What exactly is a Euclidean vector space? Getting ahead of ourselves somewhat, it is a finite-dimensional vector space over the real numbers along with an inner product. We will eventually give precise meaning to all of these notions, starting in this section with vector spaces. We will also introduce what will be for us essentially the only example of a Euclidean vector space: namely, \(\R^n\) together with its standard vector operations and the dot product. This might strike the reader as a somewhat overly restricted focus: if there are other examples of Euclidean spaces out there, why not invite them in? As it turns out, an \(\R^n\)-centered treatment of linear algebra will give us plenty of rich material to deal with in this introductory course. Furthermore, although there are other “exotic” examples of Euclidean vector spaces, they are all structurally equivalent to \(\R^n\) (for some positive integer \(n\)) with its usual vector operations and the dot product. More precisely, getting ahead of ourselves once again, any two \(n\)-dimensional Euclidean vector spaces are isomorphic to one another. (We will make sense of this notion also, in due time.) As a result, the theory of linear algebra and inner product spaces, as articulated in the specific context of \(\R^n\text{,}\) can be seen as entirely characteristic of the general theory.
Given a positive integer \(n\text{,}\) we denote by \(\R^n\) the set of all \(n\)-tuples of real numbers. Tuples are treated in more detail in Section 0.3. In short, as defined in Definition 0.3.1, an \(n\)-tuple is an ordered sequence of elements. The defining property of tuples is that they are ordered objects, as opposed to unordered. This is captured by the notion of equality between tuples: we have
You have probably encountered \(\R^2\) and \(\R^3\) as models of planar and three-dimensional space. Perhaps you have even encountered \(\R^4\) as a model of relativistic space-time. In keeping with this tradition, we will call \(\R^n\)(real) \(n\)-space.
Let \(n\) be a positive integer. The set of all real \(n\)-tuples is called real \(n\)-space (or just \(n\)-space) and is denoted \(\R^n\text{:}\) i.e.,
Let \(n\) be a positive integer. We will call the operations below the standard vector operations on \(\R^n\text{.}\)
Vector addition on \(\R^n\).
Given elements \(n\)-tuples \(\boldx=(x_1,x_2,\dots, x_n)\) and \(\boldy=(y_1,y_2,\dots, y_n)\text{,}\) we define their vector sum \(\boldx+\boldy\) as
It is a good habit, when dealing with a variety of types of mathematical operations, to have in your mind a qualitative summary of what their inputs and outputs are. For example, vector addition in \(\R^n\) takes as input a pair of \(n\)-tuples, \(\boldx\) and \(\boldy\text{,}\) and returns as output the \(n\)-tuple \(\boldx+\boldy\text{.}\) By contrast, scalar multiplication in \(\R^n\) is a sort of hybrid operation that takes as input a real number \(c\) and \(n\)-tuple \(\boldx\) and returns as output a new \(n\)-tuple \(c\boldx\text{.}\)
The mores of conventional textbook writing dictate that we provide some examples of these vector operations on \(\R^n\text{,}\) as unenlightening as they may be. Instead, we give you an interactive SageCell that will allow you to create and play with examples on your own. The cells in Sage example 1 can be evaluated by clicking the Evaluate (Sage) button, or by typing shift+return. You can experiment by editing the code in these cells and then evaluating.
To create a vector in Sage, use the vector() command. The input should be a sequence of numbers enclosed in brackets. You can make use of sequence routines to create special sequences. If you prefer the two outputs above to not be listed as a pair, you can use the print() command in sequence. (This is a peculiarity of interactive SageCells, not Sage itself.) The standard vector operations of \(\R^n\) are implemented using an intuitive syntax in Sage. Once a vector v is created in Sage, various properties of the vector can be computed using the v.foo() syntax. For example, the command v.length() returns the length of the vector v.
We now jump to the heart of the matter, which is that \(\R^n\text{,}\) together with its standard vector operations, constitutes an example of an important type of mathematical object called a vector space.
There is an element \(\boldzero\in V\) such that for all \(\boldv\in V\text{,}\) we have \(\boldzero+\boldv=\boldv+\boldzero=\boldv
\text{.}\) We call \(\boldzero\) the zero vector of \(V\text{.}\)
For all \(\boldv\in V\text{,}\) there is another element \(-\boldv\) satisfying \(-\boldv+\boldv=\boldv+(-\boldv)=\boldzero
\text{.}\) We call \(-\boldv\) the vector inverse of \(\boldv\text{.}\)
It is possible to define the notion of a vector space over number systems other than the real numbers \(\R\text{.}\) For example, by replacing every instance of \(\R\) in Definition 1.1.4 with \(\C\text{,}\) we get the definition of a complex vector space. For our purposes, we will deal almost exclusively with real vector spaces, and accordingly will not use the ‘real’ modifier unless absolutely necessary.
It is essential to understand the very general nature of Definition 1.1.4. The definition does not specify what the underlying set \(V\) of the vector space is, or what the vector operations are. Rather, it allows for any set \(V\) and any choice of operations to be called a vector space, as long as our choices satisfy the vector space axioms. In this sense, the act of establishing a vector space consists of first making a sequence of declarations (“vector addition and scalar multiplication are defined like so”, “this element is the zero vector of our space”, “we define the vector inverse of elements like so”), and then verifying that these various choices satisfy the vector axioms. Procedure 1.1.6 provides a useful model for carrying out these steps.
Define the rule that assigns to each vector \(\boldv\in V\) its vector inverse \(-\boldv\) and verify that this definition of \(-\boldv\) satisfies the identity of Axiom iv.
Note how Procedure 1.1.6 highlights the special role played by Axioms iii–iv. These are sometimes called the existential axioms, as they posit the existence of certain special elements of \(V\text{:}\) a vector satisfying the identity of Axiom iii that is denoted \(\boldzero\text{,}\) and for all vectors \(\boldv\in V\text{,}\) a vector inverse denoted \(-\boldv\) that satisfies the identity of Axiom iv. Note also that our choice of vector inverses in Axiom iv depends on our choice of the zero vector \(\boldzero\) in Axiom iii, as this appears in the defining identity of Axiom iv.
Let’s apply Procedure 1.1.6 to verify that \(\R^n\text{,}\) together with the vector operations defined in Definition 1.1.2, constitutes a vector space.
The statement itself of the theorem has already taken care of some of the steps of Procedure 1.1.6: it has identified the underlying set \(\R^n\) and proposed vector operations (steps (1)-(2)), and it has identified the zero vector and the rule for computing vector inverses (steps (3)-(4)). It remains to show that the proposed zero vectors and vector inverses satisfy the identities of Axioms iii–iv, and that Axioms (i)-(ii) and (v)-(viii) are satisfied. We first consider Axioms iii–iv.
We claim that \((0,0,\dots, 0)\) satisfies the identity of Axiom iii, and thus that \(\boldzero=(0,0,\dots, 0)\text{.}\) Indeed, for all \(\boldx\in \R^n\) we have
We claim that given any \(\boldx=(x_1,x_2,\dots, x_n)\text{,}\) the vector \((-x_1,-x_2,\dots, -x_n)\) satisfies the identity of Axiom iv, and thus that \(-\boldx=(-x_1,-x_2,\dots, -x_n)\text{.}\) Indeed, we have
As for the remaining axioms, we will verify Axioms ii and vi, and leave the rest as an exercise. As you see below, the desired identities here all boil down to a familiar property of real number arithmetic: e.g., commutativity of real number addition, real number distributivity, etc. In what follows, \(\boldx=(x_1,x_2,\dots, x_n), \boldy=(y_1,y_2,\dots, y_n), \boldz=(z_1,z_2,\dots, z_n)\) will denote arbitrary elements of \(\R^n\text{,}\) and \(c,d\) will denote arbitrary elements of \(\R\text{.}\)
It should be noted that there are (infinitely) many different ways to define a vector space structure \(\R^n\text{.}\) From now on, however, we will assume without further comment that the vector operations on \(\R^n\) are the standard ones given in Definition 1.1.2. With respect to these standard operations the zero vector and vector inverses of \(\R^n\) are as described in Theorem 1.1.7. We make this official below.
Definition1.1.8.Vector space terminology for \(\R^n\).
Fix a positive integer \(n\text{.}\) When treating \(\R^n\) as a vector space, \(n\)-tuples \(\boldx\in \R^n\) are called \(n\)-vectors.The zero vector of \(\R^n\) is defined as \(\boldzero=(0,0,\dots, 0)\text{.}\) Given an \(n\)-vector \(\boldx=(x_1,x_2,\dots, x_n)\text{,}\) its vector inverse is the vector \(-\boldx\) defined as \(-\boldx=(-x_1,-x_2,\dots, -x_n)\text{.}\)
In other words, the vector inverse of \(\boldx\) is equal to the scalar multiple \((-1)\boldx\text{.}\) As it turns out, this is not particular to the specific vector space \(\R^n\text{,}\) but rather a general property of all vector spaces, as we see in Theorem 1.1.9. In order to prove this and other properties for a general vector space \(V\text{,}\) our arguments must rely only on the vector space axioms. In particular, we cannot assume that \(V=\R^n\) along with its standard vector operations, as this is but one example of a vector space.
Again, we emphasize that \(\R^n\) is just one example of a vector space: albeit, the example that we will mostly focus on. For good measure we include a few more examples of vector spaces here. (And we will also meet a few other examples later.) We begin with the simplest of all vector spaces, the zero space. Elementary as this example is, it serves well to illustrate the axiomatic nature of Definition 1.1.4.
Let \(V=\{\bot\}\text{,}\) a set containing exactly one element. There is a unique vector space structure that can be given to \(V\text{,}\) defined as follows.
Vector operations.
Vector addition on \(V\) is defined as \(\bot+\bot=\bot\text{;}\) scalar multiplication on \(V\) is defined as \(c\cdot \bot=\bot\) for all \(c\in \R\text{.}\)
Definition 1.1.10 makes two claims: that the given operations make \(V=\{\bot\}\) into a vector space, and that this is the only way to make \(V=\{\bot\}\) into a vector space. As with all claims in mathematics, these need to be proved, but as you will see, the proof is a very light affair.
Since \(V=\{\bot\}\) only has one item, there is no choice for what vector addition \(V\times V\rightarrow V\) and scalar multiplciation \(\R\times V\rightarrow V\) can be. They must be defined in the manner given in Definition 1.1.10. Similarly, we must have \(\boldzero=\bot\) and \(-\bot=\bot\text{,}\) as once again, \(\bot\) is the only element of \(V\text{!}\) This shows that there can be at most one way of giving \(V\) a vector space structure.
It is now easy to see that these choices do indeed satisfy the vector space axioms. That \(\bot\) satisfies the identity of Axiom iii defining the zero vector \(\boldzero\) follows from the fact that for all \(\boldv\in V\) we have \(\boldv=\bot\) (since \(V=\{\bot\}\)), and thus
Similarly, to show all elements of \(V\) have vector inverses amounts to showing that \(\bot\) has a vector inverse, since this is the only element of \(V\text{.}\) It is claimed that \(-\bot=\bot\) (i.e., \(\bot\) is its own vector inverse), which follows from the fact that
Lastly, the identities of Axioms i-ii and v-viii in this setting all reduce to trivial statements of the form \(\bot+\bot=\bot+\bot\text{.}\) Consider Axiom vii, for example. For all \(\boldv,\boldw\in V\text{,}\) we have \(\boldv=\boldw=\bot\text{,}\) in which case
We also include two “exotic” examples of vector spaces. We leave verification of the vector space axioms for these spaces as an exercise. (See Exercise 1.1.3.1 and Exercise 1.1.3.2.)
Vector addition and scalar multiplication of sequences is defined term-wise, exactly as with \(\R^n\text{.}\) In other words, given sequences \((a_i)_{i=1}^\infty\) and \((b_i)_{i=1}^\infty\text{,}\) and scalar \(c\in \R\text{,}\) we define
Define vector addition on \(\R_{> 0}\) to be real number multiplication, and define scalar multiplication on \(\R_{> 0}\) to be real number exponentiation: i.e., given vectors \(\boldv=r\) and \(\boldw=s\) in \(\R_{> 0}\text{,}\) and scalar \(c\in \R\text{,}\) we define
Before returning to \(\R^n\text{,}\) we introduce another important notion from general vector spaces: linear combinations. As simple as the idea of a linear combination is, you will see that it plays a crucial role in many concepts to come.
Let \(V\) be a vector space, and let \(\boldv_1,\boldv_2,\dots, \boldv_n\) be a collection of vectors of \(V\text{.}\) A linear combination of the \(\boldv_i\) is a vector expression of the form
where \(c_i\in \R\) for all \(1\leq i\leq n\text{.}\) The scalars \(c_i\) appearing in (1.1) are called the coefficients of the linear combination. The linear combination (1.1) is trivial if \(c_i=0\) for all \(1\leq i\leq n\text{,}\) and nontrivial if \(c_i\ne 0\) for some \(1\leq i\leq n\text{.}\)
Let \(\boldv_1=(1,0,0), \boldv_2=(0,1,0), \boldv_3=(0,0,1)\text{.}\) Show that every vector in \(\R^3\) is a linear combination of the \(\boldv_i\text{.}\)
It is natural to want to rewrite a linear combination of the form \(2\boldv_1+(-1)\boldv_2\) as \(2\boldv_1-\boldv_2\text{.}\) But of course this expression doesn’t quite make sense. What we are missing is the vector difference operation.
We will only explicitly visualize (or graph) elements of \(\R^n\) for \(n=2\) and \(n=3\text{.}\) However, these special cases bring to light an important point-vector duality in how we conceive of \(n\)-tuples that carries over into higher values of \(n\text{.}\) Fix \(n=3\) for now. We will sometimes conceive of a triple \((a,b,c)\in \R^3\) as a point, in which case we will use capital letters to denote the triple (e.g., \(P=(a,b,c)\)), and will represent the point visually with respect to a coordinate system as the point in \(3\)-space reached by starting at the origin \(O=(0,0,0)\) and moving a directed distance \(a\) units in the \(x\)-direction, \(b\) units in the \(y\)-direction and \(c\) units in the \(z\)-direction.
Alternatively, when conceiving of a triple \((a,b,c)\) as a vector, we will use lowercase bold letters to denote it (e.g., \(\boldx=(a,b,c)\) or \(\boldv=(a,b,c)\)), and represent it visually as a directed line segment (i.e., an arrow). In more detail, given the 3-vector \(\boldx=(a,b,c)\text{,}\) we choose an initial point \(Q=(x_0,y_0,z_0)\) and represent \(\boldx\) as the directed line segment \(\overrightarrow{QR}\) that starts at \(Q\) and ends at the point \(R=(x_0+a,y_0+b,z_0+c)\text{,}\) the terminal point of \(\overrightarrow{QR}\text{.}\) Note that in this manner we get infinitely-many different graphical representations of \(\boldx\text{,}\) one for each choice of starting point \(Q\text{.}\) Although these are technically different arrows (they have different starting points), we consider them to be equal as vectors. You can think of each particular choice of arrow-representation \(\boldx=\overrightarrow{QR}\) as an instance or incarnation of the vector \(\boldx=(a,b,c)\text{.}\) When the initial point of our arrow representation is chosen to be the origin \(O=(0,0,0)\text{,}\) we have \(\boldx=\overrightarrow{OP}\text{,}\) where \(P=(a,b,c)\text{.}\) We call \(\overrightarrow{OP}\) the position vector of the point \(P\text{.}\)
Figure1.1.18.Vector visualization of triple \(\boldx=\overrightarrow{OP}=\overrightarrow{QR}\text{.}\) Drag \(P\) to change the vector \(\boldx\text{.}\) Drag \(Q\) to change the initial point of \(\overrightarrow{QR}\text{.}\) Made with GeoGebra.
The representation of vectors as arrows gives rise to the so-called tip-to-tail interpretation of vector addition. Let \(\boldx=(a,b,c)\) and \(\boldy=(d,e,f)\text{.}\) Starting with an initial point \(P=(x_0,y_0,z_0)\text{,}\) we can represent \(\boldx\) as \(\overrightarrow{PQ}\text{,}\) where \(Q=(x_0+a,y_0+b,z_0+c)\text{,}\) and \(\boldy=\overrightarrow{QR}\text{,}\) where \(R=(x_0+a+d,y_0+b+e,z_0+c+f)\text{.}\) But then we have
In other words, if we choose our arrow representations so that the terminal point (the tip) of \(\boldx=\overrightarrow{PQ}\) is placed at the initial point (the tail) of \(\boldy=\overrightarrow{QR}\) , then \(\boldx+\boldy\) is represented by the arrow whose initial point is \(P\text{,}\) and whose terminal point is reached by first traveling along \(\boldx\text{,}\) and then traveling along \(\boldy\text{.}\)
The tip-to-tail visualization of vector addition gives rise to a similar conceptualization of vector difference. Performing a little vector algebra on the definition
Using tip-to-tail terminology, this means if we represent \(\boldx=\overrightarrow{PQ}\) and \(\boldy=\overrightarrow{PR}\text{,}\) then \(\boldy-\boldx\) is the arrow \(\overrightarrow{QR}\) that starts at the tip of \(\boldx\) and ends at the tip of \(\boldy\text{.}\) We thus have a tip-to-tip description of vector difference.
Next consider scalar multiplication. Given a vector \(\boldx=(a,b,c)=\overrightarrow{PQ}\) and a scalar \(k\in \R\text{,}\) the scalar multiple \(k\boldx=(ka,kb,kc)\) can be represented as an arrow that starts at \(P\) and points along the line containing \(\overrightarrow{PQ}\text{.}\) As we will see in the next section, the length of the resulting arrow is multiplied by the factor \(\abs{k}\text{,}\) resulting in a stretched arrow if \(\abs{k}> 1\) and a shrunk arrow if \(\abs{k}< 1\text{.}\) Furthermore, if \(k\geq 0\text{,}\) then the arrow representing \(k\boldx\) points in the same direction as \(\overline{PQ}\text{;}\) if \(k< 0\text{,}\) it points in the opposite direction.
Show that the given set together with the given operations constitutes a vector space by verifying that the vector axioms hold. Follow Procedure 1.1.6. In particular, explicitly identify what element of the set acts as the zero vector \(\boldzero\text{,}\) and how vector inverses of elements are defined.
Define vector addition on \(\R_{> 0}\) to be real number multiplication, and define scalar multiplication on \(\R_{> 0}\) to be real number exponentiation: i.e., given vectors \(\boldv=r\) and \(\boldw=s\) in \(\R_{> 0}\text{,}\) and scalar \(c\in \R\text{,}\) we define
Prove statements (2)-(4) of Theorem 1.1.9. When treating a specific part you may assume the results of any part that has already been proven, including statement (1).
Show that the zero vector of \(V\) is unique: i.e., show that if \(\boldw\in V\) satisfies \(\boldw+\boldv=\boldv\) for all \(\boldv\in V\text{,}\) then \(\boldw=\boldzero\text{.}\)
Fix \(\boldv\in V\text{.}\) Show that the vector inverse of \(\boldv\) is unique: i.e., show that if \(\boldw+\boldv=\boldzero\text{,}\) then \(\boldw=-\boldv\text{.}\)
Let \(V\) be a vector space. Prove that either \(V=\{\boldzero\}\) (i.e., \(V\) is the zero space) or \(V\) is infinite. In other words, a vector space contains either exactly one element or infinitely many elements.
Assume \(V\) contains a nonzero vector \(\boldv\ne\boldzero\text{.}\) Show that if \(c\ne d\text{,}\) then \(c\boldv\ne d\boldv\text{.}\) You assume the results of Theorem 1.1.9.
