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Section VS Vector Spaces

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.

Subsection VS Vector Spaces

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 ‚Äú+‚ÄĚ, and (2) scalar multiplication, which combines a complex number with an element of \(V\) and is denoted by juxtaposition. Then \(V\text{,}\) along with the two operations, is a vector space over \(\complexes\) if the following ten properties hold.

AC Additive Closure

If \(\vect{u},\,\vect{v}\in V\text{,}\) then \(\vect{u}+\vect{v}\in V\text{.}\)

SC Scalar Closure

If \(\alpha\in\complexes\) and \(\vect{u}\in V\text{,}\) then \(\alpha\vect{u}\in V\text{.}\)

C Commutativity

If \(\vect{u},\,\vect{v}\in V\text{,}\) then \(\vect{u}+\vect{v}=\vect{v}+\vect{u}\text{.}\)

AA Additive Associativity

If \(\vect{u},\,\vect{v},\,\vect{w}\in V\text{,}\) then \(\vect{u}+\left(\vect{v}+\vect{w}\right)=\left(\vect{u}+\vect{v}\right)+\vect{w}\text{.}\)

Z Zero Vector

There is a vector, \(\zerovector\text{,}\) called the zero vector, such that \(\vect{u}+\zerovector=\vect{u}\) for all \(\vect{u}\in V\text{.}\)

AI Additive Inverses

If \(\vect{u}\in V\text{,}\) then there exists a vector \(\vect{-u}\in V\) so that \(\vect{u}+ (\vect{-u})=\zerovector\text{.}\)

SMA Scalar Multiplication Associativity

If \(\alpha,\,\beta\in\complexes\) and \(\vect{u}\in V\text{,}\) then \(\alpha(\beta\vect{u})=(\alpha\beta)\vect{u}\text{.}\)

DVA Distributivity across Vector Addition

If \(\alpha\in\complexes\) and \(\vect{u},\,\vect{v}\in V\text{,}\) then \(\alpha(\vect{u}+\vect{v})=\alpha\vect{u}+\alpha\vect{v}\text{.}\)

DSA Distributivity across Scalar Addition

If \(\alpha,\,\beta\in\complexes\) and \(\vect{u}\in V\text{,}\) then\((\alpha+\beta)\vect{u}=\alpha\vect{u}+\beta\vect{u}\text{.}\)

O One

If \(\vect{u}\in V\text{,}\) then \(1\vect{u}=\vect{u}\text{.}\)

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.

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 ‚Äúself-evident‚ÄĚ 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 ‚ÄĒ scalars arranged in a columnar list of fixed length. In a similar vein, you have used the symbol ‚Äú+‚ÄĚ 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 ‚Äú+‚ÄĚ 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).

In Definition¬†VS, the scalars do not have to be complex numbers. They can come from what are called in more advanced mathematics, ‚Äúfields‚ÄĚ. Examples of fields are the set of complex numbers, the set of real numbers, the set of rational numbers, and even the finite set of ‚Äúbinary numbers‚ÄĚ, \(\set{0,\,1}\text{.}\) There are many, many others. In this case we would call \(V\) a vector space over (the field) \(F\text{.}\)

A vector space is composed of three objects, a set and two operations. Some would explicitly state in the definition that \(V\) must be a nonempty set, but we can infer this from Property Z, since the set cannot be empty and contain a vector that behaves as the zero vector. Also, 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 it has 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 us look at some concrete examples now.

Subsection EVS Examples of Vector Spaces

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 nontrivial example completely. In other places, our usual thoroughness will be neglected, so grab paper and pencil and play along.

Set

\(M_{mn}\text{,}\) the set of all matrices of size \(m\times n\) and entries from \(\complexes\text{,}\) Definition VSM.

Equality

Entry-wise, Definition ME.

Vector Addition

The ‚Äúusual‚ÄĚ addition, given in Definition¬†MA.

Scalar Multiplication

The ‚Äúusual‚ÄĚ 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).

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 \(A,\,B\in M_{34}\) then we call \(A\) and \(B\) ‚Äúvectors,‚ÄĚ and we even use our previous notation for column vectors to refer to \(A\) and \(B\text{.}\) So we could legitimately write expressions like

\begin{equation*} \vect{u}+\vect{v}=A+B=B+A=\vect{v}+\vect{u} \end{equation*}

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, if we prove some general theorems about vector spaces (as we will shortly in Subsection VS.VSP), we can then instantly apply the conclusions to both \(\complex{m}\) and \(M_{mn}\text{.}\) 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.

Set

\(P_n\text{,}\) the set of all polynomials of degree \(n\) or less in the variable \(x\) with coefficients from \(\complexes\text{.}\)

Equality

\(a_0+a_1x+a_2x^2+\cdots+a_nx^n=b_0+b_1x+b_2x^2+\cdots+b_nx^n\) if and only if \(a_i=b_i\) for all \(0\leq i\leq n\text{.}\)

Vector Addition
\begin{gather*} (a_0+a_1x+a_2x^2+\cdots+a_nx^n)+(b_0+b_1x+b_2x^2+\cdots+b_nx^n)=\\ (a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+\cdots+(a_n+b_n)x^n\text{.} \end{gather*}
Scalar Multiplication
\begin{equation*} \alpha(a_0+a_1x+a_2x^2+\cdots+a_nx^n)=(\alpha a_0)+(\alpha a_1)x+(\alpha a_2)x^2+\cdots+(\alpha a_n)x^n\text{.} \end{equation*}

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.

\begin{equation*} \zerovector=0+0x+0x^2+\cdots+0x^n \end{equation*}

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

\begin{equation*} -\left(a_0+a_1x+a_2x^2+\cdots+a_nx^n\right)=(-a_0)+(-a_1)x+(-a_2)x^2+\cdots+(-a_n)x^n \end{equation*}

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

\begin{align*} \vect{u}+&(\vect{v}+\vect{w})\\ &=(a_0+a_1x+\cdots+a_nx^n)+\left((b_0+b_1x+\cdots+b_nx^n)+(c_0+c_1x+\cdots+c_nx^n)\right)\\ &=(a_0+a_1x+\cdots+a_nx^n)+((b_0+c_0)+(b_1+c_1)x+\cdots+(b_n+c_n)x^n)\\ &=(a_0+(b_0+c_0))+(a_1+(b_1+c_1))x+\cdots+(a_n+(b_n+c_n))x^n\\ &=((a_0+b_0)+c_0)+((a_1+b_1)+c_1)x+\cdots+((a_n+b_n)+c_n)x^n\\ &=((a_0+b_0)+(a_1+b_1)x+\cdots+(a_n+b_n)x^n)+(c_0+c_1x+\cdots+c_nx^n)\\ &=\left((a_0+a_1x+\cdots+a_nx^n)+(b_0+b_1x+\cdots+b_nx^n)\right)+(c_0+c_1x+\cdots+c_nx^n)\\ &=(\vect{u}+\vect{v})+\vect{w} \end{align*}

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).

Set

\(\complex{\infty}=\setparts{(c_0,\,c_1,\,c_2,\,c_3,\,\ldots)}{c_i\in\complexes,\ i\in\mathbb{N}}\text{.}\)

Equality

\((c_0,\,c_1,\,c_2,\,\ldots)=(d_0,\,d_1,\,d_2,\,\ldots)\) if and only if \(c_i=d_i\) for all \(i\geq 0\text{.}\)

Vector Addition

\((c_0,\,c_1,\,c_2,\,\ldots)+(d_0,\,d_1,\,d_2,\,\ldots)=(c_0+d_0,\,c_1+d_1,\,c_2+d_2,\,\ldots)\text{.}\)

Scalar Multiplication

\(\alpha (c_0,\,c_1,\,c_2,\,c_3,\,\ldots)=(\alpha c_0,\,\alpha c_1,\,\alpha c_2,\,\alpha c_3,\,\ldots)\text{.}\)

This should remind you of the vector space \(\complex{m}\text{,}\) 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)?

Let \(X\) be any set.

Set

\(F=\setparts{f}{f:X\rightarrow\complexes}\text{.}\)

Equality

\(f=g\) if and only if \(f(x)=g(x)\) for all \(x\in X\text{.}\)

Vector Addition

\(f+g\) is the function with outputs defined by \((f+g)(x)=f(x)+g(x)\text{.}\)

Scalar Multiplication

\(\alpha f\) is the function with outputs defined by \((\alpha f)(x)=\alpha f(x)\text{.}\)

So this is the set of all functions of one variable that take elements of the set \(X\) 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 use as \(X\text{.}\) But since we are allowing our scalars to be complex numbers, we need to specify that the range of our functions is the complex numbers. Study carefully how the definitions of the operation are made, and think about the different uses of ‚Äú+‚ÄĚ 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 \(z(x)=0\) for every input \(x\in X\text{.}\)

Vector spaces of functions are very important in mathematics and physics, where the field of scalars may be the real numbers, so the ranges of the functions can in turn also be the set of real numbers.

Here is a unique example.

Set

\(Z=\set{\vect{z}}\text{.}\)

Equality

Huh?

Vector Addition

\(\vect{z}+\vect{z}=\vect{z}\text{.}\)

Scalar Multiplication

\(\alpha\vect{z}=\vect{z}\text{.}\)

This should look pretty wild. First, just what is \(\vect{z}\text{?}\) Column vector, matrix, polynomial, sequence, function? Mineral, plant, or animal? We aren't saying! \(\vect{z}\) just is. And we have definitions of vector addition and scalar multiplication that are sufficient for an occurrence 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 us check associativity of vector addition (Property AA). For all \(\vect{u},\,\vect{v},\,\vect{w}\in Z\text{,}\)

\begin{align*} \vect{u}+(\vect{v}+\vect{w}) &=\vect{z}+(\vect{z}+\vect{z})\\ &=\vect{z}+\vect{z}\\ &=(\vect{z}+\vect{z})+\vect{z}\\ &=(\vect{u}+\vect{v})+\vect{w}\text{.} \end{align*}

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 \(\vect{z}=\zerovector\text{?}\) It appears that \(\vect{z}\) 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.

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.

Set

\(C=\setparts{(x_1,\,x_2)}{x_1,\,x_2\in\complexes}\text{.}\)

Equality

\((x_1,\,x_2)=(y_1,\,y_2)\) if and only if \(x_1=y_1\) and \(x_2=y_2\text{.}\)

Vector Addition

\((x_1,\,x_2)+(y_1,\,y_2)=(x_1+y_1+1,\,x_2+y_2+1)\text{.}\)

Scalar Multiplication

\(\alpha(x_1,\,x_2)=(\alpha x_1+\alpha-1,\,\alpha x_2+\alpha-1)\text{.}\)

Now, the first thing I hear you say is ‚ÄúYou can't do that!‚ÄĚ And my response is, ‚ÄúOh yes, I can!‚ÄĚ 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.

Property C:

\begin{align*} \vect{u}+\vect{v}&=(x_1,\,x_2)+(y_1,\,y_2)=(x_1+y_1+1,\,x_2+y_2+1)\\ &=(y_1+x_1+1,\,y_2+x_2+1)=(y_1,\,y_2)+(x_1,\,x_2)\\ &=\vect{v}+\vect{u} \end{align*}

Property AA:

\begin{align*} \vect{u}+(\vect{v}+\vect{w})&=(x_1,\,x_2)+\left((y_1,\,y_2)+(z_1,\,z_2)\right)\\ &=(x_1,\,x_2)+(y_1+z_1+1,\,y_2+z_2+1)\\ &=(x_1+(y_1+z_1+1)+1,\,x_2+(y_2+z_2+1)+1)\\ &=(x_1+y_1+z_1+2,\,x_2+y_2+z_2+2)\\ &=((x_1+y_1+1)+z_1+1,\,(x_2+y_2+1)+z_2+1)\\ &=(x_1+y_1+1,\,x_2+y_2+1)+(z_1,\,z_2)\\ &=\left((x_1,\,x_2)+(y_1,\,y_2)\right)+(z_1,\,z_2)\\ &=\left(\vect{u}+\vect{v}\right)+\vect{w} \end{align*}

Property¬†Z: The zero vector is ‚Ķ\(\zerovector=(-1,\,-1)\text{.}\) Now I hear you say, ‚ÄúNo, no, that can't be, it must be \((0,\,0)\text{!}\)‚ÄĚ. Indulge me for a moment and let us check my proposal.

\begin{equation*} \vect{u}+\zerovector=(x_1,\,x_2)+(-1,\,-1)=(x_1+(-1)+1,\,x_2+(-1)+1)=(x_1,\,x_2)=\vect{u} \end{equation*}

Feeling better? Or worse?

Property AI: For each vector, \(\vect{u}\text{,}\) we must locate an additive inverse, \(\vect{-u}\text{.}\) Here it is, \(-(x_1,\,x_2)=(-x_1-2,\,-x_2-2)\text{.}\) As odd as it may look, I hope you are withholding judgment. Check:

\begin{align*} \vect{u}+ (\vect{-u})&=(x_1,\,x_2)+(-x_1-2,\,-x_2-2)\\ &=(x_1+(-x_1-2)+1,\,-x_2+(x_2-2)+1)=(-1,\,-1)=\zerovector \end{align*}

Property SMA:

\begin{align*} \alpha(\beta\vect{u}) &=\alpha(\beta(x_1,\,x_2))\\ &=\alpha(\beta x_1+\beta-1,\,\beta x_2+\beta-1)\\ &=(\alpha(\beta x_1+\beta-1)+\alpha-1,\,\alpha(\beta x_2+\beta-1)+\alpha-1)\\ &=((\alpha\beta x_1+\alpha\beta-\alpha)+\alpha-1,\,(\alpha\beta x_2+\alpha\beta-\alpha)+\alpha-1)\\ &=(\alpha\beta x_1+\alpha\beta-1,\,\alpha\beta x_2+\alpha\beta-1)\\ &=(\alpha\beta)(x_1,\,x_2)\\ &=(\alpha\beta)\vect{u} \end{align*}

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

\begin{align*} \alpha(\vect{u}&+\vect{v})\\ &=\alpha\left((x_1,\,x_2)+(y_1,\,y_2)\right)\\ &=\alpha(x_1+y_1+1,\,x_2+y_2+1)\\ &=(\alpha(x_1+y_1+1)+\alpha-1,\,\alpha(x_2+y_2+1)+\alpha-1)\\ &=(\alpha x_1+\alpha y_1+\alpha+\alpha-1,\,\alpha x_2+\alpha y_2+\alpha+\alpha-1)\\ &=(\alpha x_1+\alpha-1+\alpha y_1+\alpha-1+1,\,\alpha x_2+\alpha-1+\alpha y_2+\alpha-1+1)\\ &=((\alpha x_1+\alpha-1)+(\alpha y_1+\alpha-1)+1,\,(\alpha x_2+\alpha-1)+(\alpha y_2+\alpha-1)+1)\\ &=(\alpha x_1+\alpha-1,\,\alpha x_2+\alpha-1)+(\alpha y_1+\alpha-1,\,\alpha y_2+\alpha-1)\\ &=\alpha(x_1,\,x_2)+\alpha(y_1,\,y_2)\\ &=\alpha\vect{u}+\alpha\vect{v} \end{align*}

Property DSA:

\begin{align*} (\alpha&+\beta)\vect{u}\\ &=(\alpha+\beta)(x_1,\,x_2)\\ &=((\alpha+\beta)x_1+(\alpha+\beta)-1,\,(\alpha+\beta)x_2+(\alpha+\beta)-1)\\ &=(\alpha x_1+\beta x_1+\alpha+\beta-1,\,\alpha x_2+\beta x_2+\alpha+\beta-1)\\ &=(\alpha x_1+\alpha-1+\beta x_1+\beta-1+1,\,\alpha x_2+\alpha-1+\beta x_2+\beta-1+1)\\ &=((\alpha x_1+\alpha-1)+(\beta x_1+\beta-1)+1,\,(\alpha x_2+\alpha-1)+(\beta x_2+\beta-1)+1)\\ &=(\alpha x_1+\alpha-1,\,\alpha x_2+\alpha-1)+(\beta x_1+\beta-1,\,\beta x_2+\beta-1)\\ &=\alpha(x_1,\,x_2)+\beta(x_1,\,x_2)\\ &=\alpha\vect{u}+\beta\vect{u} \end{align*}

Property O: After all that, this one is easy, but no less pleasing.

\begin{equation*} 1\vect{u}=1(x_1,\,x_2)=(x_1+1-1,\,x_2+1-1)=(x_1,\,x_2)=\vect{u} \end{equation*}

That is 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.

Subsection VSP Vector Space Properties

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 ‚ÄúLet \(V\) be a vector space.‚ÄĚ From this we may assume the ten properties of Definition¬†VS, and nothing more. It is 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 ‚ÄĒ 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 Proof 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.

To prove uniqueness, a standard technique is to suppose the existence of two objects (Proof Technique U). So let \(\zerovector_1\) and \(\zerovector_2\) be two zero vectors in \(V\text{.}\) Then

\begin{align*} \zerovector_1 &=\zerovector_1+\zerovector_2&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\text{ for }\zerovector_2\\ &=\zerovector_2+\zerovector_1&& \knowl{./knowl/property-C.html}{\text{Property C}}\\ &=\zerovector_2&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\text{ for }\zerovector_1\text{.} \end{align*}

This proves the uniqueness since the two zero vectors are really the same.

To prove uniqueness, a standard technique is to suppose the existence of two objects (Proof Technique U). So let \(\vect{-u}_1\) and \(\vect{-u}_2\) be two additive inverses for \(\vect{u}\text{.}\) Then

\begin{align*} \vect{-u}_1&=\vect{-u}_1+\zerovector&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\vect{-u}_1+(\vect{u}+\vect{-u}_2)&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=(\vect{-u}_1+\vect{u})+\vect{-u}_2&& \knowl{./knowl/property-AA.html}{\text{Property AA}}\\ &=\zerovector+\vect{-u}_2&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=\vect{-u}_2&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\text{.} \end{align*}

So the two additive inverses are really the same.

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.

Notice that \(0\) is a scalar, \(\vect{u}\) is a vector, so Property SC says \(0\vect{u}\) is again a vector. As such, \(0\vect{u}\) has an additive inverse, \(-(0\vect{u})\) by Property AI. Then

\begin{align*} 0\vect{u} &=\zerovector+0\vect{u}&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\left(-(0\vect{u}) + 0\vect{u}\right)+0\vect{u}&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=-(0\vect{u}) + \left(0\vect{u}+0\vect{u}\right)&& \knowl{./knowl/property-AA.html}{\text{Property AA}}\\ &=-(0\vect{u}) + (0+0)\vect{u}&& \knowl{./knowl/property-DSA.html}{\text{Property DSA}}\\ &=-(0\vect{u}) + 0\vect{u}&& \knowl{./knowl/property-ZCN.html}{\text{Property ZCN}}\\ &=\zerovector&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\text{.} \end{align*}

Here is another theorem that looks like it should be obvious, but is still in need of a proof.

Notice that \(\alpha\) is a scalar, \(\zerovector\) is a vector, so Property SC means \(\alpha\zerovector\) is again a vector. As such, \(\alpha\zerovector\) has an additive inverse, \(-(\alpha\zerovector)\) by Property AI. Then

\begin{align*} \alpha\zerovector &=\zerovector+\alpha\zerovector&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\left(-(\alpha\zerovector)+\alpha\zerovector\right)+\alpha\zerovector&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=-(\alpha\zerovector)+\left(\alpha\zerovector+\alpha\zerovector\right)&& \knowl{./knowl/property-AA.html}{\text{Property AA}}\\ &=-(\alpha\zerovector)+\alpha\left(\zerovector+\zerovector\right)&& \knowl{./knowl/property-DVA.html}{\text{Property DVA}}\\ &=-(\alpha\zerovector)+\alpha\zerovector&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\zerovector&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\text{.} \end{align*}

Here is 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 used notation like \(\vect{u}^\sharp\) for the additive inverse of \(\vect{u}\text{.}\) Then we would have written the defining property, Property AI, as \(\vect{u}+\vect{u}^\sharp=\zerovector\text{.}\) This theorem would become \(\vect{u}^\sharp=(-1)\vect{u}\text{.}\) Not really quite as pretty, is it?

We have

\begin{align*} \vect{-u}&=\vect{-u}+\zerovector&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\vect{-u}+0\vect{u}&& \knowl{./knowl/theorem-ZSSM.html}{\text{Theorem ZSSM}}\\ &=\vect{-u}+\left(1+(-1)\right)\vect{u}&& \knowl{./knowl/property-AICN.html}{\text{Property AICN}}\\ &=\vect{-u}+\left(1\vect{u}+(-1)\vect{u}\right)&& \knowl{./knowl/property-DSA.html}{\text{Property DSA}}\\ &=\vect{-u}+\left(\vect{u}+(-1)\vect{u}\right)&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=\left(\vect{-u}+\vect{u}\right)+(-1)\vect{u}&& \knowl{./knowl/property-AA.html}{\text{Property AA}}\\ &=\zerovector+(-1)\vect{u}&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=(-1)\vect{u}&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\text{.} \end{align*}

See Exercise VS.T31 for a different approach to proving this theorem.

Because of this theorem, we can now write linear combinations like \(6\vect{u}_1+(-4)\vect{u}_2\) as \(6\vect{u}_1-4\vect{u}_2\text{,}\) 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 (‚Äúthe zero vector is unique‚ÄĚ) it is an implication (‚Äúif‚Ķ, then‚Ķ‚ÄĚ) and so can be used in proofs to convert a vector equality into two possibilities, one a scalar equality and the other a vector equality. It should remind you of the situation for complex numbers. If \(\alpha,\,\beta\in\complexes\) and \(\alpha\beta=0\text{,}\) then \(\alpha=0\) or \(\beta=0\text{.}\) This critical property is the driving force behind using a factorization to solve a polynomial equation.

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 \(\alpha=0\text{.}\) 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 \(\alpha\neq 0\text{.}\) Then

\begin{align*} \vect{u} &=1\vect{u}&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=\left(\frac{1}{\alpha}\alpha\right)\vect{u}&&\alpha\neq 0,\knowl{./knowl/property-MICN.html}{\text{Property MICN}}\\ &=\frac{1}{\alpha}\left(\alpha\vect{u}\right)&& \knowl{./knowl/property-SMA.html}{\text{Property SMA}}\\ &=\frac{1}{\alpha}\left(\zerovector\right)&&\text{Hypothesis}\\ &=\zerovector&& \knowl{./knowl/theorem-ZVSM.html}{\text{Theorem ZVSM}}\text{.} \end{align*}

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.

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 \(C\text{.}\) 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 \(C\text{.}\)

Suppose \(\vect{u}\in C\text{.}\) Then, as given by Theorem ZSSM,

\begin{equation*} 0\vect{u}=0(x_1,\,x_2)=(0x_1+0-1,\,0x_2+0-1)=(-1,-1)=\zerovector \end{equation*}

And as given by Theorem ZVSM,

\begin{align*} \alpha\zerovector &=\alpha(-1,\,-1)=(\alpha(-1)+\alpha-1,\,\alpha(-1)+\alpha-1)\\ &=(-\alpha+\alpha-1,\,-\alpha+\alpha-1)=(-1,\,-1)=\zerovector\text{.} \end{align*}

Finally, as given by Theorem AISM,

\begin{align*} (-1)\vect{u} &=(-1)(x_1,\,x_2)=((-1)x_1+(-1)-1,\,(-1)x_2+(-1)-1)\\ &=(-x_1-2,\,-x_2-2)=-\vect{u}\text{.} \end{align*}

Subsection RD Recycling Definitions

When we say that \(V\) is a vector space, we then know we have a set of objects (the ‚Äúvectors‚ÄĚ), but we also know we have been provided with two operations (‚Äúvector addition‚ÄĚ and ‚Äúscalar multiplication‚ÄĚ) 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 \(\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3\in V\) then an expression like

\begin{equation*} 5\vect{u}_1+7\vect{u}_2-13\vect{u}_3 \end{equation*}

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:

Reading Questions VS Reading Questions

2.

In the crazy vector space, \(C\text{,}\) (Example CVS) compute the linear combination

\begin{equation*} 2(3,\,4)+(-6)(1,\,2)\text{.} \end{equation*}

3.

Suppose that \(\alpha\) is a scalar and \(\zerovector\) is the zero vector. Why should we prove anything as obvious as \(\alpha\zerovector=\zerovector\) such as we did in Theorem ZVSM?

Exercises VS Exercises

M10.

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

\begin{equation*} \alpha\vect{x}=\zerovector=\colvector{0\\0},\qquad\alpha\in\complexes,\ \vect{x}\in\complex{2}\text{.} \end{equation*}

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.

M11.

Let \(V\) be the set \(\complex{2}\) with the usual vector addition, but with scalar multiplication defined by

\begin{equation*} \alpha \colvector{x\\y} = \colvector{\alpha y \\ \alpha x} \end{equation*}

Determine whether or not \(V\) is a vector space with these operations.

Solution.

The set \(\complex{2}\) with the proposed operations is not a vector space since Property O is not valid. A counterexample is \(1\colvector{3\\2 } = \colvector{2\\3}\ne\colvector{3\\2 }\text{,}\) so in general, \(1\vect{u}\ne\vect{u}\text{.}\)

M12.

Let \(V\) be the set \(\complex{2}\) with the usual scalar multiplication, but with vector addition defined by

\begin{equation*} \colvector{x\\y} + \colvector{z\\w} = \colvector{y + w \\ x + z} \end{equation*}

Determine whether or not \(V\) is a vector space with these operations.

Solution.

Let us consider the existence of a zero vector, as required by Property¬†Z of a vector space. The ‚Äúregular‚ÄĚ zero vector fails, (remember that the property must hold for every vector, not just for some),

\begin{equation*} \colvector{x\\y}+\colvector{0\\0} = \colvector{y\\x}\ne\colvector{x\\y}\text{.} \end{equation*}

Is there another vector that fills the role of the zero vector? Suppose that \(\zerovector=\colvector{z_1\\z_2}\text{.}\) Then for any vector \(\colvector{x\\y}\text{,}\) we have

\begin{align*} \colvector{x\\y}+\colvector{z_1 \\z_2} &= \colvector{y + z_2\\x + z_1} = \colvector{x\\y} \end{align*}

so that \(x = y + z_2\) and \(y = x + z_1\text{.}\) This means that \(z_1 = y - x\) and \(z_2 = x - y\text{.}\) However, since \(x\) and \(y\) can be any complex numbers, there are no fixed complex numbers \(z_1\) and \(z_2\) that satisfy these equations. Thus, there is no zero vector, Property Z does not hold, and the set \(\complex{2}\) with the proposed operations is not a vector space.

M13.

Let \(V\) be the set \(M_{22}\) with the usual scalar multiplication, but with addition defined by \(A + B = \zeromatrix_{2,2}\) for all \(2 \times 2\) matrices \(A\) and \(B\text{.}\) Determine whether or not \(V\) is a vector space with these operations.

Solution.

Since scalar multiplication remains unchanged, we only need to consider the axioms that involve vector addition. Since every sum is the zero matrix, the first 4 properties hold easily. However, there is no zero vector in this set. Suppose that there was. Then there is a matrix \(Z\) so that \(A + Z = A\) for any \(2\times 2\) matrix \(A\text{.}\) However, \(A + Z = \zeromatrix_{2,2}\text{,}\) which is in general not equal to \(A\text{,}\) so Property Z fails and this set is not a vector space.

M14.

Let \(V\) be the set \(M_{22}\) with the usual addition, but with scalar multiplication defined by \(\alpha A= \zeromatrix_{2,2}\) for all \(2 \times 2\) matrices \(A\) and scalars \(\alpha\text{.}\) Determine whether or not \(V\) is a vector space with these operations.

Solution.

Since addition is unchanged, we only need to check the axioms involving scalar multiplication. The proposed scalar multiplication clearly fails Property O : \(1A = \zeromatrix_{2,2}\ne A\text{.}\) Thus, the proposed set is not a vector space.

M15.

Consider the following sets of \(3 \times 3\) matrices, where the symbol \(*\) indicates the position of an arbitrary complex number. Determine whether or not these sets form vector spaces with the usual operations of addition and scalar multiplication for matrices.

  1. All matrices of the form \(\begin{bmatrix} * & * & 1\\ * & 1 & *\\ 1 & * & * \end{bmatrix}\)

  2. All matrices of the form \(\begin{bmatrix} * & 0 & *\\ 0 & * & 0\\ * & 0 & * \end{bmatrix}\)

  3. All matrices of the form \(\begin{bmatrix} * & 0 & 0\\ 0 & * & 0\\ 0 & 0 & * \end{bmatrix}\) (These are the diagonal matrices.)

  4. All matrices of the form \(\begin{bmatrix} * & * & *\\ 0 & * & *\\ 0 & 0 & * \end{bmatrix}\) (These are the upper triangular matrices.)

Solution.

There is something to notice here that will make our job much easier: Since each of these sets are comprised of \(3\times 3\) matrices with the standard operations of addition and scalar multiplication of matrices, the last 8 properties will automatically hold. That is, we really only need to verify Property AC and Property SC.

  • This set is not closed under either scalar multiplication or addition (fails Property¬†AC and Property¬†SC). For example,

    \begin{align*} 3\begin{bmatrix} *&*&1\\ *&1&*\\ 1&*&*\end{bmatrix} = \begin{bmatrix} *&*&3\\ *&3&*\\ 3&*&*\end{bmatrix} \end{align*}
    is not a member of the proposed set.

  • This set is closed under both scalar multiplication and addition, so this set is a vector space with the standard operation of addition and scalar multiplication.

  • This set is closed under both scalar multiplication and addition, so this set is a vector space with the standard operation of addition and scalar multiplication.

  • This set is closed under both scalar multiplication and addition, so this set is a vector space with the standard operation of addition and scalar multiplication.

M20.

Explain why we need to define the vector space \(P_n\) as the set of all polynomials with degree up to and including \(n\) instead of the more obvious set of all polynomials of degree exactly \(n\text{.}\)

Solution.

Hint: The set of all polynomials of degree exactly \(n\) fails one of the closure properties of a vector space. Which one, and why?

M21.

The set of integers is denoted \(\mathbb{Z}\text{.}\) Does the set \(\mathbb{Z}^2 = \setparts{\colvector{m\\n}}{m,n\in\mathbb{Z}}\) with the operations of standard addition and scalar multiplication of vectors form a vector space?

Solution.

Additive closure will hold, but scalar closure will not. The best way to convince yourself of this is to construct a counterexample. Such as, \(\frac{1}{2}\in\complexes\) and \(\colvector{1\\0}\in\mathbb{Z}^2\text{,}\) however \(\frac{1}{2}\colvector{1\\0}=\colvector{\frac{1}{2}\\0}\not\in\mathbb{Z}^2\text{,}\) which violates Property SC. So \(\mathbb{Z}^2\) is not a vector space.

Exercise Group.

The next three problems suggest that under the right situations we can ‚Äúcancel.‚ÄĚ 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 \(\vect{u},\,\vect{v},\,\vect{w}\in V\text{.}\) If \(\vect{w}+\vect{u}=\vect{w}+\vect{v}\text{,}\) then \(\vect{u}=\vect{v}\text{.}\)

Solution.
\begin{align*} \vect{u}&=\zerovector+\vect{u}&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\left(\vect{-w}+\vect{w}\right)+\vect{u}&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=\vect{-w}+\left(\vect{w}+\vect{u}\right)&& \knowl{./knowl/property-AA.html}{\text{Property AA}}\\ &=\vect{-w}+\left(\vect{w}+\vect{v}\right)&& \text{Hypothesis}\\ &=\left(\vect{-w}+\vect{w}\right)+\vect{v}&& \knowl{./knowl/property-AA.html}{\text{Property AA}}\\ &=\zerovector+\vect{v}&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=\vect{v}&& \knowl{./knowl/property-Z.html}{\text{Property Z}} \end{align*}
T22.

Suppose \(V\) is a vector space, \(\vect{u},\,\vect{v}\in V\) and \(\alpha\) is a nonzero scalar from \(\complexes\text{.}\) If \(\alpha\vect{u}=\alpha\vect{v}\text{,}\) then \(\vect{u}=\vect{v}\text{.}\)

Solution.
\begin{align*} \vect{u}&=1\vect{u}&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=\left(\frac{1}{\alpha}\alpha\right)\vect{u}&& \alpha\neq 0\\ &=\frac{1}{\alpha}\left(\alpha\vect{u}\right)&& \knowl{./knowl/property-SMA.html}{\text{Property SMA}}\\ &=\frac{1}{\alpha}\left(\alpha\vect{v}\right)&& \text{Hypothesis}\\ &=\left(\frac{1}{\alpha}\alpha\right)\vect{v}&& \knowl{./knowl/property-SMA.html}{\text{Property SMA}}\\ &=1\vect{v}\\ &=\vect{v}&& \knowl{./knowl/property-O.html}{\text{Property O}} \end{align*}
T23.

Suppose \(V\) is a vector space, \(\vect{u}\neq\zerovector\) is a vector in \(V\) and \(\alpha,\,\beta\in\complexes\text{.}\) If \(\alpha\vect{u}=\beta\vect{u}\text{,}\) then \(\alpha=\beta\text{.}\)

Solution.
\begin{align*} \zerovector&=\alpha\vect{u} + -\left(\alpha\vect{u}\right)&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=\beta\vect{u} + -\left(\alpha\vect{u}\right)&& \text{Hypothesis}\\ &=\beta\vect{u} + (-1)\left(\alpha\vect{u}\right)&& \knowl{./knowl/theorem-AISM.html}{\text{Theorem AISM}}\\ &=\beta\vect{u} + \left((-1)\alpha\right)\vect{u}&& \knowl{./knowl/property-SMA.html}{\text{Property SMA}}\\ &=\beta\vect{u} + \left(-\alpha\right)\vect{u}\\ &=\left(\beta-\alpha\right)\vect{u}&& \knowl{./knowl/property-DSA.html}{\text{Property DSA}}\\ \end{align*}

By hypothesis, \(\vect{u}\neq\zerovector\text{,}\) so Theorem SMEZV implies

\begin{align*} 0&=\beta-\alpha\\ \alpha&=\beta \end{align*}

T30.

Suppose that \(V\) is a vector space and \(\alpha\in\complexes\) is a scalar such that \(\alpha\vect{x}=\vect{x}\) for every \(\vect{x}\in V\text{.}\) Prove that \(\alpha = 1\text{.}\) In other words, Property¬†O is not duplicated for any other scalar but the ‚Äúspecial‚ÄĚ scalar, 1. (This question was suggested by James Gallagher.)

Solution.

We have,

\begin{align*} \zerovector &=\vect{x}-\vect{x}&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=\alpha\vect{x}-\vect{x} &&\text{Hypothesis}\\ &=\alpha\vect{x}-1\vect{x}&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=(\alpha-1)\vect{x}&& \knowl{./knowl/property-DSA.html}{\text{Property DSA}}\text{.} \end{align*}

So by Theorem SMEZV we conclude that \(\alpha-1=0\) or \(\vect{x}=\zerovector\text{.}\) However, since our hypothesis was for every \(\vect{x}\in V\text{,}\) we are left with the first possibility and \(\alpha=1\text{.}\)

There is one flaw in the proof above, and as stated, the problem is not correct either. Can you spot the flaw and as a result correct the problem statement? (Hint: Example VSS).

T31.

Construct an alternate proof of Theorem AISM by demonstrating that \((-1)\vect{x}\) is the additive inverse of \(\vect{x}\text{.}\)

Solution.

We have

\begin{align*} \vect{x}+(-1)\vect{x}&=1\vect{x}+(-1)\vect{x}&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=(1+(-1))\vect{x}&& \knowl{./knowl/property-DSA.html}{\text{Property DSA}}\\ &=0\vect{x}&& \knowl{./knowl/property-AICN.html}{\text{Property AICN}}\\ &=\zerovector&& \knowl{./knowl/theorem-ZSSM.html}{\text{Theorem ZSSM}}\text{.} \end{align*}

Then, in the spirit of the proof of Theorem SS, the vector \((-1)\vect{x}\) does what it needs to do to satisfy Property AI and therefore is an additive inverse of \(\vect{x}\) (and we know the inverse is unique by Theorem AIU).