Abstract Vector Spaces

Earlier in the text we discussed \(\RR ^n\) as a vector space and introduced the notion of a subspace of \(\RR ^n\). In this section we will consider sets other than \(\RR ^n\) that have two operations and satisfy the same properties as \(\RR ^n\). Such sets, together with the operations of addition and scalar multiplication, will also be called vector spaces.

Properties of Vector Spaces

Recall that \(\RR ^n\) is said to be a vector space because

  • \(\RR ^n\) is closed under vector addition
  • \(\RR ^n\) is closed under scalar multiplication

and satisfies the following properties:

1.
Commutative Property of Addition: \(\vec {u}+\vec {v}=\vec {v}+\vec {u}\)
2.
Associative Property of Addition: \((\vec {u}+\vec {v})+\vec {w}=\vec {u}+(\vec {v}+\vec {w})\)
3.
Existence of Additive Identity: \(\vec {u}+\vec {0}=\vec {u}\)
4.
Existence of Additive Inverse: \(\vec {u}+(-\vec {u})=\vec {0}\)
5.
Distributive Property over Vector Addition: \(k(\vec {u}+\vec {v})=k\vec {u}+k\vec {v}\)
6.
Distributive Property over Scalar Addition: \((k+p)\vec {u}=k\vec {u}+p\vec {u}\)
7.
Associative Property for Scalar Multiplication: \(k(p\vec {u})=(kp)\vec {u}\)
8.
Multiplication by \(1\): \(1\vec {u}=\vec {u}\)

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

Definition of a Vector Space

Examples 1 and 2 show us that there are many times in mathematics when we encounter a set with two operations (that we call addition and scalar multiplication) such that the set is closed under the two operations, and satisfies the same eight properties as \(\RR ^n\). We will refer to such sets as vector spaces.

When scalars \(k\) and \(p\) in the above definition are restricted to real numbers, as they are in this chapter, vector space \(V\) may be referred to as a vector space over the real numbers.

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 + \ldots + a_nx^n \end{equation*}

where \(a_{0}, a_{1}, a_{2}, \ldots , a_{n}\) are real numbers called the coefficients of the polynomial. If all the coefficients are zero, the polynomial is called the zero polynomial and is denoted simply as \(0\). If \(p(x) \neq 0\), the highest power of \(x\) with a nonzero coefficient is called the degree of \(p(x)\) denoted as \(\mbox {deg}(p(x))\). The coefficient itself is called the leading coefficient of \(p(x)\). Hence \(\mbox {deg}(3 + 5x) = 1\), \(\mbox {deg}(1 + x + x^{2}) = 2\), and \(\mbox {deg}(4) = 0\). (The degree of the zero polynomial is not defined.)

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

\begin{align*} p(x) &= a_0 + a_1x + a_2x^2 + \ldots \\ q(x) &= b_0 + b_1x + b_2x^2 + \ldots \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, \(a_{0} = b_{0}\), \(a_{1} = b_{1}\), \(a_{2} = b_{2}\), and so on. In particular, \(a_{0} + a_{1}x + a_{2}x^{2} + \ldots = 0\) means that \(a_{0} = 0\), \(a_{1} = 0\), \(a_{2} = 0\), \(\ldots \).

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

\begin{align*} p(x) + q(x) &= (a_0 + b_0) + (a_1 + b_1)x + (a_2 + b_2)x^2 + \ldots \\ kp(x) &= ka_0 + (ka_1)x + (ka_2)x^2 + \ldots \end{align*}

Set \(Y\) in Example 8 is not a vector space, but if we make a slight modification, we can make it into a vector space.

Subspaces

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.

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.

Observe that Properties 1, 2, 5, 6, 7 and 8 hold for all elements of \(V\). Thus, these properties will hold for all elements of \(U\). We say that these properties are inherited from \(V\).

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

\[0\vec {u}=(0+0)\vec {u}=0\vec {u}+0\vec {u}\]
Adding the additive inverse of \(0\vec {u}\) to both sides gives us
\[0\vec {u}+(-0\vec {u})=(0\vec {u}+0\vec {u})+(-0\vec {u})\]
By Properties 2 and 4
\[\vec {0}=0\vec {u}+(0\vec {u}+(-0\vec {u}))\]
By Properties 3 and 4
\[\vec {0}=0\vec {u}+\vec {0}=0\vec {u}\]

Because \(U\) is closed under scalar multiplication \(0\vec {u}=\vec {0}\) is in \(U\).

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

\[\vec {u}+(-1)\vec {u}=1\vec {u}+(-1)\vec {u}=(1+(-1))\vec {u}=0\vec {u}=\vec {0}\]
Thus \((-1)\vec {u}\) is the additive inverse of \(\vec {u}\). By closure, \((-1)\vec {u}\) is in \(U\).

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 \(a\). For example, if \(p(x) = 5 - 6x + 2x^{2}\), then the evaluation of \(p(x)\) at \(a = 2\) is \(p(2) = 5 - 12 + 8 = 1\). If \(p(a) = 0\), the number \(a\) is called a root of \(p(x)\).

Linear Combinations and Span

See Practice Problem .

Practice Problems

Given the set of all points in \(\mathbb {R}^2\), define the operations of addition and scalar multiplication as follows:

Addition: \((a, b)+(c, d)=(a+d, b+c)\)
Scalar Multiplication: \(k(a, b)=(ka, kb)\)

Is addition, as defined above, commutative?

Yes No
Given the set of all points in \(\mathbb {R}^2\), define the operations of addition and scalar multiplication as follows:

Addition: \((a, b)+(c, d)=(0, b+d)\)
Scalar Multiplication: \(k(a, b)=(ka, kb)\)

Is addition, as defined above, commutative?

Yes No

Is there an additive identity?

Yes No
Given the set of all points in \(\mathbb {R}^2\), define the operations of addition and scalar multiplication as follows:

Addition: \((a, b)+(c, d)=(a+c, b+d)\)
Scalar Multiplication: \(k(a, b)=(a, kb)\)

Is the Distributive Property over Vector Addition satisfied?

Yes No

Is the Distributive Property over Scalar Addition satisfied?

Yes No
Let \(\mathcal {F}\) be the set of all real-valued functions whose domain is all real numbers. Define addition and scalar multiplication as follows:
\[(f+g)(x)=f(x)+g(x)\quad (cf)(x)=cf(x)\]
Verify that \(\mathcal {F}\) is a vector space.
Recall that for a fixed matrix \(A\) in \(\mathbb {M}_{n,n}\), the set \(C_A\) of all \(n\times n\) matrices that commute with \(A\) under matrix multiplication is a subspace of \(\mathbb {M}_{n,n}\). (Example 16)

Describe all elements of \(C_I\), where \(I\) is a \(3\times 3\) identity matrix.

Is the subset of all invertible \(n\times n\) matrices a subspace of \(\mathbb {M}_{n,n}\)?
Yes No

Prove your claim.

Is the subset of all symmetric \(n\times n\) matrices a subspace of \(\mathbb {M}_{n,n}\)?
Yes No

Prove your claim.

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,
\[Z=\{B : BY=YB \mbox { for all } Y \in \mathbb {M}_{n,n}\}\]

Is \(Z\) a subspace of \(\mathbb {M}_{n,n}\)?

List several elements of \(\mbox {span}\left (\begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}\right )\). Suggest a spanning set for \(\mathbb {M}_{2,2}\).
Find \(\mbox {span}(1, x, x^2, x^3)\).
Let \(V\) be a vector space. Let \(S\) be any subset of \(V\). Prove that \(U=\mbox {span}(S)\) is a subspace of \(V\). (Theorem 24)

Text Source

The discussion on polynomials was adapted from Section 6.1 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 331-332

Example Source

Examples 18 and 22 were adapted from Examples 6.2.4 and 6.2.7 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 340-341

Exercise Source

Practice Problems -?? is adopted from Problems 9.1.1-9.1.4 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, p. 469.