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

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1 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, will transfertoatruestatement 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 {v 1,...,v n } is linearly independent in V if and only if the set {L(v 1 ),...,L(v n )} 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 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{v 1,...,v n } and {w 1,...,w n } for V and W, respectively. Foranyvectorv in V, wewritev = a 1 v a n v n,and define L(v) =L(a 1 v a n v n )=a 1 w a n w n. 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 R 2,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 R n. 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

2 28 CHAPTER 4. ISOMORPHISM AND COORDINATES One special case of this is when we look at isomorphisms V V. Suchanisomorphism is called a change of coordinates. If S = {v 1,...,v n } is a basis for V, wesaythen-tuple (a 1,...,a n ) is the coordinate vector of v with respect to S if v = a 1 v + + a n v n.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 t 2 + 3t + 2withrespecttothebasisS = {t 2 + 1, t + 1, t 1}. Wesett 2 + 3t + 2 = a(t 2 + 1)+b(t + 1)+c(t 1). Collectingliketermsgives t 2 + 3t + 2 = at 2 +(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 t 2 + 3t + 2 = 1(t 2 + 1)+2(t + 1)+ 1(t 1), sothat[t 2 + 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 V, the coordinate vector [v] S is an element of R n. Proposition For any basis S for an n dimensional vector space V, thecorrespondence v [v] S is an isomorphism from V to R n. Corollary Every n dimensional vector space over a R is isomorphic to R n.

3 Chapter 5 Linear Maps R n R m Since every finite-dimensional vector space over R is isomorphic to R n, 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 R n. Since our ultimate goal is to understand linear maps V W, wewillfocusoureffortsonunderstandinglinearmapsr n R m,without worrying about expressing things in abstract terms. Remark. Unlike any previous section, we focus specifically on R n in this chapter. To emphasize the distinction, we use x to denote an arbitrary vector in R n. 5.1 Linear maps from R n 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. For the next step, we allow our domain to have multiple dimensions, but insist that our target space be R. Wewilldiscoverthatlinear maps L : R n R are already familiar to us. Theorem If L : R n R is a linear map, then there is some vector m such that L(x) =a x. Proof. For j = 1,..., n, wesete j equal to the jth standard basis vector in R n. Set a = (a 1,...,a n ),whereeacha j = L(e j ),andconsideranarbitraryvectorx =(x 1,...,x n ) in R n.wecompute L(x) =L(x 1 e x n e n )=x 1 L(e 1 )+ + x n L(e n )=x 1 a x n a n = 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 in factourdefinitionof 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 R n to R has the form f (x 1,,...,x n )=a 1 x a n x n 29

4 30 CHAPTER 5. LINEAR MAPS R N R M for some fixed real numbers a 1,...,a n. It just so happens that we have a name for this 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 R n. 5.2 Linear Maps R n R m 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 function f : R 2 R 3,say,wethinkofitas f (x, y) =(f 1 (x, y), f 2 (x, y), f 3 (x, y)), whereeach component function f j is a map R 2 R 1. We thus expect to find that linear maps from R n to R m are those whose component functions are linear maps from R n to R, which we saw in the last section are just dot products. This is the content of the following. Theorem The function L: R n R m is linear if and only if each component function L j : R n R is linear. Proof. Omitted. Thus any linear map R n R m 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 R n.forexample,thesamevectorinr 3 can be represented as 3i + 2j 4k, 3, 2, 4, (3, 2, 4), [3, 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, [ 1 2 ] = = 4. Note that we are also avoiding commas in the row vector.

5 5.3. MATRICES 31 Now suppose L is an arbitrary linear map from R n to R. Thengiveninputvectorx, L(x) is the dot product a x for some fixed vector a. Thuswemaywrite x 1 x n x 1 L. = [ ] a 1 a 2 a n.. Now suppose L is a linear map from R n to R m,andtheith component functions is the dot product with a i.thewecanwrite x 1 a 11 a 12 a 1n x 1 a 1 x L. = a 21 a 22 a 2n.... = a 2 x.. x n a m1 a m2 a mn x n a m x Thus we can think of any linear map from R n to R m as multiplication by a matrix, assuming we define multiplication in exactly this way. Definition If A =(a ij ) 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 : R n R m 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 Suppose L : R n R m and T : R m R p are linear maps. Then the composition T L : R n R p 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 R 2 to R 2,asdoesT, and suppose L dots with a =(a 1, a 2, ) and b =(b 1, b 2 ) while T dots with c =(c 1, c 2 ) and d =(d 1, d 2 ).Then T L(x) =T ([ ]) a x = T b x ([ ]) a1 x 1 + a 2 = b 1 x 1 + b 2 [ c1 a = 1 + c 2 b 1 c 1 a 2 + c 2 b 2 d 1 a 1 + d 2 b 1 d 1 a 2 + d 2 b 2 x n [ ] c1 (a 1 x 1 + a 2 )+c 2 (b 1 x 1 + b 2 ) d 1 (a 1 x 1 + a 2 )+d 2 (b 1 x 1 + b 2 ) ][ x1 ]

6 32 CHAPTER 5. LINEAR MAPS R N R M [ ] [ ] a1 a Thus if L is multiplication by A = 2 c1 c and T is multiplication by B = 2, b 1 b 2 d 1 d 2 then T L is multiplication by C =(c ij ),wherec ij is the dot product of the ith row of B with the jth row of A. Inotherwords,wehave [ ][ ] [ ] c1 c 2 a1 a 2 c1 a = 1 + c 2 b 1 c 1 a 2 + c 2 b 2. d 1 d 2 b 1 b 2 d 1 a 1 + d 2 b 1 d 1 a 2 + d 2 b 2 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, the rows of A 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 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 map and the identity. It is not to hard to see that the zero map R n R m can be represented as multiplication by the zero matrix 0 m n.theidentitymapr n R m is represented by the aptly named identity matrix I m n,whichhas1sonitsmaindiagonaland0selsewhere. Notethatit follows that IA = AI = A for approriately sized I, whilea0 = 0A = 0, forappropriatelysized 0.