 0\mathbf{v} + 0\mathbf{v}\right) + (-0\mathbf{v}) \qquad &\text{Axiom 4} \\ \mathbf{0} &= 0\mathbf{v} + \left( 0\mathbf{v} + (-0\mathbf{v})\right) \qquad &\text{Axiom 2} \\ \mathbf{0} &= 0\mathbf{v} + \mathbf{0} \qquad &\text{Axiom 4} \\ \mathbf{0} &= 0\mathbf{v} \qquad &\text{Axiom 3} \end{align*} $$
   

Notice that we proved Theorem 3 for all vector spaces using the axioms in the definition of a vector space!

Proofs for Properties 2-4 are left as an exercise for the reader.

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4.2.10 Examples of Vector Spaces

We need to recognize vector spaces in our mathematical models and create mathematical models utilizing vector spaces. This chapter is a lot of work, and worth every minute. Mastery of vector spaces and subspaces will make the rest of the course much easier to understand.

Summary

Some of the important vector spaces to know

$\mathbb{R} =\,$ the vector space of real numbers

$\mathbb{R}^2 =\,$ the vector space of 2-tuples

$\mathbb{R}^3 =\,$ the vector space of 3-tuples

$\mathbb{R}^n =\,$ the vector space of $n$-tuples

$P =\,$ the vector space of all polynomials

$P_n=\,$ the vector space of polynomials of degree $\le\ n$

$\mathbb{R}^{m\times n} = M_{m,n} = M_{m,n}(\mathbb{R}) =\,$ the vector space of $m\times n$ real-valued matrices

$\mathbb{R}^{n\times n} = M_{n,n} = M_{n,n}(\mathbb{R}) =\,$ the vector space of all square $n\times n$ real-valued matrices

$\mathbb{C}^{m\times n} = M_{m,n}(\mathbb{C}) =\,$ the vector space of all $m\times n$ complex-valued matrices

$\mathbb{C}^{n\times n} = M_{n,n}(\mathbb{C}) =\,$ the vector space of all square $n\times n$, complex-valued matrices

$C(-\infty, \infty) =\,$ the vector space of all continuous functions defined on the real number line

$C[a,b] =\,$ the vector space of continuous functions defined on the closed interval $[a,b]$, where $a\neq b$

$C^1[a,b] =\,$ the vector space of continuously differentiable functions on the interval $[a,b]$, where $a\neq b$

$C^2[a,b] =\,$ the vector space of twice continuously differentiable functions on the interval $[a,b]$, where $a\neq b$

$C^n[a,b] =\,$ the vector space of $n$ times, continuously differentiable functions on the interval $[a,b]$, where $a\neq b$

$C^{\infty}[a,b] =\,$ the vector space of functions on the interval $[a,b]$, where $a\neq b$, whose $k^{\text{th}}$ derivative exists and is continuous for every positive integer $k$

$\mathbb{R}^{2n} =\,$ the vector space of $\,$ states of an object in orbit about a gravity source (typically $\mathbb{R}^6$ in our 3D-world)

$\mathcal{H} =\,$ the vector space of $\,$ states of a quantum system

$\mathcal{L}\left(\mathscr{H}\right) =\,$ the vector space of linear operators on a Hilbert Space.

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