Introduction to Vector Spaces

Section 4.1 Introduction to Vector Spaces

There are two operations on vectors in $\R^n\text{,}$ namely, vector addition and scalar multiplication. We learned that addition and scalar multiplication satisfy a number of nice properties (see Theorem 2.1.22). These properties give $\R^n$ an algebraic structure.

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We begin this section by introducing another property, called closure. Adding closure to the properties we studied earlier allows us to show that $\R^n$ satisfies all of the properties of a vector space, which we will formally define in this section.

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Subsection 4.1.1 Closure

Definition 4.1.1.

A set $V$ is said to be closed under addition if for each element $\mathbf{u} \in V$ and $\mathbf{v} \in V$ the sum $\mathbf{u}+\mathbf{v}$ is also in $V\text{.}$

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Definition 4.1.2.

A set $V$ is said to be closed under scalar multiplication if for each element $\mathbf{v} \in V$ and for each scalar $k \in \R$ the product $k\mathbf{v}$ is also in $V\text{.}$

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Here are a few examples and non-examples to understand these properties.

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Example 4.1.3.

Let $E$ be the set of positive even integers. Then $E$ is closed under addition, because the sum of two even integers is again an even integer.

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Example 4.1.4.

Let $D$ be the set of positive odd integers. Then $D$ is not closed under addition, for the sum of two odd integers need not be an odd integer (in fact, it will always be even).

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Example 4.1.5.

Let $X_1$ be defined as the set of vectors in $\mathbb{R}^2$ in the first quadrant (or on an axis on the boundary of the first quadrant). Written in symbols:

\begin{equation*} X_1=\left\{\begin{bmatrix} x_1\\x_2\end{bmatrix} \in \mathbb{R}^2 : x_1 \ge 0, \ x_2 \ge 0 \right\} \end{equation*}

Show that $X_1$ is closed under addition, but $X_1$ is not closed under scalar multiplication.

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Answer.

Suppose $\mathbf{v}_1=[a,b]$ and $\mathbf{v}_2=[c,d]$ are in $X_1\text{.}$ This means that $a, b, c, d\geq 0\text{.}$ But then we have $a+c, b+d\geq 0\text{.}$ Therefore

\begin{equation*} \mathbf{v}_1+\mathbf{v}_2=\begin{bmatrix}a+c\\b+d\end{bmatrix} \end{equation*}

is also in $X_1\text{.}$ We conclude that $X_1$ is closed under addition. $X_1$ is not closed under scalar multiplication because

\begin{equation*} (-1)\mathbf{v}_1=\begin{bmatrix}-a\\-b\end{bmatrix} \end{equation*}

is not in $X_1\text{.}$The figure below helps us see that the sum of any two vectors in $X_1$ also lies in $X_1\text{.}$

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$X1 Sum Example graphed$

The next picture pictures how negative multiples of a vector does not list in $X_1\text{.}$

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$X1 Scalar Example graphed$

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Example 4.1.6.

Let $X_1$ be as in Example 4.1.5. Let

\begin{equation*} X_3=\left\{\begin{bmatrix} x_1\\x_2\end{bmatrix} \in \mathbb{R}^2 : x_1 \le 0, \ x_2 \le 0 \right\} \end{equation*}

In other words, $X_3$ is the set of vectors in $\R^2$ in the third quadrant (or on an axis on the boundary of the third quadrant). Let $Y$ consist of $X_1$ and $X_3$ combined. In other words, $Y$ is the set of vectors in $\R^2$ that are either in the first quadrant, the third quadrant, or lie along one of the axes, as shown below.

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$Y pictured$

Then $Y$ is closed under scalar multiplication, but $Y$ is not closed under addition.

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Subsection 4.1.2 Properties of Vector Spaces

So, we have now seen that $\R^n$ is

closed under vector addition (why?),
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closed under scalar multiplication (why?),
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and satisfies the following properties:

Commutative Property of Addition: $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}.$
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Associative Property of Addition: $(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w}).$
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Existence of Additive Identity: $\mathbf{u}+\mathbf{0}=\mathbf{u}.$
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Existence of Additive Inverse: $\mathbf{u}+(-\mathbf{u})=\mathbf{0}.$
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Distributive Property over Vector Addition: $k(\mathbf{u}+\mathbf{v})=k\mathbf{u}+k\mathbf{v}.$
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Distributive Property over Scalar Addition: $(k+p)\mathbf{u}=k\mathbf{u}+p\mathbf{u}.$
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Associative Property for Scalar Multiplication: $k(p\mathbf{u})=(kp)\mathbf{u}.$
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Multiplication by $1\text{:}$ $1\mathbf{u}=\mathbf{u}.$
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Remark 4.1.7.

All scalars in this text are assumed to be real numbers. Considering other scalars, such as complex numbers, is interesting and leads to the definition of complex vector spaces. We will not discuss complex vector spaces in this text, but you may encounter them in a future course or your professional career.

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In the next two examples we will explore two sets other than $\R^n$ endowed with vector addition and scalar multiplication and satisfying the same properties.

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Example 4.1.8.

Let $\mathbb{M}_{m,n}$ be the set of all $m\times n$ matrices. Matrix addition and scalar multiplication were defined in Section 3.1. Observe that the sum of two $m\times n$ matrices is also an $m\times n$ matrix. Likewise, a scalar multiple of an $m\times n$ matrix is an $m\times n$ matrix. Thus

$\mathbb{M}_{m,n}$ is closed under matrix addition;
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$\mathbb{M}_{m,n}$ is closed under scalar multiplication.
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Next, Theorem 3.1.5 and Theorem 3.1.8 give us the following properties of matrix addition and scalar multiplication. Note that these properties are analogous to the eight vector properties above.

Commutative Property of Addition: $\quad A+B=B+A\text{.}$
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Associative Property of Addition: $\quad (A+B)+C=A+(B+C)\text{.}$
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Existence of Additive Identity: $\quad A+O=A \ $ where $O$ is the $m \times n$ zero matrix.
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Existence of Additive Inverse: $\quad A+(-A)=O\text{.}$
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Distributive Property over Matrix Addition: $\quad k(A+B)=kA+kB\text{.}$
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Distributive Property over Scalar Addition: $\quad (k+p)A=kA+pA\text{.}$
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Associative Property for Scalar Multiplication: $\quad k(pA)=(kp)A\text{.}$
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Multiplication by $1\text{:}$ $\quad 1A=A\text{.}$
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Because $\mathbb{M}_{m,n}$ with these two operations satisfies all of these properties, it is a vector space.

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Example 4.1.9.

Consider the set $\mathbb{L}$ of all linear functions. This set includes all polynomials of degree $1$ and degree $0\text{.}$ We will use addition and scalar multiplication of polynomials as the two operations, and show that $\mathbb{L}$ is closed under those operations and satisfies eight properties analogous to those of vectors of $\R^n\text{.}$

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Answer.

Elements of $\mathbb{L}$ are functions $f$ given by

\begin{equation*} f(x)=mx+b. \end{equation*}

(Note that either $m$ or $b$ or both can be equal to zero.)

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Given $f_1$ and $f_2$ in $\mathbb{L}\text{,}$ it is easy to verify that $f_1+f_2$ is also in $\mathbb{L}\text{.}$ This gives us closure under function addition. For any scalar $k\text{,}$ we have

\begin{equation*} kf(x)=k(mx+b)=(km)x+(kb). \end{equation*}

Therefore $kf$ is in $\mathbb{L}\text{,}$ and $\mathbb{L}$ is closed under scalar multiplication. We now proceed to formulate eight properties analogous to those of vectors of $\R^n\text{.}$

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Let $f_1\text{,}$ $f_2$ and $f_3$ be elements of $\mathbb{L}$ given by $f_1(x)=m_1 x + b_1\text{,}$ $f_2(x)=m_2 x + b_2\text{,}$ and $f_3(x)=m_3 x + b_3\text{.}$ Let $k$ and $p$ be scalars.

Commutative Property of Addition: $f_1+f_2=f_2+f_1.$
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This property holds because

\begin{align*} f_1(x) + f_2(x) \amp = (m_1 x + b_1) + (m_2 x + b_2) \\ \amp = (m_2 x + b_2) + (m_1 x + b_1) \\ \amp = f_2(x) + f_1(x). \end{align*}

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Associative property of Addition:

\begin{equation*} (f_1 + f_2) + f_3 = f_1 + (f_2 + f_3). \end{equation*}

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This property is easy to verify and is left to the reader.
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Existence of additive identity:

\begin{equation*} f_1 + f_0 = f_1 \end{equation*}

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The additive identity $f_0$ is given by $f_0(x)=0\text{.}$ Note that $f_0$ is a vector in the space $\mathbb{L}\text{.}$
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Existence of additive inverse:

\begin{equation*} f_1 + (-f_1) = f_0. \end{equation*}

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The additive inverse of $f_1$ is a function $-f_1$ given by $-f_1(x)=-mx+(-b)\text{.}$ Note that $-f_1$ is in $\mathbb{L}\text{.}$
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Distributive Property over Vector Addition:

\begin{equation*} k(f_1+f_2)=kf_1+kf_2. \end{equation*}

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This property holds because

\begin{align*} k(f_1(x) + f_2(x)) \amp = k((m_1 x + b_1) + (m_2 x + b_2)) \\ \amp = k(m_1 x + b_1) + k(m_2 x + b_2) \\ \amp = k f_1(x) + k f_2(x). \end{align*}

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Distributive property over scalar addition:

\begin{equation*} (k+p)f_1=kf_1+pf_1. \end{equation*}

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This property holds because

\begin{align*} (k+p)f_1(x)\amp = (k+p)(m_1 x + b_1) \\ \amp =k(m_1 x + b_1) + p(m_1 x + b_1) \\ \amp = k f_1(x) + p f_1(x). \end{align*}

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Associative property for scalar multiplication: $(k(pf_1))=(kp)f_1.$
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This property holds because

\begin{align*} k(p(f_1(x)))\amp =k(p(m_1 x + b_1)) \\ \amp =k(p m_1 x +p b_1) \\ \amp =kp m_1 x +kp b_1 \\ \amp = (kp) m_1 x + (kp) b_1 \\ \amp = (kp)(m_1 x + b_1) \\ \amp =(kp)f_1(x). \end{align*}

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Multiplication by $1$
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This follows from

\begin{equation*} 1 f_1=f_1. \end{equation*}

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Because $\mathbb{L}$ with these two operations satisfies all of these properties, it is a vector space.

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Subsection 4.1.3 Definition of a Vector Space

Looking at Example 4.1.8 and Example 4.1.9 shows us that a set with two operations (which we will call addition and scalar multiplication) such that the set is closed under the two operations, and satisfies the same eight properties as $\R^n\text{.}$ We will refer to such a set with its two operations as a vector space.

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Definition 4.1.10.

Let $V$ be a nonempty set. Suppose that elements of $V$ can be added together and can be multiplied by scalars. The set $V\text{,}$ together with these operations of addition and scalar multiplication, is called a vector space provided that

$V$ is closed under addition,
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$V$ is closed under scalar multiplication
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and the following properties hold for $\mathbf{u}\text{,}$ $\mathbf{v}$ and $\mathbf{w}$ in $V$ and scalars $k$ and $p\text{:}$

Commutative Property of Addition: $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}.$
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Associative Property of Addition:\quad $(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w}).$
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Existence of Additive Identity: $\mathbf{u}+\mathbf{0}=\mathbf{u}.$
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Existence of Additive Inverse: $\mathbf{u}+(-\mathbf{u})=\mathbf{0}.$
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Distributive Property over Vector Addition: $k(\mathbf{u}+\mathbf{v})=k\mathbf{u}+k\mathbf{v}.$
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Distributive Property over Scalar Addition: $(k+p)\mathbf{u}=k\mathbf{u}+p\mathbf{u}.$
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Associative Property for Scalar Multiplication: $k(p\mathbf{u})=(kp)\mathbf{u}.$
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Multiplication by $1\text{:}$ $1\mathbf{u}=\mathbf{u}.$
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We will refer to elements of $V$ as vectors.

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When scalars $k$ and $p$ in the above definition are required to be real numbers, as they usually are in this text, the vector space $V$ may be called a real vector space or vector space over the real numbers. In Chapter 6, we will sometimes consider complex numbers and that means that our typical vector space will be $\mathbb{C}^n$ and our scalars will be complex numbers; no change in the definition above is needed, other than allowing scalars to be complex numbers, for a vector space over the complex numbers.

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We have already encountered two real vectors spaces in Example 4.1.8 and Example 4.1.9, namely $\mathbb{M}_{m,n}$ and $\mathbb{L}\text{.}$

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Sets of polynomials provide an important source of examples, so we review some basic facts. A polynomial with real coefficients in $x$ is an expression

\begin{equation*} p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \end{equation*}

where $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ are real numbers called the coefficients of the polynomial.

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If all the coefficients are zero, the polynomial is called the zero polynomial and is denoted simply as $0\text{.}$

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If $p(x) \neq 0\text{,}$ the highest power of $x$ with a nonzero coefficient is called the degree of $p(x)$ and denoted as $\mbox{deg}(p(x))\text{.}$ The degree of the zero polynomial is not defined. The coefficient of this highest power of $x$ is called the leading coefficient of $p(x)\text{.}$ Hence $\mbox{deg}(3 + 5x) = 1$ and its leading coefficient is $5\text{,}$ $\mbox{deg}(1 + 3x - x^{2}) = 2$ and its leading coefficient is $-1\text{,}$ and the constant polynomial $p(x)=4$ had degree 0 and leading coefficient 4.

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To simplify our calculations, for the polynomial $p(x) = a_0 + a_1x + \cdots + a_nx^n\text{,}$ we define $a_{n+1}, a_{n+2}, \ldots$ to all be zero. This means $a_i$ will make sense for all non-negative values of $i\text{.}$ For example, $p(x)$ is the zero polynomial exactly when $a_i=0$ for all non-negative integers $i\text{.}$ Because $p(x)$ is a polynomial, there are only finitely many $i$ with $a_i \neq 0$ and, when $p(x)$ is not the zero polynomial, the largest such index $i$ is the degree of $p(x)\text{.}$

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Let $\mathbb{P}$ denote the set of all polynomials and suppose that

\begin{align*} p(x) \amp = a_0 + a_1x + a_2x^2 + \cdots + a_mx^m\\ q(x) \amp = b_0 + b_1x + b_2x^2 + \cdots + b_nx^n \end{align*}

are two polynomials in $\mathbb{P}$ (possibly of different degrees). Then $p(x)$ and $q(x)$ are called equal (written $p(x) = q(x)$) if and only if all the corresponding coefficients are equal--- that is, one has $a_{0} = b_{0}\text{,}$ $a_{1} = b_{1}\text{,}$ $a_{2} = b_{2}\text{,}$ and so on. In particular, $a_{0} + a_{1}x + \ldots + a_mx^m = 0$ means $a_{0} = 0\text{,}$ $a_{1} = 0\text{,}$ $a_{2} = 0\text{,}$ $\ldots\text{.}$

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The set $\mathbb{P}$ has vector addition and scalar multiplication operations defined as follows: if $p(x)$ and $q(x)$ are as before and $k$ is a real number,

\begin{align*} p(x) + q(x) \amp = (a_0 + b_0) + (a_1 + b_1)x + (a_2 + b_2)x^2 + \cdots + (a_j+b_j) x^j\\ kp(x) \amp = ka_0 + (ka_1)x + (ka_2)x^2 + \cdots + (k a_m) x^m \end{align*}

where $j$ is the larger of $m$ or $n\text{.}$ Notice that the degree of $p(x)+q(x)$ is at most the larger of the degrees of $p(x)$ and $q(x)$ and the degree of $k(p(x))$ is exactly the degree of $p(x)\text{,}$ provided $k \ne 0\text{.}$

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This is a ton of terminology, so we need some examples to understand it.

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Example 4.1.11.

$\mathbb{P}$ is a vector space.

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Answer.

It is easy to see that the sum of two polynomials is again a polynomial, and that a scalar multiple of a polynomial is a polynomial. Thus, $\mathbb{P}$ is closed under addition and scalar multiplication. The other eight vector space properties can be verified (do one yourself), and we conclude that $\mathbb{P}$ is a vector space.

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Example 4.1.12.

Let $Y$ be the set of all degree two polynomials in $x\text{.}$ In other words,

\begin{equation*} Y=\left \lbrace ax^2+bx+c : a \ne 0 \right \rbrace. \end{equation*}

We claim that $Y$ is not a vector space.

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Answer.

Observe that $Y$ is not closed under addition. To see this, let $y_1 = 2x^2+3x+4$ and let $y_2=-2x^2\text{.}$ Then $y_1$ and $y_2$ are both elements of $Y\text{.}$ However, $y_1+y_2 = 3x+4$ is not an element of $Y\text{,}$ as it is only a degree one polynomial. We require the coefficient $a$ of $x^2$ to be nonzero for a polynomial to be in $Y\text{,}$ and this is not the case for $y_1+y_2\text{.}$ As an exercise, check the remaining vector space properties one-by-one to see which properties hold and which do not.

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The set $Y$ in Example 4.1.12 is not a vector space, but if we make a slight modification, we can make it into a vector space.

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Example 4.1.13.

Let $\mathbb{P}^2$ be the set of polynomials of degree two or less. In other words,

\begin{equation*} \mathbb{P}^2=\left \lbrace ax^2+bx+c : a,b,c \in \mathbb{R} \right \rbrace. \end{equation*}

Note that $\mathbb{P}^2$ contains the zero polynomial (let $a=b=c=0$). Unlike set $Y$ in Example 4.1.12, $\mathbb{P}^2$ is closed under polynomial addition and scalar multiplication. It is easy to verify that all vector space properties hold, so $\mathbb{P}^2$ is a vector space.

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Example 4.1.14.

Let $n$ be a natural number. Define $\mathbb{P}^n$ to be the set of polynomials of degree $n$ or less than $n\text{,}$ then by reasoning similar to Example 4.1.13, $\mathbb{P}^n$ is a vector space.

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Subsection 4.1.4 Subspaces

Definition 4.1.15.

A nonempty subset $U$ of a vector space $V$ is called a subspace of $V\text{,}$ provided that $U$ is itself a vector space when given the same addition and scalar multiplication as $V\text{.}$

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An example to showcase this is in order.

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Example 4.1.16.

In Example 4.1.13 we demonstrated that $\mathbb{P}^2$ is a vector space. From Example 4.1.11 we know that $\mathbb{P}$ is a vector space. But $\mathbb{P}^2$ is a subset of $\mathbb{P}\text{,}$ and uses the same operations of polynomial addition and scalar multiplication. Therefore $\mathbb{P}^2$ is a subspace of $\mathbb{P}\text{.}$

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Checking all ten properties to verify that a subset of a vector space is a subspace can be cumbersome. Fortunately we have the following theorem.

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Theorem 4.1.17. Subspace Test.

Let $U$ be a nonempty subset of a vector space $V\text{.}$ If $U$ is closed under the operations of addition and scalar multiplication of $V\text{,}$ then $U$ is a subspace of $V\text{.}$

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Proof.

To prove that closure is a sufficient condition for $U$ to be a subspace, we will need to show that closure under addition and scalar multiplication of $V$ guarantees that the remaining eight properties are satisfied automatically.

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Observe that Item 1, Item 2, Item 5, Item 6, Item 7 and Item 8 hold for all elements of $V\text{.}$ Thus, these properties will hold for all elements of $U\text{.}$ We say that these properties are inherited from $V\text{.}$

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To prove Item 3 we need to show that $\mathbf{0}\text{,}$ which we know to be an element of $V\text{,}$ is contained in $U\text{.}$ Let $\mathbf{u}$ be an element of $U$ (recall that $U$ is nonempty). We will show that $0\mathbf{u}=\mathbf{0}$ in $V\text{.}$ Then, by closure under scalar multiplication, we will be able to conclude that $0\mathbf{u}=\mathbf{0}$ must be in $U\text{.}$

\begin{equation*} 0\mathbf{u}=(0+0)\mathbf{u}=0\mathbf{u}+0\mathbf{u}. \end{equation*}

Adding the additive inverse of $0\mathbf{u}$ to both sides gives us

\begin{equation*} 0\mathbf{u}+(-0\mathbf{u})=(0\mathbf{u}+0\mathbf{u})+(-0\mathbf{u}). \end{equation*}

Thanks to Item 2 and Item 4.

\begin{equation*} \mathbf{0}=0\mathbf{u}+(0\mathbf{u}+(-0\mathbf{u})). \end{equation*}

By Item 3 and Item 4

\begin{equation*} \mathbf{0}=0\mathbf{u}+\mathbf{0}=0\mathbf{u}. \end{equation*}

Because $U$ is closed under scalar multiplication $0\mathbf{u}=\mathbf{0}$ is in $U\text{.}$ We know that every element of $U\text{,}$ being an element of $V\text{,}$ has an additive inverse in $V\text{.}$ We need to show that the additive inverse of every element of $U$ is contained in $U\text{.}$ Let $\mathbf{u}$ be any element of $U\text{.}$ We will show that $(-1)\mathbf{u}$ is the additive inverse of $\mathbf{u}\text{.}$ Then by closure, $(-1)\mathbf{u}$ will have to be contained in $U\text{.}$ To show that $(-1)\mathbf{u}$ is the additive inverse of $\mathbf{u}\text{,}$ we must show that $\mathbf{u}+(-1)\mathbf{u}=\mathbf{0}\text{.}$ We compute:

\begin{equation*} \mathbf{u}+(-1)\mathbf{u}=1\mathbf{u}+(-1)\mathbf{u}=(1+(-1))\mathbf{u}=0\mathbf{u}=\mathbf{0}. \end{equation*}

Thus $(-1)\mathbf{u}$ is the additive inverse of $\mathbf{u}\text{.}$ By closure, $(-1)\mathbf{u}$ is in $U\text{.}$

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Let us apply the theorem in a couple of examples.

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Example 4.1.18.

Let $V$ be the set of vectors in $\R^3$ on the $y-$axis. Then $V$ is a subspace of $\R^3\text{.}$

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Answer.

To verify this, note that any vector in $V$ is of the form

\begin{equation*} \begin{bmatrix}0\\y\\0 \end{bmatrix}\text{.} \end{equation*}

If we multiply such a vector by a scalar $c\text{,}$ we get a vector of the form

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

which is clearly still in $V\text{.}$ This proves $V$ is closed under scalar multiplication. Next, note that

\begin{equation*} \begin{bmatrix}0\\y_1\\0 \end{bmatrix} + \begin{bmatrix}0\\y_2\\0 \end{bmatrix}= \begin{bmatrix}0\\y_1 + y_2\\0\end{bmatrix}, \end{equation*}

so $V$ is closed under addition.

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Example 4.1.19.

Let $A$ be a fixed matrix in $\mathbb{M}_{n,n}\text{.}$ Show that the set $C_A$ of all $n\times n$ matrices that commute with $A$ under matrix multiplication is a subspace of $\mathbb{M}_{n,n}\text{.}$

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Answer.

The set $C_A$ consists of all $n\times n$ matrices $X$ such that $AX=XA\text{.}$ First, observe that $C_A$ is not empty because $I_n$ is an element. Now we need to show that $C_A$ is closed under matrix addition and scalar multiplication. Suppose that $X_1$ and $X_{2}$ lie in $C_A\text{.}$ Then $AX_1 = X_1A$ and $AX_{2} = X_{2}A\text{.}$ Then

\begin{equation*} A(X_1 + X_2) = AX_1 + AX_2 = X_1A + X_2A + (X_1 + X_2)A. \end{equation*}

Therefore $(X_1+X_2)$ commutes with $A\text{.}$ Thus $(X_1+X_2)$ is in $C_A\text{.}$ We conclude that $C_A$ is closed under matrix addition. Now suppose $X$ is in $C_A\text{.}$ Let $k$ be a scalar, then

\begin{equation*} A(kX)= k(AX) = k(XA) = (kX)A. \end{equation*}

Therefore $(kX)$ commutes with $A\text{.}$ We conclude that $(kX)$ is in $C_A\text{,}$ and $C_A$ is closed under scalar multiplication. Hence $C_A$ is a subspace of $\mathbb{M}_{n,n}\text{.}$

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Remark 4.1.20.

Suppose $p(x)$ is a polynomial and $a$ is a number. Then the number $p(a)$ obtained by replacing $x$ by $a$ in the expression for $p(x)$ is called the evaluation of $p(x)$ at $a\text{.}$ For example, if $p(x) = 5 - 6x + 2x^{2}\text{,}$ then the evaluation of $p(x)$ at $a = 2$ is

\begin{equation*} p(2) = 5 - 12 + 8 = 1. \end{equation*}

If $p(a) = 0\text{,}$ the number $a$ is called a root of $p(x)\text{.}$

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To get used to the new terminology, let us look at an example in the context of polynomials.

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Example 4.1.21.

Consider the set $U$ of all polynomials in $\mathbb{P}$ that have $3$ as a root:

\begin{equation*} U = \lbrace p(x) \in \mathbb{P} : p(3) = 0 \rbrace. \end{equation*}

Show that $U$ is a subspace of $\mathbb{P}\text{.}$

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Answer.

Observe that $U$ is not empty because $r(x)=x-3$ is an element of $U\text{.}$ Suppose $p(x)$ and $q(x)$ lie in $U\text{.}$ Then $p(3) = 0$ and $q(3) = 0\text{.}$ We have

\begin{equation*} (p + q)(x) = p(x) + q(x) \end{equation*}

for all $x\text{,}$ so

\begin{equation*} (p + q)(3) = p(3) + q(3) = 0 + 0 = 0, \end{equation*}

and $U$ is closed under addition. The verification that $U$ is closed under scalar multiplication is similar.

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The following important results provides us with a quick way to determine that some subsets are not subspaces.

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Theorem 4.1.22.

If $W$ is a subspace of a vector space $V\text{,}$ then the zero vector $\mathbf{0}$ is in $W\text{.}$

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Proof.

Since a subspace is nonempty, there is some vector $\mathbf{w}$ in $W\text{.}$ Then $0 \mathbf{w} = \mathbf{0}$ is in $W$ because $W$ is closed under scalar multiplication.

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Consider lines in $\R^n\text{.}$ Applying Theorem 4.1.22 shows that the only lines in $\R^n$ that are subspaces are those that pass through the origin. The same holds true for planes and hyperplanes. For example, the plane $z=3$ in $\R^3$ is not a subspace of $\R^3\text{.}$ It turns out that any plane containing the origin is a subspace.

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Be warned that containing the zero vector is a necessary condition but not a sufficient condition to be a subspace. That is, if a subset does not contain the zero vector, then it is not a subspace. But if a subset does contain the zero vector, it may or may not be a subspace. For example, the subset of $\R^2$ given by $\{ (x,y) : y=x^2 \}\text{,}$ the graph of the parabola $y=x^2\text{,}$ contains the zero vector of $\R^2\text{,}$ but is not a subspace since it is not closed under addition. (Don’t belive me? Just check!)

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Theorem 4.1.23.

If $W$ is a subspace of a vector space $V\text{,}$ then for any vector $\mathbf{w} \in W\text{,}$ the additive inverse, $-\mathbf{w}\text{,}$ is also in $W\text{.}$

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The proof is similar to what was done for the previous theorem and is left as an exercise.

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Subsection 4.1.5 Linear Combinations and Span

Just as we have seen in Section 2.2, linear combination and span are fundamental to understanding vector spaces. These concepts will be used extensively in the next section and in the rest of the text.

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Definition 4.1.24.

Let $V$ be a vector space and let $\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_n$ be vectors in $V\text{.}$ A vector $\mathbf{v}$ is said to be a linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2,\ldots, \mathbf{v}_n$ if

\begin{equation*} \mathbf{v}=a_1\mathbf{v}_1+ a_2\mathbf{v}_2+\cdots + a_n\mathbf{v}_n, \end{equation*}

for some scalars $a_1, a_2, \ldots ,a_n\text{.}$

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Definition 4.1.25.

Let $V$ be a vector space and let $\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p$ be vectors in $V\text{.}$ The set of all linear combinations of

\begin{equation*} \mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p \end{equation*}

is called the span of $\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p$ and we write it as

\begin{equation*} \mbox{span}(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p) \end{equation*}

Given some subset $S$ we say that vectors $\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p$ span $S$ if every vector in $S$ is a linear combination of the $v_1,\ldots,v_p\text{.}$ That is,

\begin{equation*} S = \mbox{span}(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p) \end{equation*}

Any vector in $S$ is said to be in the span of $\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\text{.}$ The set

\begin{equation*} \{\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\} \end{equation*}

is called a spanning set for $S\text{.}$

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We revisit the situation for some specific polynomials.

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Example 4.1.26.

Consider $p_{1} = 1 + x + 4x^{2}$ and $p_{2} = 1 + 5x + x^{2}$ in $\mathbb{P}^{2}\text{.}$ Determine whether $p_{1}$ and $p_{2}$ lie in

\begin{equation*} \mbox{span}\{1 + 2x - x^{2}, 3 + 5x + 2x^{2}\}. \end{equation*}

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Answer.

For $p_{1}\text{,}$ we want to determine if $a$ and $b$ exist such that

\begin{equation*} p_1 = a(1 + 2x - x^2) + b(3 + 5x + 2x^2). \end{equation*}

Expanding the right hand side gives us:

\begin{equation*} a+2ax-ax^2+3b+5bx+2bx^2. \end{equation*}

Combining like terms, we get:

\begin{equation*} (a+3b)+(2a+5b)x+(-a+2b)x^2. \end{equation*}

Setting this equal to $p_{1} = 1 + x + 4x^{2}$ and equating coefficients of powers of $x$ gives us a system of equations

\begin{equation*} 1 = a + 3b,\quad 1 = 2a + 5b, \quad \mbox{ and } \quad 4 = -a + 2b. \end{equation*}

This system has the solution $a = -2$ and $b = 1\text{,}$ so $p_{1}$ is indeed in $\mbox{span}\{1 + 2x - x^{2}, 3 + 5x + 2x^{2}\}\text{.}$ Turning to $p_{2} = 1 + 5x + x^{2}\text{,}$ we are looking for $a$ and $b$ such that

\begin{equation*} p_{2} = a(1 + 2x - x^{2}) + b(3 + 5x + 2x^{2}). \end{equation*}

Again equating coefficients of powers of $x$ gives equations $1 = a + 3b\text{,}$ $5 = 2a + 5b\text{,}$ and $1 = -a + 2b\text{.}$ But in this case there is no solution, so $p_{2}$ is not in $\mbox{span}\{1 + 2x - x^{2}, 3 + 5x + 2x^{2}\}\text{.}$

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Theorem 4.1.27.

Let $V$ be a vector space. Let $S$ be any subset of $V\text{.}$ Then $U=\mbox{span}(S)$ is a subspace of $V\text{.}$

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Proof.

We leave this an exercise for the reader, namely Exercise 4.1.6.22.

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This theorem is used often, as it gives a quick way to show a subset of a vector space is a subspace. If we can show the subset is the span of some set of vectors, then we know the subset is a subspace without further calculation. Going back to Example Example 4.1.13, a quick way to show $\mathbb{P}^2$ is a subspace of $\mathbb{P}$ is by observing that $\mathbb{P}^2=\mbox{span}(1,x,x^2)\text{.}$

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Example 4.1.28.

Verify that every line, plane, and hyperplane through the origin in $\R^n$ is a subspace of $\R^n\text{.}$

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Answer.

This is a direct consequence of Theorem 4.1.27. For example, consider the line through the origin in $\R^3$ spanned by the vector $\mathbf{v} = [1,2,3]\text{.}$ The span of this vector is the set of all scalar multiples of $\mathbf{v}\text{,}$ which is a subspace of $\R^3\text{.}$ Similarly, any plane or hyperplane through the origin can be expressed as the span of two or more vectors, respectively, and thus is also a subspace.

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Exercises 4.1.6 Exercises

Exercise Group.

Let $Y^+$ be the set of all vectors $[x,y]$ in $\mathbb{R}^2$ whose $y$ components are non-negative.

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1.

Is $Y^+$ closed under vector addition?

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Answer.

Yes.

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2.

Is $Y^+$ closed under scalar multiplication?

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Answer.

No.

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Exercise Group.

Let $X$ be the set of all vectors in $\R^3$ that lie on either the $x$-axis, the $y$-axis, or the $z$-axis.

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3.

Is $X$ closed under vector addition?

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Answer.

No.

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4.

Is $X$ closed under scalar multiplication?

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Answer.

Yes.

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Exercise Group.

For each figure below, determine whether the set $V$ of vectors shown in the figure is closed under vector addition and scalar multiplication. Justify your responses.

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5.

$V$ consists of all vectors in $\mathbb{R}^3$ in a slanted half-plane which has the $x$-axis as a boundary.

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$The graph of $V\text{,}$ a (half) plane in $\R^3$ bounded by a line.$

Is $V$ closed under scalar multiplication? What about addition?

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Answer.

It is closed under addition but not under scalar multiplication.

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6.

$V$ consists of all vectors along the line, as shown.

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$The graph of the line $y=2x\text{.}$$

Is $V$ closed under scalar multiplication? What about addition?

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Answer.

It is closed under both addition and scalar multiplication.

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Exercise Group.

Is the set of all points in $\mathbb{R}^2$ a vector space under the given definitions of addition and scalar multiplication? In each case be specific about which vector space properties hold and which properties fail.

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7.

Addition:

\begin{equation*} (a, b)+(c, d)=(a+d, b+c) \end{equation*}

and scalar Multiplication:

\begin{equation*} k(a, b)=(ka, kb). \end{equation*}

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8.

Addition:

\begin{equation*} (a, b)+(c, d)=(0, b+d) \end{equation*}

and scalar Multiplication:

\begin{equation*} k(a, b)=(ka, kb). \end{equation*}

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9.

Addition:

\begin{equation*} (a, b)+(c, d)=(a+c, b+d) \end{equation*}

and scalar Multiplication:

\begin{equation*} k(a, b)=(a, kb). \end{equation*}

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10.

Addition:

\begin{equation*} (a, b)+(c, d)=(a-c, b-d) \end{equation*}

and scalar Multiplication:

\begin{equation*} k(a, b)=(ka, kb). \end{equation*}

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11.

Let $\mathcal{F}$ be the set of all real-valued functions whose domain is all real numbers. Define addition and scalar multiplication as follows:

\begin{equation*} (f+g)(x)=f(x)+g(x)\quad (cf)(x)=cf(x). \end{equation*}

Verify that $\mathcal{F}$ is a vector space.

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12.

A differential equation is an equation that contains derivatives. Consider the differential equation:

\begin{equation} f''+f=0.\tag{4.1.1} \end{equation}

A solution to such an equation is a function.

Verify that $f(x)=\sin x$ is a solution to (4.1.1).
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Is $f(x)=2\sin x$ a solution?
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Is $f(x)=\cos x$ a solution?
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Is $f(x)=\sin x+\cos x$ a solution?
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Let $S$ be the set of all solutions to (4.1.1). Prove that $S$ is a vector space.
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13.

In this problem we will check that the set $\mathbb{C}$ of all complex numbers is in fact a real vector space. Let $z_1 = a_1 + b_1 i$ be a complex number. Similarly, let $z_2 = a_2 + b_2 i\text{,}$ $z_3 = a_3 + b_3 i$ be complex numbers, and let $k$ and $p$ be real number scalars. Check that complex numbers are closed under addition and multiplication, and that they satisfy each of the vector space properties.

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14.

Prove Theorem 4.1.23.

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15.

Let $A$ be an $m \times n$ matrix. Let $V$ be the subset of $\R^n$ consisting of all vectors $\mathbf{x}$ such that $A \mathbf{x} = \mathbf{0}\text{.}$ Prove that $V$ is a subspace of $\R^n\text{.}$ (Note that this subspace is called the null space of the matrix $A$ and we will denote it $\mbox{null}(A)\text{.}$)

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16.

Refer to Example 4.1.19 and describe all elements of $C_I\text{,}$ where $I$ is a $3\times 3$ identity matrix.

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17.

Is the subset of all invertible $n\times n$ matrices a subspace of $\mathbb{M}_{n,n}\text{?}$ Prove your claim.

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18.

Is the subset of all symmetric $n\times n$ matrices a subspace of $\mathbb{M}_{n,n}\text{?}$ (eee Definition 3.1.18.) Prove your claim.

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19.

Let $Z$ be a subset of $\mathbb{M}_{n,n}$ that consists of $n\times n$ matrices that commute with every matrix in $\mathbb{M}_{n,n}$ under matrix multiplication. In other words,

\begin{equation*} Z= \lbrace B : BY=YB \mbox{ for all } Y \in \mathbb{M}_{n,n} \rbrace. \end{equation*}

Is $Z$ a subspace of $\mathbb{M}_{n,n}\text{?}$

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