Chapter 2

Linear Maps

2.1 Definition

We consider Rn to be the set of all ordered n-tuples of real numbers. We will refer to the elements of Rn as vectors,anddenotethembyv, w,etc.Wedonotassumeanyadditional geometric or algebraic structure to be associated with Rn.Forexample,R2 is the set of all ordered pairs (x, y),withx, y ∈ R.Thereisnoplane,noorigin,nocoordinateaxes, no lines, no angles, no dot product, no cross product, no vector addition, etc. For various reasons (which you can for now simply consider to be a personal chal- lenge), we will approach our subject from an entirely algebraic point of view, with no recourse to geometry. Functions of the form f (x)=mx have lots of nice algebraic properties. Because real number multiplication distributes over addition, we have that m(x + y)=mx + my,sothatf (x + y)=f (x)+ f (y).Becauserealnumbermultiplication is associative and commutative, we have that m(ax)=a(mx),sothatf (ax)=af(x).

Remark. The vast majority of functions do not have either of these properties. For example, if g(x)=x2,theng(ax)=(ax)2 = a2x2,whichformostvaluesofa is not equal (as a function) to ag(x)=ax2.Similarlywehavethatg(x + y)=(x + y)2 = x2 + 2xy + y2, which is not equal (as a function) to g(x)+g(y)=x2 + y2.

We therefore define a function L: Rm → Rn to be linear if it satisfies the following properties:

(i) L(v + w)=L(v)+L(w)

(ii) L(av)=aL(v)

where v and w are elements of Rm and a is any real number. Looking at this definition, we see that we need to be able to add elements of Rm, and we need to be able to multiply elements of Rm by real numbers. We define these operations as follows:

(i) (x1,...,xn)+(y1,...,yn)=(x1 + y1,...,xn + yn)

5 6 CHAPTER 2. LINEAR MAPS

(ii) a(x1,...,xn)=(ax1,...,axn) You may recognize this as vector addition and scalar multiplication from vector calculus, physics, or some other course. These are the only algebraic operations we will need for the purpose of studying linear maps. Because we are now adding elements of Rn to one another, the element (0, . . . , 0) gains a special importance. We let 0 denote this element, which we call the zero vector of Rn. One can define Cn just as we did Rn,asorderedn-tuples of complex numbers. We define linear maps L: Cm → Cn just as above, where vector addition and scalar multi- plication are also defined as above. Because we want most of ourdiscussionstoapply equally to both Rn and Cn (and other sets later), we will use the letters V, W,etc.,to n n denote one of R or C (and later others). The zero vector of V will be denoted 0V .The reason for the choice of the letter V is that we are thinking of Rn and Cn as vector spaces. Aformaldefinitionwillbegivenlater,butitsufficesfornowto say that we are thinking of Rn and Cn as vector spaces when we are thinking of them as sets whose elements we know how to add and how to multiply by scalars, while explicitly ignoring anything else we may know about the set (like the geometry of the set, concepts like length, angle, operations like cross products, etc.). Here are some examples of linear functions: Example. The map L: R3 → R2 which projects onto the xy-plane is linear. The equation for this map is L(x, y, z)=(x, y,0). Example. The map L: R2 → R2 which rotates about the origin by an angle θ is linear. The equation for this map is L(x, y)=(x cos θ − y sin θ, y cos θ + x sin θ). Example. The map L: R2 → R2 that scales the entire plane by a factor of λ ∈ R is linear. The equation for this map is L(x, y)=(λx, λy). Example. The map L: R2 → R2 with formula L(x, y)=(x + y, y) is linear. You should explore the geometry of this map. Example. The conjugation map L: C → C defined by L(a + bi)=a − bi is linear. Example. The map L: R2 → R2 that shifts everything by a fixed vector (a, b) is not linear (unless a = 0 = b). The equation for this map is L(x, y)=(x + a, y + b).Youcanseethat it is not linear by the fact that L(0, 0)=(a, b) $=(0, 0). Homework. Chapter 5: #9, 49. We will often use the term map for a function between vector spaces. The reason for this will be seen later. We will also use the term linear transformation,especiallywhen the domain and target space are the same set. Thus our purpose is study linear maps L: V → W between vector spaces V and W.ThesetsRn and Cn are not the only vector spaces, but they are the most important examples of such for our purposes, so they are all we will worry about until later. 2.2. BASIC PROPERTIES OF LINEAR MAPS 7

Remark. As mentioned earlier, we are now in the somewhat unfortunate situation that f (x)=mx + b is not alinearfunction!Thissituationisresolvedbygivingsuchfunctions anewname,andthengettingoverit.FunctionsoftheformA(x)=mx + b are affine (or ‘linear plus constant’). We will not have much use for these sorts of functions in this course.

Remark. We took as motivation for defining linear functions from R to R those func- tions of the form f (x)=mx.Certainlythesesatisfythedefiningpropertiesforlinear functions, but are these really the only functions R → R satisfying these properties? The answer is yes, as the following proposition shows.

Proposition 2.1.1. The function L: R → R is linear if and only if L(x)=mx for some m ∈ R.

Proof. Suppose that L is linear. If L(x)=0forallx ∈ R,thenL(x)=0x,andthetheorem is confirmed. Thus we may assume that there is some a ∈ R so that L(a)=b $= 0. Then by the second defining property of linearity we have L(0a)=0L(a)=0 $= b,sothat a $= 0. Then for all x ∈ R we have

x x x b L(x)=L a = L(a)= b = x, a a a a ! " so that L(x)=mx with m = b/a. For the other implication we have that L(x + y)=m(x + y)=mx + my = L(x)+L(y) and L(ax)=m(ax)=a(mx)=aL(x).

2.2 Basic Properties of Linear Maps

Linear maps are defined so as to send sums to sums and scalar multiples to scalar multiples. It is no surprise, then, that linear maps send sumsofscalarmultiplestosums of scalar multiples.

Proposition 2.2.1. Suppose L : V →W is a linear map between vector spaces V and W.Forany v1,...,vn in V and any scalars a1,...,an,wehavethat

L(a1v1 + ···+ anvn)=a1L(v1)+···+ an L(vn).

Expressions of the form a1v1 + ···+ anvn are called linear combinations.Thusthis proposition can be summarized as saying that linear maps sendlinearcombinationsto linear combinations. One special case of this deserves its own statement.

Proposition 2.2.2. Suppose L: V →W is a linear map. Then L(0V )=0W .

Proof. We compute: L(0V )=L(00V )=0L(0V )=0W . 8 CHAPTER 2. LINEAR MAPS

Homework. Chapter 5: #11, 50.

Note that we have made strong use of the fact that for any element x of a vector space V we have that 0x = 0V. This is obvious for the particular vector spaces we are considering (Rn and Cn), but it will be important later on to remember that we used this fact in proving this theorem.

2.2.1 One-to-One

One of the most basic things one can ask about a function or map is whether it is one-to- one or not. Recall that a function F : X → Y is called one-to-one if F(x1)=F(x2) implies that x1 = x2.Inotherwords,iftheoutputsarethesame,thentheinputsmust have been the same. Alternatively, one can say that F is one-to-one if different inputs necessarily produce different outputs. As an example, consider the function f : R→ R defined by f (x)=2x3 − 3/2x2 − 12x. This function is not one-to-one in general. For instance, the different inputs 3.5 and −1 produce the same output 7. On the other hand, the only input that produces 32 is 4. So the function is one-to-one at x = 4, but not one-to-one at 3.5 or −1. (In fact, one can easily determine by graphing that the function fails to be one-to-one for all values of x between −2.5 and 3.5, but is one-to-one at all other x-values.) It turns out that linear functions are simpler than this example.

Proposition 2.2.3. Suppose L: V → W is a linear map. If L(v1)=L(v2) for some v1 $= v2, then for every u there is some v $= u such that L(u)=L(v).InparticularL(z)=0W for some z $= 0V.

Proof. Suppose L(v1)=L(v2) for some v1 $= v2.Setz = v1 − v2,andnotethatsince v1 $= v2,wehavethatz $= 0V.ButthennotethatL(z)=L(v1 − v2)=L(v2) − L(v2)= 0W.Thusforanyu ∈ V we have that L(u + z)=L(u)+L(z)=L(u)+0W = L(u).

Note that we have used the fact that v + 0V = v for any v ∈ V.Again,thisisobvious for the vector spaces we’ve been considering, but it will be useful later on to remember that we used this fact. This proposition says that if L fails to be one-to-one anywhere, then it fails to be one-to-one everywhere. Thus to check if a linear map is one-to-one, we can choose to focus on trying to find non-zero solutions to the equation L(v)=0W.

Example. Consider the linear map L: R2 → R2 defined by L(x, y)=(2x − y, y − 2x).To determine if this map is one-to-one, we look at the equation L(x, y)=(0, 0).Inother words, we solve (2x − y, y − 2x)=(0, 0).Forthistobetruewemusthavethat2x − y = 0 and y − 2x = 0. Both of these equations hold as long as y = 2x.Thus,forinstance,we see that L(3, 6)=(0, 0).Thefunctionisnotone-to-one. 2.2. BASIC PROPERTIES OF LINEAR MAPS 9

Example. Consider the linear map L: R2 → R2 defined by L(x, y)=(2x + y, x + y).The equation L(x, y)=(0, 0) becomes the system of equations

2x + y = 0 #x + y = 0

The second equation says that y = −x.Substitutingthisintothefirstequationwefind that 2x +(−x)=0, so that x = 0. The second equation then says that y = 0, also. Thus the only solution to L(x, y)=(0, 0) is (0, 0).Itfollowsthatthismapisone-to-one.

We now observe that solutions to the ‘one-to-one’ equation have a particularly nice property.

Proposition 2.2.4. Suppose L: V → W is a linear map, and suppose L(v1)=0W and L(v2)= 0W.ThenL(av1 + bv2)=0W. Proof. We compute

L(av1 + bv2)=aL(v1)+bL(v2)=a0W + b0W = 0W.

Note that we have used the facts that a0V = 0V and 0V + 0V = 0V. It follows that solutions to the equation L(v)=0W are closed under linear combinations. This means that any linear combination of solutions is still asolution.SubsetsofV with this property are important enough to be given a special name.WesaythatasubsetW of a vector space V is a subspace if it is closed under linear combinations.

Example. Let W be the subset of R3 consisting of those vectors of the form (a, a, a).In other words, these are the vectors all of whose entries are equal. This is a subspace of R3.

Example. Let P be any plane through the origin in R3,andletW be the set of vectors (x, y, z) that lie in this plane. This is a subspace of R3.

Example. The zero vector 0V is a subspace of the vector space V. Example. Every vector space is a subspace of itself.

Homework. Chapter 4: #77.

Solutions to equations of the form L(v)=0W are important enough to warrant their own name also. We call the set of solutions to this equation the kernel of L.Withthis terminology, the preceeding proposition states simply thatthekernelofalinearmapis a subspace of its domain.

Example. The kernel of the projection map L(x, y, z)=(x, y,0) is the set of vectors of the form (0, 0, z) for any real number z. 10 CHAPTER 2. LINEAR MAPS

Example. Consider the map L: R3 → R3 defined by L(x, y, z)=(x − y + z,2x − 2y + 3z,3x − 3y + 4z).SolvingtheequationL(x, y, z)=(0, 0, 0) leads to the system of equa- tions x − y + z = 0 2x − 2y + 3z = 0 3x − 3y + 4z = 0

From the first equation, we see thatz = y − x.Substitutingthisintothesecondandthird equations we find that −x + y = 0 #−x + y = 0

This implies that x = y.Thusaslongaswechoose(x, y, z) so that x = y and z = y − x = 0, we will be in the kernel. Hence the kernel is the set of all vectors of the form (a, a,0).

Example. Consider the map L: R3 → R2 defined by L(x, y, z)=(x − y + z,0).Solving the equation L(x, y, z)=(0, 0) leads to the equation x − y + z = 0. Thus as long as we choose (x, y, z) to satisfy this equation, we will be in the kernel. Hence the kernel is the set of all vectors of the form (a, b, b − a).Notethatwecanalsowritethisasthesetofall vectors of the form av1 + bv2, where v1 =(1, 0, −1) and v2 =(0, 1, 1).

Homework. As in the last examples, describe the kernels of the followinglinearmaps: (a) L(x, y, z)=(x + y + z, x − y − z) (b) L(x, y, z, w)=(x − y + z + w,2x − 2y + 3z + 4w,3x − 3y + 4z + 5w)

In the last example above, we found that the kernel was equal tothesetofallex- pressions of the form av1 + bv2 for certain vectors v1 and v2.Inotherwords,thekernel turns out to be the set of all linear combinatinos of a pair of vectors. On the other hand, we also know that the kernel is a subspace. It turns out that these two facts are related. In order to see this, we need a definition. The span of a set of elements {v1,...,vn} in a vector space V is the set of all linear combinations of these elements. We have an immediate consequence.

Proposition 2.2.5. The span of any set of elements of a vector space V is a subspace of V.

Proof. PROOF.

Thus spans are subspaces. More to the point, many of the subspaces we will talk about will turn out to be the span of a finite set of vectors. So wecanreplacethe somewhat vague instructions, ‘describe the kernel’ by the more specific ‘find a spanning set.’