Discover Linear Algebra

The vector space \(\R^n\).

One set of prototypical examples of vector spaces are the collections of vectors we have been studying in Chapters 12–15: \(\R^2\text{,}\) \(\R^3\text{,}\) and the higher-dimensional spaces \(\R^n\text{,}\) \(n\ge 4\text{.}\) In these spaces,

• adding vectors or scalar multiplying a vector results in a vector in the same space, satisfying Axiom A 1 and Axiom S 1;
• the zero vector is \(\zerovec = (0,0,\dotsc,0)\) as usual;
• the negative of a vector is the parallel vector of the same length in the opposite direction; and
• we know that the rest of the axioms hold true from our knowledge of vector algebra in these spaces (Proposition 12.5.1).

In Discovery 16.3.e, we discovered that even the collection of real numbers itself can be considered as a vector space. We might think of this space as \(\R^1\text{,}\) and visualize its vectors as directed line segments lying along the real number line.

The vector space \(\matrixring_{m \times n}(\R)\).

Another set of prototypical examples of vectors spaces are the collections of matrices of given dimensions, \(\matrixring_{m \times n}(\R)\text{.}\) But these matrix spaces represent our first expansion of the word vector to include other kinds of objects — since all ten axioms hold true here, we can justifiably refer to any matrix, of any size, as a vector. In these spaces,

• adding matrices or scalar multiplying a matrix does not change its dimensions, so these operations always result in a vector in the same space, satisfying Axiom A 1 and Axiom S 1;
• the zero vector is the zero matrix of the appropriate size;
• the negative of a vector is the matrix of the same dimensions where all the entries are the negatives of those of the original matrix; and
• we know that the rest of the axioms hold true from our knowledge of matrix algebra (Proposition 4.5.1).

Spaces of polynomials.

In Discovery 16.3, we also explored some new examples of vector spaces consisting of polynomials as vectors. First, we considered the collection \(\poly(\R)\) of polynomials with real coefficients of arbitrary degree in Discovery 16.3.c. Here are some observations on the vector space axioms for this space.

• We add polynomials algebraically, by adding like terms. For example,
  \begin{equation*} (5x^3 + 3x^2 + 2x - 1) + (6x^{101} - 3x^3 + x + 1) = 6x^{101} + 2x^3 + 3x^2 + 3x. \end{equation*}
  Clearly, the result of adding polynomials is another polynomial, satisfying Axiom A 1.
• We scalar multiply a polynomial by distributing the scalar across the addition of the polynomials terms. For example,
  \begin{equation*} -2(6x^{101} - 3x^3 + x + 1) = 12x^{101} + 6x^3 - 2x - 2. \end{equation*}
  The result of multiplying a polynomial by a scalar is another polynomial, satisfying Axiom S 1.
• The zero vector is the constant (i.e. degree zero) polynomial \(p(x) = 0\text{.}\)
• The negative of a vector is the polynomial of the same degree where all the coefficients are the negatives of those of the the original polynomial.
• The rest of the axioms are familiar rules of algebra involving polynomial expressions.

Next, we considered the collection \(\poly_2(\R)\) of polynomials with real coefficients of maximum degree \(2\) (Discovery 16.3.d). Everything here works the same as in \(\poly(\R)\text{,}\) except that we need to reconsider Axiom A 1 and Axiom S 1. We define vector addition and scalar multiplication for polynomials as before, but we need to make sure that the result of each of these operations is always equal to another in the collection of objects. But neither of the operations can increase the degree of a polynomial, so their results will always again be a polynomial of degree \(2\) or less.

The space of functions.

This instance of a vector space is a generalization of the space of polynomials. We let \(F(D)\) represent the collection of all functions, not just polynomials, defined on a domain \(D\) of real numbers. Our first task is define how the two operations will work.

To create a new “sum” function out of two old functions, or to create a new “scaled” version of an old function, we first need understand how to create new functions. To define a new function, we must describe the input-output process. If we have two functions, \(f\) and \(g\text{,}\) then these functions have an already defined input-output process, but addition must somehow take these two processes and create a new one that is the sum of the old. The natural thing to do would be to add outputs of the two old process. That is, we define the sum function \(f+g\) by the input-output rule

For example, if function \(f\) produces output \(5\) at input \(3\) (i.e. \(f(3)=5\)), and function \(g\) produces output \(2\) at input \(3\) (i.e. \(g(3)=2\)), then the sum function \(f+g\) will produce output \(7\) at input \(3\text{:}\)

Similarly, to scale a function we should scale its outputs. That is,

For example, if function \(f\) produces output \(5\) at input \(3\) (i.e. \(f(3)=5\)), then the scaled function \(\sqrt{2}f\) will produce output \(5\sqrt{2}\) at input \(3\text{:}\)

Both of these processes result in a new function with the same domain as the old, so Axiom A 1 and Axiom S 1 are satisfied.

In this space, the zero vector is the zero function, whose outputs are always zero: \(\zerovec(x) = 0\) for all \(x\text{.}\) And the negative of a function is obtained by negating all of its outputs: \((-f)(x) = -f(x)\text{.}\)

See Subsection 16.5.2 for examples of carrying out the verification of some of the vector axioms in this space.

The trivial vector space.

In Discovery 16.5, we explored the possibility of a vector space with just one vector in it. But Axiom A 4 requires that every vector space have a zero vector, so that single vector inside must be it. And when we add this vector to itself or try to scale this vector, the condition that “the result is always equal to another in the collection” in both Axiom A 1 and Axiom S 1 requires that the result is actually always equal to that one vector, because there are no other vectors in the collection to choose from.

This simple vector space consisting of just a zero vector is called the zero vector space or the trivial vector space.

A weird instance of a vector space.

To emphasize the fact that the words vector, addition, and the phrase scalar multiplication can potentially mean anything, let's consider a weird example.

We'll take our collection of objects to be the collection of positive numbers. (But our scalars can still be any number, whether positive, negative, or zero.) To help distinguish between a vector and a scalar, we'll put brackets around a number if it is to mean a vector (as if we were considering \(\R^1\)). And to make sure we don't get mixed up with ordinary addition and multiplication of numbers, we'll use the symbols \(\oplus\) and \(\odot\) to mean vector addition and scalar multiplication, respectively.

To define vector addition in this space, we'll actually use ordinary multiplication of numbers. That is, for numbers \(a,b>0\text{,}\) we will add the vectors \((a),(b)\) according to the rule

This definition satisfies Axiom A 1 because multiplying two positive numbers results in another positive number.

To define scalar multiplication in this space, we'll use exponentiation. That is, for number \(a>0\) and scalar \(k\) (also a number), we will scale the vector \((a)\) by the scale factor \(k\) according to the rule

with the usual conventions that \(a^0 = 1\) and \(\inv{a} = 1/a\text{.}\) This definition satisfies Axiom S 1 because a power of a positive number results in another positive number.

What is the zero vector in this space? For Axiom A 4 to hold true, we need a vector \((z)\) so that

for all other vectors \((a)\text{.}\) But we can't use \(z=0\text{,}\) because the vectors in our space must all be positive numbers. Inserting the definition of \(\oplus\) as ordinary multiplication, we need a positive number \(z\) so that

for all \(a>0\text{.}\) But we only get \(az=a\) for all \(a>0\) when \(z=1\text{.}\) So in this weird space, the zero vector is the number one.

What is the negative of a vector in this space? Given positive \(a>0\text{,}\) the negative of vector \((a)\) can't be \((-a)\text{,}\) because all our vectors have to be positive numbers. To repeat, in this case, \(-(a)\) is not equal to \((-a)\). We know that every vector in this space is represented by a single positive number. That is, the negative of \((a)\) must be equal to \((b)\) for some positive number \(b\text{.}\) To satisfy Axiom A 5, this negative vector must satisfy

So we need \(b>0\) so that

Inserting the definition of \(\oplus\) as ordinary multiplication, and inserting \(\zerovec = (1)\) from above, we see we need

That is, we need \(ab = 1\text{,}\) so that \(b=1/a\) (which is positive since \(a\) is positive). So we have

In this weird space, the negative (i.e. additive inverse) of a vector corresponds to the reciprocal (i.e. multiplicative inverse) of the positive number representing that vector.

In Subsection 16.5.1, we will verify some of the other vector space axioms for algebra in this space.