From A First Course in Linear Algebra

Version 2.00

© 2004.

Licensed under the GNU Free Documentation License.

http://linear.ups.edu/

In this section we present a formal definition of a vector space, which will lead to
an extra increment of abstraction. Once defined, we study its most basic
properties.

Here is one of the two most important definitions in the entire course.

Definition VS

Vector Space

Suppose that $V$
is a set upon which we have defined two operations: (1) vector addition, which combines
two elements of $V$
and is denoted by “+”, and (2) scalar multiplication,
which combines a complex number with an element of
$V$ and is denoted by
juxtaposition. Then $V$,
along with the two operations, is a vector space if the following ten properties
hold.

- AC Additive Closure

If $u,\phantom{\rule{0em}{0ex}}v\in V$, then $u+v\in V$. - SC Scalar Closure

If $\alpha \in {\u2102}^{}$ and $u\in V$, then $\alpha u\in V$. - C Commutativity

If $u,\phantom{\rule{0em}{0ex}}v\in V$, then $u+v=v+u$. - AA Additive Associativity

If $u,\phantom{\rule{0em}{0ex}}v,\phantom{\rule{0em}{0ex}}w\in V$, then $u+\left(v+w\right)=\left(u+v\right)+w$. - Z Zero Vector

There is a vector, $0$, called the zero vector, such that $u+0=u$ for all $u\in V$. - AI Additive Inverses

If $u\in V$, then there exists a vector $-u\in V$ so that $u+\left(-u\right)=0$. - SMA Scalar Multiplication Associativity

If $\alpha ,\phantom{\rule{0em}{0ex}}\beta \in {\u2102}^{}$ and $u\in V$, then $\alpha \left(\beta u\right)=\left(\alpha \beta \right)u$. - DVA Distributivity across Vector Addition

If $\alpha \in {\u2102}^{}$ and $u,\phantom{\rule{0em}{0ex}}v\in V$, then $\alpha \left(u+v\right)=\alpha u+\alpha v$. - DSA Distributivity across Scalar Addition

If $\alpha ,\phantom{\rule{0em}{0ex}}\beta \in {\u2102}^{}$ and $u\in V$, then $\left(\alpha +\beta \right)u=\alpha u+\beta u$. - O One

If $u\in V$, then $1u=u$.

The objects in $V$ are called vectors, no matter what else they might really be, simply by virtue of being elements of a vector space. $\u25b3$

Now, there are several important observations to make. Many of these will be easier to understand on a second or third reading, and especially after carefully studying the examples in Subsection VS.EVS.

An axiom is often a “self-evident” truth. Something so fundamental that we all agree it is true and accept it without proof. Typically, it would be the logical underpinning that we would begin to build theorems upon. Some might refer to the ten properties of Definition VS as axioms, implying that a vector space is a very natural object and the ten properties are the essence of a vector space. We will instead emphasize that we will begin with a definition of a vector space. After studying the remainder of this chapter, you might return here and remind yourself how all our forthcoming theorems and definitions rest on this foundation.

As we will see shortly, the objects in $V$ can be anything, even though we will call them vectors. We have been working with vectors frequently, but we should stress here that these have so far just been column vectors — scalars arranged in a columnar list of fixed length. In a similar vein, you have used the symbol “+” for many years to represent the addition of numbers (scalars). We have extended its use to the addition of column vectors and to the addition of matrices, and now we are going to recycle it even further and let it denote vector addition in any possible vector space. So when describing a new vector space, we will have to define exactly what “+” is. Similar comments apply to scalar multiplication. Conversely, we can define our operations any way we like, so long as the ten properties are fulfilled (see Example CVS).

A vector space is composed of three objects, a set and two operations. However, we usually use the same symbol for both the set and the vector space itself. Do not let this convenience fool you into thinking the operations are secondary!

This discussion has either convinced you that we are really embarking on a new level of abstraction, or they have seemed cryptic, mysterious or nonsensical. You might want to return to this section in a few days and give it another read then. In any case, let’s look at some concrete examples now.

Our aim in this subsection is to give you a storehouse of examples to work with, to become comfortable with the ten vector space properties and to convince you that the multitude of examples justifies (at least initially) making such a broad definition as Definition VS. Some of our claims will be justified by reference to previous theorems, we will prove some facts from scratch, and we will do one non-trivial example completely. In other places, our usual thoroughness will be neglected, so grab paper and pencil and play along.

Example VSCV

The vector space ${\u2102}^{m}$

Set: ${\u2102}^{m}$, all column
vectors of size $m$,
Definition VSCV.

Equality: Entry-wise, Definition CVE.

Vector Addition: The “usual” addition, given in Definition CVA.

Scalar Multiplication: The “usual” scalar multiplication, given in
Definition CVSM.

Does this set with these operations fulfill the ten properties? Yes. And by design all we need to do is quote Theorem VSPCV. That was easy. $\u22a0$

Example VSM

The vector space of matrices, ${M}_{mn}$

Set: ${M}_{mn}$, the set of
all matrices of size $m\times n$
and entries from ${\u2102}^{}$,
Example VSM.

Equality: Entry-wise, Definition ME.

Vector Addition: The “usual” addition, given in Definition MA.

Scalar Multiplication: The “usual” scalar multiplication, given in
Definition MSM.

Does this set with these operations fulfill the ten properties? Yes. And all we need to do is quote Theorem VSPM. Another easy one (by design). $\u22a0$

So, the set of all matrices of a fixed size forms a vector space. That entitles us to call a matrix a vector, since a matrix is an element of a vector space. For example, if $A,\phantom{\rule{0em}{0ex}}B\in {M}_{3,4}$ then we call $A$ and $B$ “vectors,” and we even use our previous notation for column vectors to refer to $A$ and $B$. So we could legitimately write expressions like

$$u+v=A+B=B+A=v+u$$ |

This could lead to some confusion, but it is not too great a danger. But it is worth comment.

The previous two examples may be less than satisfying. We made all the relevant definitions long ago. And the required verifications were all handled by quoting old theorems. However, it is important to consider these two examples first. We have been studying vectors and matrices carefully (Chapter V, Chapter M), and both objects, along with their operations, have certain properties in common, as you may have noticed in comparing Theorem VSPCV with Theorem VSPM. Indeed, it is these two theorems that motivate us to formulate the abstract definition of a vector space, Definition VS. Now, should we prove some general theorems about vector spaces (as we will shortly in Subsection VS.VSP), we can instantly apply the conclusions to both ${\u2102}^{m}$ and ${M}_{mn}$. Notice too how we have taken six definitions and two theorems and reduced them down to two examples. With greater generalization and abstraction our old ideas get downgraded in stature.

Let us look at some more examples, now considering some new vector spaces.

Example VSP

The vector space of polynomials,
${P}_{n}$

Set: ${P}_{n}$, the set of all
polynomials of degree $n$
or less in the variable $x$
with coefficients from ${\u2102}^{}$.

Equality:

$${a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots +{a}_{n}{x}^{n}={b}_{0}+{b}_{1}x+{b}_{2}{x}^{2}+\cdots +{b}_{n}{x}^{n}\text{ifandonlyif}{a}_{i}={b}_{i}\text{for}0\le i\le n$$ |

Vector Addition:

$$\begin{array}{lll}\hfill \left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots +{a}_{n}{x}^{n}\right)+\left({b}_{0}+{b}_{1}x+{b}_{2}{x}^{2}+\cdots +{b}_{n}{x}^{n}\right)=& \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \left({a}_{0}+{b}_{0}\right)+\left({a}_{1}+{b}_{1}\right)x+\left({a}_{2}+{b}_{2}\right){x}^{2}+\cdots +\left({a}_{n}+{b}_{n}\right){x}^{n}& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$Scalar Multiplication:

$$\alpha \left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots +{a}_{n}{x}^{n}\right)=\left(\alpha {a}_{0}\right)+\left(\alpha {a}_{1}\right)x+\left(\alpha {a}_{2}\right){x}^{2}+\cdots +\left(\alpha {a}_{n}\right){x}^{n}$$ |

This set, with these operations, will fulfill the ten properties, though we will not work all the details here. However, we will make a few comments and prove one of the properties. First, the zero vector (Property Z) is what you might expect, and you can check that it has the required property.

$$0=0+0x+0{x}^{2}+\cdots +0{x}^{n}$$ |

The additive inverse (Property AI) is also no surprise, though consider how we have chosen to write it.

$$-\left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots +{a}_{n}{x}^{n}\right)=\left(-{a}_{0}\right)+\left(-{a}_{1}\right)x+\left(-{a}_{2}\right){x}^{2}+\cdots +\left(-{a}_{n}\right){x}^{n}$$ |

Now let’s prove the associativity of vector addition (Property AA). This is a bit tedious, though necessary. Throughout, the plus sign (“+”) does triple-duty. You might ask yourself what each plus sign represents as you work through this proof.

$$\begin{array}{llll}\hfill u+& \left(v+w\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({a}_{0}+{a}_{1}x+\cdots +{a}_{n}{x}^{n}\right)+\left(\left({b}_{0}+{b}_{1}x+\cdots +{b}_{n}{x}^{n}\right)+\left({c}_{0}+{c}_{1}x+\cdots +{c}_{n}{x}^{n}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({a}_{0}+{a}_{1}x+\cdots +{a}_{n}{x}^{n}\right)+\left(\left({b}_{0}+{c}_{0}\right)+\left({b}_{1}+{c}_{1}\right)x+\cdots +\left({b}_{n}+{c}_{n}\right){x}^{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({a}_{0}+\left({b}_{0}+{c}_{0}\right)\right)+\left({a}_{1}+\left({b}_{1}+{c}_{1}\right)\right)x+\cdots +\left({a}_{n}+\left({b}_{n}+{c}_{n}\right)\right){x}^{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left({a}_{0}+{b}_{0}\right)+{c}_{0}\right)+\left(\left({a}_{1}+{b}_{1}\right)+{c}_{1}\right)x+\cdots +\left(\left({a}_{n}+{b}_{n}\right)+{c}_{n}\right){x}^{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left({a}_{0}+{b}_{0}\right)+\left({a}_{1}+{b}_{1}\right)x+\cdots +\left({a}_{n}+{b}_{n}\right){x}^{n}\right)+\left({c}_{0}+{c}_{1}x+\cdots +{c}_{n}{x}^{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left({a}_{0}+{a}_{1}x+\cdots +{a}_{n}{x}^{n}\right)+\left({b}_{0}+{b}_{1}x+\cdots +{b}_{n}{x}^{n}\right)\right)+\left({c}_{0}+{c}_{1}x+\cdots +{c}_{n}{x}^{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(u+v\right)+w\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Notice how it is the application of the associativity of the (old) addition of complex numbers in the middle of this chain of equalities that makes the whole proof happen. The remainder is successive applications of our (new) definition of vector (polynomial) addition. Proving the remainder of the ten properties is similar in style and tedium. You might try proving the commutativity of vector addition (Property C), or one of the distributivity properties (Property DVA, Property DSA). $\u22a0$

Example VSIS

The vector space of infinite sequences

Set: ${\u2102}^{\infty}=\left\{\left.\left({c}_{0},\phantom{\rule{0em}{0ex}}{c}_{1},\phantom{\rule{0em}{0ex}}{c}_{2},\phantom{\rule{0em}{0ex}}{c}_{3},\phantom{\rule{0em}{0ex}}\dots \right)\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{c}_{i}\in {\u2102}^{},\phantom{\rule{0ex}{0ex}}i\in \mathbb{N}\right\}$.

Equality:

$$\left({c}_{0},\phantom{\rule{0em}{0ex}}{c}_{1},\phantom{\rule{0em}{0ex}}{c}_{2},\phantom{\rule{0em}{0ex}}\dots \right)=\left({d}_{0},\phantom{\rule{0em}{0ex}}{d}_{1},\phantom{\rule{0em}{0ex}}{d}_{2},\phantom{\rule{0em}{0ex}}\dots \right)\text{ifandonlyif}{c}_{i}={d}_{i}\text{forall}i\ge 0$$ |

Vector Addition:

$$\left({c}_{0},\phantom{\rule{0em}{0ex}}{c}_{1},\phantom{\rule{0em}{0ex}}{c}_{2},\phantom{\rule{0em}{0ex}}\dots \right)+\left({d}_{0},\phantom{\rule{0em}{0ex}}{d}_{1},\phantom{\rule{0em}{0ex}}{d}_{2},\phantom{\rule{0em}{0ex}}\dots \right)=\left({c}_{0}+{d}_{0},\phantom{\rule{0em}{0ex}}{c}_{1}+{d}_{1},\phantom{\rule{0em}{0ex}}{c}_{2}+{d}_{2},\phantom{\rule{0em}{0ex}}\dots \right)$$ |

Scalar Multiplication:

$$\alpha \left({c}_{0},\phantom{\rule{0em}{0ex}}{c}_{1},\phantom{\rule{0em}{0ex}}{c}_{2},\phantom{\rule{0em}{0ex}}{c}_{3},\phantom{\rule{0em}{0ex}}\dots \right)=\left(\alpha {c}_{0},\phantom{\rule{0em}{0ex}}\alpha {c}_{1},\phantom{\rule{0em}{0ex}}\alpha {c}_{2},\phantom{\rule{0em}{0ex}}\alpha {c}_{3},\phantom{\rule{0em}{0ex}}\dots \right)$$ |

This should remind you of the vector space ${\u2102}^{m}$, though now our lists of scalars are written horizontally with commas as delimiters and they are allowed to be infinite in length. What does the zero vector look like (Property Z)? Additive inverses (Property AI)? Can you prove the associativity of vector addition (Property AA)? $\u22a0$

Example VSF

The vector space of functions

Set: $F=\left\{\left.f\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}f:{\u2102}^{}\to {\u2102}^{}\right\}$.

Equality: $f=g$ if
and only if $f\left(x\right)=g\left(x\right)$
for all $x\in {\u2102}^{}$.

Vector Addition: $f+g$ is the
function with outputs defined by $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$.

Scalar Multiplication: $\alpha f$
is the function with outputs defined by
$\left(\alpha f\right)\left(x\right)=\alpha f\left(x\right)$.

So this is the set of all functions of one variable that take a complex number to a complex number. You might have studied functions of one variable that take a real number to a real number, and that might be a more natural set to study. But since we are allowing our scalars to be complex numbers, we need to expand the domain and range of our functions also. Study carefully how the definitions of the operation are made, and think about the different uses of “+” and juxtaposition. As an example of what is required when verifying that this is a vector space, consider that the zero vector (Property Z) is the function $z$ whose definition is $z\left(x\right)=0$ for every input $x$.

While vector spaces of functions are very important in mathematics and physics, we will not devote them much more attention. $\u22a0$

Here’s a unique example.

Example VSS

The singleton vector space

Set: $Z=\left\{z\right\}$.

Equality: Huh?

Vector Addition: $z+z=z$.

Scalar Multiplication: $\alpha z=z$.

This should look pretty wild. First, just what is $z$? Column vector, matrix, polynomial, sequence, function? Mineral, plant, or animal? We aren’t saying! $z$ just is. And we have definitions of vector addition and scalar multiplication that are sufficient for an occurence of either that may come along.

Our only concern is if this set, along with the definitions of two operations, fulfills the ten properties of Definition VS. Let’s check associativity of vector addition (Property AA). For all $u,\phantom{\rule{0em}{0ex}}v,\phantom{\rule{0em}{0ex}}w\in Z$,

$$\begin{array}{llll}\hfill u+\left(v+w\right)& =z+\left(z+z\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =z+z\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(z+z\right)+z\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(u+v\right)+w\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$What is the zero vector in this vector space (Property Z)? With only one element in the set, we do not have much choice. Is $z=0$? It appears that $z$ behaves like the zero vector should, so it gets the title. Maybe now the definition of this vector space does not seem so bizarre. It is a set whose only element is the element that behaves like the zero vector, so that lone element is the zero vector. $\u22a0$

Perhaps some of the above definitions and verifications seem obvious or like splitting hairs, but the next example should convince you that they are necessary. We will study this one carefully. Ready? Check your preconceptions at the door.

Example CVS

The crazy vector space

Set: $C=\left\{\left.\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\in {\u2102}^{}\right\}$.

Vector Addition: $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)=\left({x}_{1}+{y}_{1}+1,\phantom{\rule{0em}{0ex}}{x}_{2}+{y}_{2}+1\right)$.

Scalar Multiplication: $\alpha \left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=\left(\alpha {x}_{1}+\alpha -1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\alpha -1\right)$.

Now, the first thing I hear you say is “You can’t do that!” And my response is, “Oh yes, I can!” I am free to define my set and my operations any way I please. They may not look natural, or even useful, but we will now verify that they provide us with another example of a vector space. And that is enough. If you are adventurous, you might try first checking some of the properties yourself. What is the zero vector? Additive inverses? Can you prove associativity? Ready, here we go.

Property AC, Property SC: The result of each operation is a pair of complex numbers, so these two closure properties are fulfilled.

$$\begin{array}{llll}\hfill u+v& =\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)=\left({x}_{1}+{y}_{1}+1,\phantom{\rule{0em}{0ex}}{x}_{2}+{y}_{2}+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({y}_{1}+{x}_{1}+1,\phantom{\rule{0em}{0ex}}{y}_{2}+{x}_{2}+1\right)=\left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)+\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =v+u\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $$\begin{array}{llll}\hfill u+\left(v+w\right)& =\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left(\left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)+\left({z}_{1},\phantom{\rule{0em}{0ex}}{z}_{2}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left({y}_{1}+{z}_{1}+1,\phantom{\rule{0em}{0ex}}{y}_{2}+{z}_{2}+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({x}_{1}+\left({y}_{1}+{z}_{1}+1\right)+1,\phantom{\rule{0em}{0ex}}{x}_{2}+\left({y}_{2}+{z}_{2}+1\right)+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({x}_{1}+{y}_{1}+{z}_{1}+2,\phantom{\rule{0em}{0ex}}{x}_{2}+{y}_{2}+{z}_{2}+2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left({x}_{1}+{y}_{1}+1\right)+{z}_{1}+1,\phantom{\rule{0em}{0ex}}\left({x}_{2}+{y}_{2}+1\right)+{z}_{2}+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({x}_{1}+{y}_{1}+1,\phantom{\rule{0em}{0ex}}{x}_{2}+{y}_{2}+1\right)+\left({z}_{1},\phantom{\rule{0em}{0ex}}{z}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)\right)+\left({z}_{1},\phantom{\rule{0em}{0ex}}{z}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(u+v\right)+w\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Property Z: The zero vector is …$0=\left(-1,\phantom{\rule{0em}{0ex}}-1\right)$. Now I hear you say, “No, no, that can’t be, it must be $\left(0,\phantom{\rule{0em}{0ex}}0\right)$!” Indulge me for a moment and let us check my proposal.

$$u+0=\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left(-1,\phantom{\rule{0em}{0ex}}-1\right)=\left({x}_{1}+\left(-1\right)+1,\phantom{\rule{0em}{0ex}}{x}_{2}+\left(-1\right)+1\right)=\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=u$$ |

Feeling better? Or worse?

Property AI: For each vector, $u$, we must locate an additive inverse, $-u$. Here it is, $-\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=\left(-{x}_{1}-2,\phantom{\rule{0em}{0ex}}-{x}_{2}-2\right)$. As odd as it may look, I hope you are withholding judgment. Check:

$$u+\left(-u\right)=\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left(-{x}_{1}-2,\phantom{\rule{0em}{0ex}}-{x}_{2}-2\right)=\left({x}_{1}+\left(-{x}_{1}-2\right)+1,\phantom{\rule{0em}{0ex}}-{x}_{2}+\left({x}_{2}-2\right)+1\right)=\left(-1,\phantom{\rule{0em}{0ex}}-1\right)=0$$ |

Property DVA: If you have hung on so far, here’s where it gets even wilder. In the next two properties we mix and mash the two operations.

$$\begin{array}{llll}\hfill \alpha \left(u+v\right)& =\alpha \left(\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\alpha \left({x}_{1}+{y}_{1}+1,\phantom{\rule{0em}{0ex}}{x}_{2}+{y}_{2}+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha \left({x}_{1}+{y}_{1}+1\right)+\alpha -1,\phantom{\rule{0em}{0ex}}\alpha \left({x}_{2}+{y}_{2}+1\right)+\alpha -1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha {x}_{1}+\alpha {y}_{1}+\alpha +\alpha -1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\alpha {y}_{2}+\alpha +\alpha -1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha {x}_{1}+\alpha -1+\alpha {y}_{1}+\alpha -1+1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\alpha -1+\alpha {y}_{2}+\alpha -1+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left(\alpha {x}_{1}+\alpha -1\right)+\left(\alpha {y}_{1}+\alpha -1\right)+1,\phantom{\rule{0em}{0ex}}\left(\alpha {x}_{2}+\alpha -1\right)+\left(\alpha {y}_{2}+\alpha -1\right)+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha {x}_{1}+\alpha -1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\alpha -1\right)+\left(\alpha {y}_{1}+\alpha -1,\phantom{\rule{0em}{0ex}}\alpha {y}_{2}+\alpha -1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\alpha \left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\alpha \left({y}_{1},\phantom{\rule{0em}{0ex}}{y}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\alpha u+\alpha v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $$\begin{array}{llll}\hfill \left(\alpha +\beta \right)u& =\left(\alpha +\beta \right)\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left(\alpha +\beta \right){x}_{1}+\left(\alpha +\beta \right)-1,\phantom{\rule{0em}{0ex}}\left(\alpha +\beta \right){x}_{2}+\left(\alpha +\beta \right)-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha {x}_{1}+\beta {x}_{1}+\alpha +\beta -1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\beta {x}_{2}+\alpha +\beta -1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha {x}_{1}+\alpha -1+\beta {x}_{1}+\beta -1+1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\alpha -1+\beta {x}_{2}+\beta -1+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\left(\alpha {x}_{1}+\alpha -1\right)+\left(\beta {x}_{1}+\beta -1\right)+1,\phantom{\rule{0em}{0ex}}\left(\alpha {x}_{2}+\alpha -1\right)+\left(\beta {x}_{2}+\beta -1\right)+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\alpha {x}_{1}+\alpha -1,\phantom{\rule{0em}{0ex}}\alpha {x}_{2}+\alpha -1\right)+\left(\beta {x}_{1}+\beta -1,\phantom{\rule{0em}{0ex}}\beta {x}_{2}+\beta -1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\alpha \left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)+\beta \left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\alpha u+\beta u\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Property O: After all that, this one is easy, but no less pleasing.

$$1u=1\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=\left({x}_{1}+1-1,\phantom{\rule{0em}{0ex}}{x}_{2}+1-1\right)=\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=u$$ |

That’s it, $C$ is a vector space, as crazy as that may seem.

Notice that in the case of the zero vector and additive inverses, we only had to propose possibilities and then verify that they were the correct choices. You might try to discover how you would arrive at these choices, though you should understand why the process of discovering them is not a necessary component of the proof itself. $\u22a0$

Subsection VS.EVS has provided us with an abundance of examples of vector spaces, most of them containing useful and interesting mathematical objects along with natural operations. In this subsection we will prove some general properties of vector spaces. Some of these results will again seem obvious, but it is important to understand why it is necessary to state and prove them. A typical hypothesis will be “Let $V$ be a vector space.” From this we may assume the ten properties of Definition VS, and nothing more. Its like starting over, as we learn about what can happen in this new algebra we are learning. But the power of this careful approach is that we can apply these theorems to any vector space we encounter — those in the previous examples, or new ones we have not yet contemplated. Or perhaps new ones that nobody has ever contemplated. We will illustrate some of these results with examples from the crazy vector space (Example CVS), but mostly we are stating theorems and doing proofs. These proofs do not get too involved, but are not trivial either, so these are good theorems to try proving yourself before you study the proof given here. (See Technique P.)

First we show that there is just one zero vector. Notice that the properties only require there to be at least one, and say nothing about there possibly being more. That is because we can use the ten properties of a vector space (Definition VS) to learn that there can never be more than one. To require that this extra condition be stated as an eleventh property would make the definition of a vector space more complicated than it needs to be.

Theorem ZVU

Zero Vector is Unique

Suppose that $V$ is a vector
space. The zero vector, $0$,
is unique. $\square $

Proof To prove uniqueness, a standard technique is to suppose the existence of two objects (Technique U). So let ${0}_{1}$ and ${0}_{2}$ be two zero vectors in $V$. Then

$$\begin{array}{llllllll}\hfill {0}_{1}& ={0}_{1}+{0}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{for}{0}_{2}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={0}_{2}+{0}_{1}\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property C}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={0}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{for}{0}_{1}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$This proves the uniqueness since the two zero vectors are really the same. $\u25a0$

Theorem AIU

Additive Inverses are Unique

Suppose that $V$ is a
vector space. For each $u\in V$,
the additive inverse, $-u$,
is unique. $\square $

Proof To prove uniqueness, a standard technique is to suppose the existence of two objects (Technique U). So let ${-u}_{1}$ and ${-u}_{2}$ be two additive inverses for $u$. Then

$$\begin{array}{llllllll}\hfill {-u}_{1}& ={-u}_{1}+0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={-u}_{1}+\left(u+{-u}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left({-u}_{1}+u\right)+{-u}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0+{-u}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={-u}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So the two additive inverses are really the same. $\u25a0$

As obvious as the next three theorems appear, nowhere have we guaranteed that the zero scalar, scalar multiplication and the zero vector all interact this way. Until we have proved it, anyway.

Theorem ZSSM

Zero Scalar in Scalar Multiplication

Suppose that $V$ is a
vector space and $u\in V$.
Then $0u=0$.
$\square $

Proof Notice that $0$ is a scalar, $u$ is a vector, so Property SC says $0u$ is again a vector. As such, $0u$ has an additive inverse, $-\left(0u\right)$ by Property AI.

$$\begin{array}{llllllll}\hfill 0u& =0+0u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-\left(0u\right)+0u\right)+0u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(0u\right)+\left(0u+0u\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(0u\right)+\left(0+0\right)u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property DSA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(0u\right)+0u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property ZCN}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $\u25a0$Here’s another theorem that looks like it should be obvious, but is still in need of a proof.

Theorem ZVSM

Zero Vector in Scalar Multiplication

Suppose that $V$ is a
vector space and $\alpha \in {\u2102}^{}$.
Then $\alpha 0=0$.
$\square $

Proof Notice that $\alpha $ is a scalar, $0$ is a vector, so Property SC means $\alpha 0$ is again a vector. As such, $\alpha 0$ has an additive inverse, $-\left(\alpha 0\right)$ by Property AI.

$$\begin{array}{llllllll}\hfill \alpha 0& =0+\alpha 0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-\left(\alpha 0\right)+\alpha 0\right)+\alpha 0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(\alpha 0\right)+\left(\alpha 0+\alpha 0\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(\alpha 0\right)+\alpha \left(0+0\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property DVA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(\alpha 0\right)+\alpha 0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $\u25a0$Here’s another one that sure looks obvious. But understand that we have chosen to use certain notation because it makes the theorem’s conclusion look so nice. The theorem is not true because the notation looks so good, it still needs a proof. If we had really wanted to make this point, we might have defined the additive inverse of $u$ as ${u}^{\u266f}$. Then we would have written the defining property, Property AI, as $u+{u}^{\u266f}=0$. This theorem would become ${u}^{\u266f}=\left(-1\right)u$. Not really quite as pretty, is it?

Theorem AISM

Additive Inverses from Scalar Multiplication

Suppose that $V$ is a
vector space and $u\in V$.
Then $-u=\left(-1\right)u$.
$\square $

Proof

$$\begin{array}{llllllll}\hfill -u& =-u+0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-u+0u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Theorem ZSSM}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-u+\left(1+\left(-1\right)\right)u\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-u+\left(1u+\left(-1\right)u\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property DSA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-u+\left(u+\left(-1\right)u\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property O}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-u+u\right)+\left(-1\right)u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0+\left(-1\right)u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property AI}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-1\right)u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property Z}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $\u25a0$ Because of this theorem, we can now write linear combinations like
$6{u}_{1}+\left(-4\right){u}_{2}$

as $6{u}_{1}-4{u}_{2}$,
even though we have not formally defined an operation called vector
subtraction. Our next theorem is a bit different from several of the others in the
list. Rather than making a declaration (“the zero vector is unique”) it is an
implication (“if…, then…”) and so can be used in proofs to convert a vector
equality into two possibilities, one a scalar equlaity and the other a vector
equality. It should remind you of the situation for complex numbers. If
$\alpha ,\phantom{\rule{0em}{0ex}}\beta \in \u2102$ and
$\alpha \beta =0$, then
$\alpha =0$ or
$\beta =0$. This
critical property is the driving force behind using a factorization to solve a
polynomial equation.

Theorem SMEZV

Scalar Multiplication Equals the Zero Vector

Suppose that $V$ is a
vector space and $\alpha \in {\u2102}^{}$.
If $\alpha u=0$, then
either $\alpha =0$
or $u=0$.
$\square $

Proof We prove this theorem by breaking up the analysis into two cases. The first seems too trivial, and it is, but the logic of the argument is still legitimate.

Case 1. Suppose $\alpha =0$. In this case our conclusion is true (the first part of the either/or is true) and we are done. That was easy.

Case 2. Suppose $\alpha \ne 0$.

$$\begin{array}{llllllll}\hfill u& =1u\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property O}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\frac{1}{\alpha}\alpha \right)u\phantom{\rule{2em}{0ex}}& \hfill & \alpha \ne 0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{\alpha}\left(\alpha u\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Property SMA}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{\alpha}\left(0\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{Hypothesis}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \text{}\text{Theorem ZVSM}\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So in this case, the conclusion is true (the second part of the either/or is true) and we are done since the conclusion was true in each of the two cases. $\u25a0$

Example PCVS

Properties for the Crazy Vector Space

Several of the above theorems have interesting demonstrations when applied to the crazy
vector space, $C$
(Example CVS). We are not proving anything new here, or learning anything we did not know
already about $C$.
It is just plain fun to see how these general theorems apply in a specific instance.
For most of our examples, the applications are obvious or trivial, but not with
$C$.

Suppose $u\in C$.

Then, as given by Theorem ZSSM,

$$0u=0\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=\left(0{x}_{1}+0-1,\phantom{\rule{0em}{0ex}}0{x}_{2}+0-1\right)=\left(-1,-1\right)=0$$ |

And as given by Theorem ZVSM,

$$\begin{array}{llll}\hfill \alpha 0& =\alpha \left(-1,\phantom{\rule{0em}{0ex}}-1\right)=\left(\alpha \left(-1\right)+\alpha -1,\phantom{\rule{0em}{0ex}}\alpha \left(-1\right)+\alpha -1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-\alpha +\alpha -1,\phantom{\rule{0em}{0ex}}-\alpha +\alpha -1\right)=\left(-1,\phantom{\rule{0em}{0ex}}-1\right)=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Finally, as given by Theorem AISM,

$$\begin{array}{llll}\hfill \left(-1\right)u& =\left(-1\right)\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=\left(\left(-1\right){x}_{1}+\left(-1\right)-1,\phantom{\rule{0em}{0ex}}\left(-1\right){x}_{2}+\left(-1\right)-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-{x}_{1}-2,\phantom{\rule{0em}{0ex}}-{x}_{2}-2\right)=-u\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $\u22a0$

When we say that $V$ is a vector space, we then know we have a set of objects (the “vectors”), but we also know we have been provided with two operations (“vector addition” and “scalar multiplication”) and these operations behave with these objects according to the ten properties of Definition VS. One combines two vectors and produces a vector, the other takes a scalar and a vector, producing a vector as the result. So if ${u}_{1},\phantom{\rule{0em}{0ex}}{u}_{2},\phantom{\rule{0em}{0ex}}{u}_{3}\in V$ then an expression like

$$5{u}_{1}+7{u}_{2}-13{u}_{3}$$ |

would be unambiguous in any of the vector spaces we have discussed in this
section. And the resulting object would be another vector in the vector space. If
you were tempted to call the above expression a linear combination, you would be
right. Four of the definitions that were central to our discussions in Chapter V
were stated in the context of vectors being column vectors, but were purposely
kept broad enough that they could be applied in the context of any vector space.
They only rely on the presence of scalars, vectors, vector addition and scalar
multiplication to make sense. We will restate them shortly, unchanged,
except that their titles and acronyms no longer refer to column vectors,
and the hypothesis of being in a vector space has been added. Take the
time now to look forward and review each one, and begin to form some
connections to what we have done earlier and what we will be doing in
subsequent sections and chapters. Specifically, compare the following pairs of
definitions:

Definition LCCV and Definition LC

Definition SSCV and Definition SS

Definition RLDCV and Definition RLD

Definition LICV and Definition LI

- Comment on how the vector space ${\u2102}^{m}$ went from a theorem (Theorem VSPCV) to an example (Example VSCV).
- In the crazy vector space, $C$,
(Example CVS) compute the linear combination
$$2\left(3,\phantom{\rule{0em}{0ex}}4\right)+\left(-6\right)\left(1,\phantom{\rule{0em}{0ex}}2\right).$$ - Suppose that $\alpha $ is a scalar and $0$ is the zero vector. Why should we prove anything as obvious as $\alpha 0=0$ such as we did in Theorem ZVSM?

M10 Define a possibly new vector space by beginning with the set and vector addition from ${\u2102}^{2}$ (Example VSCV) but change the definition of scalar multiplication to

$$\begin{array}{llllllll}\hfill \alpha x& =0=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \alpha \in {\u2102}^{},\phantom{\rule{0ex}{0ex}}x\in {\u2102}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Prove that the first nine properties required for a vector space hold, but Property O does not hold.

This example shows us that we cannot expect to be able to derive
Property O as a consequence of assuming the first nine properties. In other
words, we cannot slim down our list of properties by jettisoning the last
one, and still have the same collection of objects qualify as vector spaces.

Contributed by Robert Beezer

T10 Prove each of the ten properties of Definition VS for each of the following
examples of a vector space:

Example VSP

Example VSIS

Example VSF

Example VSS

Contributed by Robert Beezer

The next three problems suggest that under the right situations we can “cancel.” In practice, these techniques should be avoided in other proofs. Prove each of the following statements.

T21 Suppose that $V$
is a vector space, and $u,\phantom{\rule{0em}{0ex}}v,\phantom{\rule{0em}{0ex}}w\in V$.
If $w+u=w+v$,
then $u=v$.

Contributed by Robert Beezer Solution [836]

T22 Suppose $V$
is a vector space, $u,\phantom{\rule{0em}{0ex}}v\in V$
and $\alpha $ is a nonzero
scalar from ${\u2102}^{}$.
If $\alpha u=\alpha v$,
then $u=v$.

Contributed by Robert Beezer Solution [836]

T23 Suppose $V$
is a vector space, $u\ne 0$
is a vector in $V$
and $\alpha ,\phantom{\rule{0em}{0ex}}\beta \in {\u2102}^{}$.
If $\alpha u=\beta u$,
then $\alpha =\beta $.

Contributed by Robert Beezer Solution [837]

T21 Contributed by Robert Beezer Statement [835]

T22 Contributed by Robert Beezer Statement [835]

T23 Contributed by Robert Beezer Statement [835]