Vector Spaces

Section 3.1 Vector Spaces

Objectives

The definition of a vector space. This has 10 properties, most of them are familiar.
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Examples of sets of "vectors" some of which are vectors and some of which are not.
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A vector space as we will see is a set of vectors with a certain collection of properties. This notion arose from vectors of $\mathbb{R}^n\text{,}$ however, we will see that they generalize.

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Definition 3.1.1. Vector Space.

Let $V$ be a nonempty set of elements with $\vec{x}, \vec{y}, \vec{z} \in V$ and $c,k \in \mathbb{R}\text{.}$ The set $V$ is called a real vector space and the elements of $V$ are called vectors if in $V$ there are defined two algebraic operations called vector addition and denoted $+$ and scalar multiplication and denoted $\cdot$ that satisfy the following:

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Vector Addition: 🔗

The operation $+$ satisfies:

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(Closure): $\vec{x}+\vec{y} \in V\text{.}$
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(Commutativity:) $\vec{x} +\vec{y} = \vec{y}+\vec{x}$
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(Associativity:) $(\vec{x} + \vec{y}) + \vec{z} = \vec{x} + (\vec{y} + \vec{z})$
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(Identity Element:) There exist an element $\vec{0} \in V$ that satisfies: $\vec{x} + \vec{0} = \vec{x}\text{.}$
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(Inverse Elements:) For every $\vec{x} \in V\text{,}$ there exists an element denoted $-\vec{x}$ such that $\vec{x} + (-\vec{x}) = \vec{0}\text{.}$
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Scalar Multiplication: 🔗

The real numbers $c$ and $k$ are called scalars. The operations $+$ and $\cdot$ also satisfy:

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(Closure:) $c \cdot \vec{x} \in V\text{.}$
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(Distributivity:) $c \cdot (\vec{x} + \vec{y}) = c\cdot \vec{x} + c\cdot \vec{y}$
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(Distributivity:) $(c+k) \cdot \vec{x} = c\cdot \vec{x} + k\cdot \vec{x}$
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(Associativity:) $c\cdot (k\cdot \vec{x}) = (ck)\cdot \vec{x}$
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(Identity): For every $\vec{x} \in V\text{,}$ $1 \cdot \vec{x} = \vec{x}\text{.}$
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We will show that the familiar sets $\mathbb{R}^2, \mathbb{R}^3,$ and $\mathbb{R}^n$ are vectors spaces. However, it is interesting that there are many other common sets of mathematical objects that are also vectors spaces.

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Subsection 3.1.1 Examples of Vector Spaces

Let’s begin with the set of all vectors of length 2, which we typically write as $\mathbb{R}^2\text{.}$

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Lemma 3.1.2.

Show that $\mathbb{R}^2$ together with the standard notions of vector addition and scalar multiplication is a vector space.

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Proof.

Let

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\begin{align*} \vec{x} \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \amp \vec{y} \amp = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \amp \vec{z} \amp = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \end{align*}

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where $x_1, x_2, y_1, y_2, z_1, z_2 \in \mathbb{R}$ and $c,k \in \mathbb{R}$ as well as vector addition and scalar multiplication as

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\begin{align*} \vec{x} + \vec{y} \amp = \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix} \amp c \cdot \vec{x} \amp = \begin{bmatrix} c x_1 \\ c x_2 \end{bmatrix} \end{align*}

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We now show that all 10 properties of a vector space hold:

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Additive Closure: Since

\begin{align*} \vec{x} + \vec{y} \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix} \end{align*}

are elements of $\mathbb{R}^2\text{,}$ the closure by addition holds.

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Additive Commutativity: For all $\vec{x}, \vec{y}, \vec{z} \in V\text{,}$

\begin{align*} \vec{x} + \vec{y} \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix}\\ \amp = \begin{bmatrix} y_1 + x_1 \\ y_2 + x_2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} + \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \vec{y} + \vec{x} \end{align*}

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Additive Associativity:

\begin{align*} (\vec{x} + \vec{y}) + \vec{z} \amp = \biggl( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \biggr) + \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\\ \amp = \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix} + \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\\ \amp = \begin{bmatrix} (x_1 + y_1) + z_1 \\ (x_2 + y_2) + z_2 \end{bmatrix}\\ \amp = \begin{bmatrix} x_1 + (y_1 + z_1) \\ x_2 + (y_2 + z_2) \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+ \begin{bmatrix} y_1 + z_1 \\ y_2 + z_2 \end{bmatrix}\\ \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \biggl( \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} + \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \biggr) = \vec{x} + (\vec{y} + \vec{z}) \end{align*}

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Additive Identity Element: The zero element $\vec{0}$ is

\begin{equation*} \vec{0} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{equation*}

and it satisfies,

\begin{align*} \vec{0} + \vec{x} \amp = \begin{bmatrix} 0 \\ 0 \end{bmatrix} + \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 + x_1 \\ 0 + x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \vec{x} \end{align*}

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Additive Inverse Element: The inverse element $-\vec{x}$ is given as

\begin{equation*} -\vec{x} = \begin{bmatrix} -x_1 \\ -x_2 \end{bmatrix} \end{equation*}

and the following satisfies:

\begin{align*} \vec{x} + (-\vec{x}) \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} -x_1 \\ -x_2 \end{bmatrix} = \begin{bmatrix} x_1 + (-x_1) \\ x_2 + (-x_2) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \vec{0} \end{align*}

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Multiplicative Closure: For $c \in \mathbb{R}\text{,}$

\begin{align*} c \cdot \vec{x} \amp = c \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} cx_1 \\ cx_2 \end{bmatrix} \end{align*}

which is in $\mathbb{R}^2\text{,}$ so scalar multiplication is closed.

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Distributivity over vectors: Let $c \in \mathbb{R}$ and

\begin{align*} c \cdot (\vec{x} + \vec{y}) \amp = c \cdot \biggl( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \biggr) = c\cdot \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix} = \begin{bmatrix} c (x_1 +y_1) \\ c (x_2 + y_2) \end{bmatrix} \\ \amp = \begin{bmatrix} c x_1 + c y_1 \\ c x_2 + c y_2 \end{bmatrix} = \begin{bmatrix} c x_1 \\ c x_2 \end{bmatrix} + \begin{bmatrix} c y_1 \\ c y_2 \end{bmatrix} = c \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + c \cdot \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \\ \amp = c\cdot \vec{x} + c\cdot \vec{y} \end{align*}

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Distributivity over scalars: Let $c$ and $k$ be elements of $\mathbb{R}\text{.}$

\begin{align*} (c + k)\cdot \vec{x} \amp = (c+k) \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} (c+k) x_1 \\ (c+k) x_2 \end{bmatrix} = \begin{bmatrix} c x_1 + k x_1 \\ c x_2 + k x_2 \end{bmatrix} \\ \amp = \begin{bmatrix} c x_1 \\ c x_2 \end{bmatrix} + \begin{bmatrix} k x_1 \\ k x_2 \end{bmatrix} = c \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + k \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = c \cdot \vec{x} + k \cdot \vec{x} \end{align*}

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Scalar Multiplicative Associativity: Let $c$ and $k$ be elements of $\mathbb{R}\text{.}$

\begin{align*} c\cdot (k\cdot \vec{x}) \amp = c \cdot \biggl( k \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \biggr) = c \cdot \begin{bmatrix} k x_1 \\ k x_2 \end{bmatrix} = \begin{bmatrix} c (k x_1) \\ c (kx_2) \end{bmatrix} = \begin{bmatrix} (c k) x_1 \\ (ck) x_2 \end{bmatrix} \\ \amp = (ck) \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = (ck) \cdot \vec{x} \end{align*}

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Multiplicative Identity element:

\begin{equation*} 1 \cdot \vec{x} = 1\cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1 \cdot x_1 \\ 1 \cdot x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \vec{x} \end{equation*}

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The next example shows that quadratic functions also satisfy these properties, hence the set of all polynomials of degree 2 form a vector space.

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Example 3.1.3.

Show than the set, $\mathcal{P}_2\text{,}$ is a vector space with vector addition defined as the sum of two polynomials and scalar multiplication as the multiplication of a polynomial by a constant.

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Solution.

In this case scalars are elements in $\mathbb{R}$ and the vector space $V$ is the set of all polynomials of degree at most $2\text{.}$ Let

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\begin{align*} \vec{x} \amp = a_0 + a_1 x + a_2 x^2 \\ \vec{y} \amp = b_0 + b_1 x + b_2 x^2 \\ \vec{z} \amp = c_0 + c_1 x + c_2 x^2 \end{align*}

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with $a_0,a_1,a_2, b_0,b_1,b_2, c_0, c_1, c_2 \in \mathbb{R}$

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Vector addition is simply polynomial addition for example:

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\begin{align*} \vec{x} + \vec{y} \amp = (a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x + b_2 x^2) \\ \amp = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2. \end{align*}

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and scalar multiplication is given by

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\begin{equation*} r \vec{x} = r(a_0 + a_1 x + a_2 x^2) = (ra_0) + (ra_1) x + (ra_2) x^2 \end{equation*}

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Additive Closure: Show that $\vec{x}+\vec{y}$ is in $\mathcal{P}_2\text{,}$ the set of polynomials of degree 2.

\begin{align*} \vec{x} + \vec{y} \amp = (a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x+ b_2 x^2) \\ \amp = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2 \end{align*}

which is a polynomial of degree 2, so this is in $\mathcal{P}_2\text{.}$

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Additive Commutativity:

\begin{align*} \vec{x} + \vec{y} \amp = (a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x +b_2 x^2) \\ \amp = (b_0 + b_1 x + b_2 x^2) + (a_0 + a_1 x + a_2 x^2) \\ \amp = \vec{y} + \vec{x} \end{align*}

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Additive Associativity:

\begin{align*} (\vec{x} + \vec{y}) + \vec{z} \amp = (a_0 + a_1 x + a_2 x^2+ b_0 + b_1 x + b_2 x^2) \\ \amp \qquad+ (c_0 + c_1 x + c_2 x^2) \\ \amp = a_0 + a_1 x + a_2 x^2 \\ \amp \qquad + (b_0 + b_1 x + b_2 x^2) + (c_0 + c_1 x + c_2 x^2) \\ \amp = \vec{x} + (\vec{y} + \vec{z}) \end{align*}

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Additive Identity: The zero element $\vec{0}$ is the function 0.

\begin{align*} \vec{0} + \vec{x} \amp = 0 + a_0 + a_1 x + a_2 x^2 \\ \amp = a_0 + a_1 x + a_2 x^2 = \vec{x} \end{align*}

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Additive Inverse: The inverse element $-\vec{x}$ is $-a_0 - a_1 x-a_2x^2\text{,}$ and

\begin{align*} \vec{x} + (-\vec{x}) \amp = (a_0 + a_1 x + a_2 x^2) + (-a_0 - a_1 x-a_2x^2) \\ \amp = (a_0 -a_0) + (a_1 -a_1) x + (a_2 -a_2) x^2 = 0 = \vec{0}. \end{align*}

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Closure under scalar multiplication:

\begin{equation*} r \vec{x} = r(a_0 + a_1 x + a_2 x^2) = (ra_0) + (ra_1) x + (ra_2) x^2 \end{equation*}

since this is a polynomial of degree 2, then $r \vec{x} \in \mathcal{P}_2\text{.}$

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Scalar Multiple Distributivity: Let $r, s \in \mathbb{R}\text{,}$

\begin{align*} (r + s) \vec{x} \amp = (r+s) (a_0 + a_1 x + a_2 x^2) \\ \amp = r (a_0 + a_1 x + a_2 x^2) + s (a_0 + a_1 x + a_2 x^2) = r \vec{x} + s \vec{x} \end{align*}

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Vector Distributivity: Let $r \in \mathbb{R}\text{,}$

\begin{align*} r (\vec{x} + \vec{y}) \amp = r ((a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x + b_2 x^2)) \\ \amp =r (a_0 + a_1 x + a_2 x^2) + r (b_0 + b_1 x + b_2 x_2) \\ \amp = r \vec{x} + r \vec{y} \end{align*}

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Scalar Multiple Associativity: Let $r, s \in \mathbb{R}\text{,}$

\begin{align*} (rs) \vec{x} \amp = (rs) (a_0 + a_1 x + a_2 x^2) = r (s(a_0 + a_1 x + a_2 x^2)) \\ \amp = r (s \vec{x}) \end{align*}

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Scalar Multiple Identity:

\begin{equation*} 1 \cdot \vec{x} = 1 (a_0 + a_1 x + a_2 x^2) = a_0 + a_1 x + a_2 x^2 = \vec{x}. \end{equation*}

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The following are also vector spaces. There are not proofs associated with these.

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${\cal P}_n\text{,}$ the set of polynomials of degree $n\text{.}$
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The set of vectors that consist of $\mathbb{R}^3$ with the standard definition of vector addition and scalar multiplication.
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The set of vectors that consist of $\mathbb{R}^n$ with the standard definition of vector addition and scalar multiplication.
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The set of all $m \times n$ matrices, $\mathcal{M}_{m \times n}$ with standard definition of matrix addition and scalar multiplication.
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In the next example, we will also show some other interesting sets are vector spaces. To save a bunch of time and effort, in some cases, we don’t need to show all 10 properties. We can skip most of properties, if the set $V$ is a subset of a known vector space. The reason for this is that properties 2-5 and 7-10 will automatically hold because the vectors are already in a vector space. We will do this formally in the next section.

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Example 3.1.4.

Show that all lines that pass through the origin, or the set

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\begin{equation*} V = \{ ax + by = 0\; | \; x, y \in \mathbb{R} \qquad \text{with $a$ and $b$ are not both 0.} \} \end{equation*}

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is a vector space.

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Solution.

First, let’s rewrite the vector space as

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\begin{equation*} V = \{ \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \; | \; \text{$au_1+bu_2=0$, with $a$ and $b$ both not 0.}\} \end{equation*}

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because this is the same set of points, but it is written in terms of a vector, which makes things easier to show. As stated above, since $V$ is a subset of $\mathbb{R}^2\text{,}$ we don’t need to prove properties 2--5, 7--10, which hold because we already showed that these properties show. Instead, to save time and effort, we only prove properties 1 and 6 of definition of a vector space in Definition 3.1.1.

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Let $\vec{u}, \vec{v} \in V\text{.}$ Therefore the two equations $au_1+bu_2=0$ and $av_1+bv_2=0$ are satisfied. Now check the sum,
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\begin{equation*} \vec{u} + \vec{v} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix} \end{equation*}

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and need to show that this is in $V\text{.}$ Since
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\begin{align*} a(u_1 + v_1) + b(u_2+v_2) \amp = au_1 + av_1 + bu_2 + bv_2\\ \amp =(au_1 + bu_2) + (av_1 + bv_2) = 0 + 0 = 0 \end{align*}

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then $\vec{u}+\vec{v} \in V\text{.}$
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We need to show that $r \cdot \vec{u} \in V\text{,}$ so that the vector $r\cdot\vec{u}$ satisfies the property that the line passes through the origin.
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\begin{equation*} r \cdot \vec{u} = \begin{bmatrix} r u_1 \\ r u_2 \end{bmatrix} \end{equation*}

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and since $au_1 + bu_2 = 0\text{,}$ then
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\begin{equation*} r \cdot 0 = r(au_1 + bu_2) = a (ru_1) + b (r u_2) =0 \end{equation*}

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therefore $r \cdot \vec{u} \in V\text{.}$
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So this is a vector space.

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The next example shows that polynomials (specifically quadratics) are vector spaces. Since we haven’t shown anything about polynomials yet, we need to show all 10 properties of Definition 3.1.1 hold.

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Example 3.1.5.

Show than the set, $\mathcal{P}_2\text{,}$ the set of all polynomials of degree $2$ is a vector space with vector addition defined as the sum of two polynomials and scalar multiplication as the multiplication of a polynomial by a constant.

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Solution.

In this case scalars are elements in $\mathbb{R}$ and the vector space $V$ is the set of all polynomials of degree at most $2\text{.}$ Let

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\begin{align*} \vec{x} \amp = a_0 + a_1 x + a_2 x^2 \\ \vec{y} \amp = b_0 + b_1 x + b_2 x^2 \\ \vec{z} \amp = c_0 + c_1 x + c_2 x^2 \end{align*}

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Vector addition is simply polynomial addition for example:

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\begin{align*} \vec{x} + \vec{y} \amp = (a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x + b_2 x^2) \\ \amp = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2. \end{align*}

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and scalar addition is given by

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\begin{equation*} r \vec{x} = r(a_0 + a_1 x + a_2 x^2) = (ra_0) + (ra_1) x + (ra_2) x^2 \end{equation*}

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Additive Closure: Show that $\vec{x}+\vec{y}$ is in $\mathcal{P}_2\text{,}$ the set of polynomials of degree 2.

\begin{align*} \vec{x} + \vec{y} \amp = (a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x+ b_2 x^2) \\ \amp = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2 \end{align*}

which is a polynomial of degree 2, so this is in $\mathcal{P}_2\text{.}$

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Additive Commutativity:

\begin{align*} \vec{x} + \vec{y} \amp = (a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x +b_2 x^2) \\ \amp = (b_0 + b_1 x + b_2 x^2) + (a_0 + a_1 x + a_2 x^2) \\ \amp = \vec{y} + \vec{x} \end{align*}

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Additive Associativity:

\begin{align*} (\vec{x} + \vec{y}) + \vec{z} \amp = (a_0 + a_1 x + a_2 x^2+ b_0 + b_1 x + b_2 x^2) \\ \amp \qquad+ (c_0 + c_1 x + c_2 x^2) \\ \amp = a_0 + a_1 x + a_2 x^2 \\ \amp \qquad + (b_0 + b_1 x + b_2 x^2) + (c_0 + c_1 x + c_2 x^2) \\ \amp = \vec{x} + (\vec{y} + \vec{z}) \end{align*}

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Additive Identity: The zero element $\vec{0}$ is the function 0.

\begin{align*} \vec{0} + \vec{x} \amp = 0 + a_0 + a_1 x + a_2 x^2 \\ \amp = a_0 + a_1 x + a_2 x^2 = \vec{x} \end{align*}

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Additive Inverse: The inverse element $-\vec{x}$ is $-a_0 - a_1 x-a_2x^2\text{,}$ and

\begin{align*} \vec{x} + (-\vec{x}) \amp = (a_0 + a_1 x + a_2 x^2) + (-a_0 - a_1 x-a_2x^2) \\ \amp = (a_0 -a_0) + (a_1 -a_1) x + (a_2 -a_2) x^2 = 0 = \vec{0}. \end{align*}

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Closure under scalar multiplication:

\begin{equation*} r \vec{x} = r(a_0 + a_1 x + a_2 x^2) = (ra_0) + (ra_1) x + (ra_2) x^2 \end{equation*}

since this is a polynomial of degree 2, then $r \vec{x} \in \mathcal{P}_2\text{.}$

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Scalar Multiple Distributivity: Let $r, s \in \mathbb{R}\text{,}$

\begin{align*} (r + s) \vec{x} \amp = (r+s) (a_0 + a_1 x + a_2 x^2) \\ \amp = r (a_0 + a_1 x + a_2 x^2) + s (a_0 + a_1 x + a_2 x^2) = r \vec{x} + s \vec{x} \end{align*}

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Vector Distributivity: Let $r \in \mathbb{R}\text{,}$

\begin{align*} r (\vec{x} + \vec{y}) \amp = r ((a_0 + a_1 x + a_2 x^2) + (b_0 + b_1 x + b_2 x^2)) \\ \amp =r (a_0 + a_1 x + a_2 x^2) + r (b_0 + b_1 x + b_2 x_2) \\ \amp = r \vec{x} + r \vec{y} \end{align*}

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Scalar Multiple Associativity: Let $r, s \in \mathbb{R}\text{,}$

\begin{align*} (rs) \vec{x} \amp = (rs) (a_0 + a_1 x + a_2 x^2) = r (s(a_0 + a_1 x + a_2 x^2)) \\ \amp = r (s \vec{x}) \end{align*}

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Scalar Multiple Identity:

\begin{equation*} 1 \cdot \vec{x} = 1 (a_0 + a_1 x + a_2 x^2) = a_0 + a_1 x + a_2 x^2 = \vec{x}. \end{equation*}

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The above example shows that there are examples of vectors and vector spaces that don’t seem like vectors in $\mathbb{R}^n\text{.}$ The following are also vector spaces:

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The set $\{a \sin x + b \cos x \; | \; a, b \in \mathbb{R} \}\text{.}$
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The set $\{ f \; | \; f: \mathbb{R} \rightarrow \mathbb{R} \}$ of functions of a single variable.
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The set of functions that satisfy $f'+f=0\text{.}$
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Solutions of homogeneous differential equations, which is a generalization of functions that satisfy $f'+f=0\text{.}$
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Note that above there are sets of functions that are vector spaces. Similar to the 2nd degree polynomials, the “vector addition” is given by $(f+g) (x)= f(x) + g(x)$ and the “scalar multiplication” is $(r \cdot f) (x) = r \cdot f(x)\text{.}$

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Subsection 3.1.2 Is every set a vector space?

In short, no. For example, consider the following:

\begin{equation*} V = \left\{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \; | \; x_1 \geq 0, x_2 \geq 0 \right\} \end{equation*}

which is the first quadrant of the plane. To check if this is a vector space, we don’t need to check all 10 properties. This is because since $V$ is a subset of $\mathbb{R}^2\text{,}$ most of the properties (commutative, associative, distributive, for examples) will work just as they did in $\mathbb{R}^2\text{.}$ The only two that we do need to check are the closure properties.

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If $\vec{u}, \vec{v} \in V\text{,}$ then

\begin{align*} \vec{u} + \vec{v} \amp = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix} \end{align*}

and since $u_1, u_2, v_1, v_2$ are all nonnegative, then this vector is in $V\text{.}$

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Let $r \in \mathbb{R}\text{,}$ then

\begin{equation*} r \vec{u} = \begin{bmatrix} r u_1 \\ r u_2 \end{bmatrix} \end{equation*}

but if $r \lt 0$ then $r \vec{u} \not \in V\text{,}$ so $V$ is not a vector space.

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Example 3.1.6.

Is the set

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\begin{equation*} U=\left\{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \; | \; x_1^2+x_2^2=1\right\} \end{equation*}

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a vector space? If not what properties does it violate?

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Solution.

This set is a circle of radius one centered at the origin. Again, like above, $U$ is a subset of $\mathbb{R}^2$ and thus the properties to check are the closure properties.

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If we let

\begin{align*} \vec{x} \amp= \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\amp \vec{y} \amp = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \end{align*}

both be in $U\text{,}$ then

\begin{align*} \vec{x}+\vec{y} \amp = \begin{bmatrix} x_1 + y_1 \\ x_2 +y_2 \end{bmatrix} \end{align*}

Is this in $U\text{.}$ To check sum the squares of the individual elements.

\begin{align*} (x_1+y_1)^2 + (x_2+y_2)^2 \amp = x_1^2+ 2x_1y_1 + y_1^2+x_2^2+ 2x_2y_2 + y_1^2 \\ \amp = 2 + 2x_1y_1 + 2 x_2 y_2 \end{align*}

which is only 1 in special circumstances and not in general, so this is not a vector space.

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The additive identity requires that the zero vector is an element of the vector space. However, since

\begin{equation*} \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{equation*}

is not in $U\text{,}$ then this property is also not satisfied.

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This property is also not satisfied for the same reason as #1. Geometrically if $\vec{x}$ is in $U$ then $r\vec{x}$ is on a circle of radius $r\text{,}$ so won’t be in $U$ in general.
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Note 3.1.7.

We have an alternative way of looking at the circle that will be a vector space. We can still define the vectors to be on the unit circle, however, this time addition will be defined as the vector whose angle is the sum of the angles, where angle is defined from the positive $x$-axis. Scalar multiplication will be the scalar multiplication of the angle. There are some other details that are needed to formalize this as a vector space.

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