Chapter 4 Isomorphism and Coordinates

Chapter 4

Isomorphism and Coordinates

Recall that a vector space isomorphism is a linear map that is both one-to-one and onto. Such a map preserves every aspect of the vector space structure. In other words, if L: V → W is an isomorphism, then any true statement you can say about V using abstract vector notation, vector addition, and scalar multiplication, willtransfertoatruestatement about W when L is applied to the entire statement. We make this more precise with some examples.

Example. If L: V → W is an isomorphism, then the set {v1,...,vn} is linearly independent in V if and only if the set {L(v1),...,L(vn)} is linearly independent in W.The dimension of the subspace spanned by the ﬁrst set equals the dimension of the subset spanned by the second set. In particular, the dimension of V equals that of W.

This last statement about dimension is only one part of a more fundamental fact.

Theorem 4.0.1. Suppose V is a ﬁnite-dimensional vector space. Then V is isomorphic to W if and only if dim V = dim W.

Proof. Suppose that V and W are isomorphic, and let L: V → W be an isomorphism. Then L is one-to-one, so dim ker L = 0. Since L is onto, we also have dim imL = dim W. Plugging these into the rank-nullity theorem for L shows then that dim V = dim W. Now suppose that dim V = dim W = n,andchoosebases{v1,...,vn} and {w1,...,wn} for V and W,respectively.Foranyvectorv in V,wewritev = a1v1 + ···+ anvn,and deﬁne L(v)=L(a1v1 + ···+ anvn)=a1w1 + ···+ anwn. We claim that L is linear, one-to-one, and onto. (Proof omitted.)

In particular, and 2–dimensional real vector space is necessarily isomorphic to R2,for example. This helps to explain why so many problems in these other spaces ended up reducing to solving systems of equations just like those we saw in Rn. Looking at the proof, we see that isomorphisms are constructed by sending bases to bases. In particular, there is a different isomorphism V → W for each choice of basis for V and for W.

27 28 CHAPTER 4. ISOMORPHISM AND COORDINATES

One special case of this is when we look at isomorphisms V → V.Suchanisomor- phism is called a change of coordinates. If S = {v1,...,vn} is a basis for V,wesaythen-tuple (a1,...,an) is the coordinate vector of v with respect to S if v = a1v + ···+ anvn.Wedenotethisvectoras[v]S. Example. Find the coordinates for (1, 3) with respect to the basis S = {(1, 1), (−1, 1)}. We set (1, 3)=a(1, 1)+b(−1, 1),whichleadstotheequationsa − b = 1 and a + b = 3. This system has solution a = 2, b = 1. Thus (1, 3)=2(1, 1)+1(−1, 1), so that [(1, 3)]S = (2, 1).

Example. Find the coordinates for t2 + 3t + 2withrespecttothebasisS = {t2 + 1, t + 1, t − 1}.Wesett2 + 3t + 2 = a(t2 + 1)+b(t + 1)+c(t − 1).Collectingliketermsgives t2 + 3t + 2 = at2 +(b + c)t +(a + b − c).Thisleadstothesystemofequations

a = 1 b + c = 3 a + b − c = 2  The solution is a = 1, b = 2, c = 1. Thus we have t2 + 3t + 2 = 1(t2 + 1)+2(t + 1)+ 2 1(t − 1),sothat[t + 3t + 2]S =(1, 2, 1). Note that for any vector v in an n–dimensional vector space V and for any basis S for n V, the coordinate vector [v]S is an element of R . Proposition 4.0.2. For any basis S for an n–dimensional vector space V,thecorrespondence n v #→ [v]S is an isomorphism from V to R . Corollary 4.0.3. Every n–dimensional vector space over a R is isomorphic to Rn. Chapter 5

Linear Maps Rn → Rm

Since every ﬁnite-dimensional vector space over R is isomorphic to Rn, any problem we have in such a vector space that can be expressed entirely in terms of vector operations can be tranferred to one in Rn.Sinceourultimategoalistounderstandlinearmaps V → W,wewillfocusoureffortsonunderstandinglinearmapsRn → Rm,without worrying about expressing things in abstract terms. Remark. Unlike any previous section, we focus speciﬁcally on Rn in this chapter. To emphasize the distinction, we use x to denote an arbitrary vector in Rn.

5.1 Linear maps from Rn to R

We’ve already seen above that the linear maps R → R are precisely those of the form L(x)=ax for some real number a.Forthenextstep,weallowourdomaintohave multiple dimensions, but insist that our target space be R.Wewilldiscoverthatlinear maps L: Rn → R are already familiar to us. Theorem 5.1.1. If L: Rn → R is a linear map, then there is some vector m such that L(x)=a · x. n Proof. For j = 1, . . . , n,wesetej equal to the jth standard basis vector in R .Seta = (a1,...,an),whereeachaj = L(ej),andconsideranarbitraryvectorx =(x1,...,xn) in Rn.Wecompute

L(x)=L(x1e1 + ···+ xnen)=x1L(e1)+···+ xnL(en)=x1a1 + ···+ xnan = x · a.

Remark. Wait, didn’t we say that we weren’t going to think about dot products? Then we would be studying inner product spaces rather than vector spaces! Yes, and that’s still true. Within a given vector space, we will not be performing any dot products, and so in particular will never speak of length or angle. And infactourdeﬁnitionof linear map did not use the notion of dot product; it used only vector addition and scalar multiplication. What we’ve shown is that every linear map from Rn to R has the form

f (x1, x2,...,xn)=a1x1 + ···+ anxn

29 30 CHAPTER 5. LINEAR MAPS RN → RM

for some ﬁxed real numbers a1,...,an.Itjustsohappensthatwehaveanameforthis type of operation, and we call it the dot product, but this is just a convenient way to explain what linear maps do; we’re not studying the algebraic or geometric properties of the dot product in Rn.

5.2 Linear Maps Rn → Rm

One of the ﬁrst things you learn in vector calculus is that functions with multiple outputs can be thought of as a list of functions with one output. Thus given an arbitrary func- 2 3 tion f : R → R ,say,wethinkofitasf (x, y)=(f1(x, y), f2(x, y), f3(x, y)),whereeach 2 1 component function fj is a map R → R .Wethusexpecttoﬁndthatlinearmapsfrom Rn to Rm are those whose component functions are linear maps from Rn to R, which we saw in the last section are just dot products. This is the content of the following.

Theorem 5.2.1. The function L: Rn → Rm is linear if and only if each component function n Lj : R → R is linear.

Proof. Omitted.

Thus any linear map Rn → Rm is built up from a bunch of dot products in each component. In the next section we will make use of this fact to come up with a nice way to present linear maps.

5.3 Matrices

There are many ways to write vectors in Rn.Forexample,thesamevectorinR3 can be represented as

3 3i + 2j − 4k, $3, 2, −4%, (3, 2, −4), [3, 2, −4], 2 .   −4   We will focus on these last two for the time being. In particular, whenever we have a dot product x · y of two vectors x and y (in that order), we will write the ﬁrst as a row in square brackets and the second as a column in square brackets. Thus we have, for example, 2 123 3 = 2 + 6 − 12 = −4.   −4 ) *   Note that we are also avoiding commas in the row vector. 5.3. MATRICES 31

Now suppose L is an arbitrary linear map from Rn to R.Thengiveninputvectorx, L(x) is the dot product a · x for some ﬁxed vector a.Thuswemaywrite

x1 x1 x x  2  2 L . = a1 a2 ··· an . . . .     x  ) * x   n  n     Now suppose L is a linear map from Rn to Rm,andtheith component functions is the dot product with ai.Thewecanwrite

x1 a11 a12 ··· a1n x1 a1 · x x2 a21 a22 ··· a2n x2 a2 · x L  .  =  . . .   .  =  .  . . . . ··· . . .         x  a a ··· a  x  a · x  n  m1 m2 mn  n  m          Thus we can think of any linear map from Rn to Rm as multiplication by a matrix, assuming we deﬁne multiplication in exactly this way.

Deﬁnition 5.3.1. If A =(aij) is an m × n matrix and x is an n × 1columnvector,the product Ax is deﬁned to be the m × 1 column vector whose ith entry is the dot product of the ith row of A with x.

Thus we are led to the fortuitous observation that every linear map L: Rn → Rm has the form L(x)=Ax for some m × n matrix A.ThuslinearmapsfromR to itself are just multiplication by a 1 × 1matrix;i.e.,multiplicationbyaconstant.Thisagreeswith what we saw earlier. We now note an important fact about compositions of linear maps.

Theorem 5.3.2. Suppose L: Rn → Rm and T : Rm → Rp are linear maps. Then the composition T ◦ L: Rn → Rp is a linear map.

Suppose L is represented by the m × n matrix A and T is represented by the p × m matrix B.BecauseT ◦ L is also linear, it is represented by some p × n matrix C.Wenow show how to construct C from A and B. We begin with a motivating example. Suppose L maps from R2 to R2,asdoesT,and suppose L dots with a =(a1, a2, ) and b =(b1, b2) while T dots with c =(c1, c2) and d =(d1, d2).Then

a · x a x + a x c (a x + a x )+c (b x + b x ) T ◦ L(x)=T = T 1 1 2 2 = 1 1 1 2 2 2 1 1 2 2 b · x b x + b x d (a x + a x )+d (b x + b x ) 34 56 34 1 1 2 256 4 1 1 1 2 2 2 1 1 2 2 5 c a + c b c a + c b x = 1 1 2 1 1 2 2 2 1 d a + d b d a + d b x 4 1 1 2 1 1 2 2 254 25 32 CHAPTER 5. LINEAR MAPS RN → RM

a a c c Thus if L is multiplication by A = 1 2 and T is multiplication by B = 1 2 , b b d d 4 1 25 4 1 25 then T ◦ L is multiplication by C =(cij),wherecij is the dot product of the ith row of B with the jth row of A.Inotherwords,wehave

c c a a c a + c b c a + c b 1 2 1 2 = 1 1 2 1 1 2 2 2 . d d b b d a + d b d a + d b 4 1 254 1 25 4 1 1 2 1 1 2 2 25 This may seem a strange way to deﬁne the product of two matrices, but since we’re thinking of matrices as representing linear maps, it only makes sense that the product of two should be the matrix of the composition, so the deﬁnition is essentially forced upon us.

Remark. According to this deﬁnition, we cannot just multiply any two matrices. Their sizes have to match up in a nice way. In particular, for the dot products to make sense in computing AB,therowsofA have to have just as many elements as the columns of B. In short, the product AB is deﬁned as long as A is m × p and B is p × n,inwhichcase the product is m × n.

Proposition 5.3.3. Matrix multiplication is associative when it is deﬁned. In other words, for any matrices A, B, and C we have A(BC)=(AB)C, as long as all the individual products in this identity are deﬁned.

Proof. It is straightforward, though incredibly tedious, to prove this directly using our algebraic deﬁnition of matrix multiplication. What is far easier, however, is simply to note that function composition is always associative, when it’s deﬁned. The result follows.

There are some particularly special linear maps: the zero mapandtheidentity.Itis not to hard to see that the zero map Rn → Rm can be represented as multiplication by the n m zero matrix 0m×n.TheidentitymapR → R is represented by the aptly named identity matrix Im×n,whichhas1sonitsmaindiagonaland0selsewhere.Notethatit follows that IA = AI = A for approriately sized I,whileA0 = 0A = 0,forappropriatelysized 0.