L E C T U R E 2 1 : M AT R I X T R A N S F O R M AT I O N S. Monday, November 14

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1 L E C T U R E 2 1 : M AT R I X T R A N S F O R M AT I O N S Monday, November 14 In this lecture we want to consider functions between vector spaces. Recall that a function is simply a rule that takes an input from a domain set and outputs exactly one object from a range set. In general, a function does not take into account any algebraic structure that a set might have, it just maps sets to sets. So rather than simply considering functions from one vector space to another, we want to consider functions that preserve the algebraic structure of our vector spaces. We call such functions linear transformations. These functions show up in many areas of mathematics. For example, the operations of differentiation and integration that are studied in calculus are linear transformations. In probability, a set of real-valued independent random variables dened on a particular probability space is a vector space, and one often considers linear transformations on this space. 1 linear transformations Let s begin by dening what a linear transformation between vector spaces is. When reading this, keep in mind the intuitive denition of a function between vector spaces that preserves the algebraic structure of the vector space. Denition 1.1. A linear transformation from a vector space V to a vector space W is a function T : V Ñ W that satises the following properties: piq Tpx ` yq Tpxq ` Tpyq for all x, y P V. piiq Tpaxq atpxq for all x P V and a P R. Now we see what was meant by a function that preserves the algebraic structure of the vector space. Conditions piq and piiq tell us that the image (or range) of T, which is just TpVq, is also a vector space. Let s look at some examples to get a better feel for this denition. 1

2 Example 1. We dene the zero transformation T 0 : V Ñ W to be the function taking every vector in V to the zero vector in W. That is, T 0 pxq 0 for all x P V. To see that this is a linear transformation, let x, y P V and let a P R. Then we have Tpx ` yq 0 0 ` 0 Tpxq ` Tpyq Tpaxq 0 a 0 atpxq. Hence, T is a linear transformation. Note that in order to prove that a function is a linear transformation we can check both conditions - that the function respects addition and scalar multiplication - at the same time. Specically, given a function T : V Ñ W between vector spaces V and W, in order to verify that T is a linear transformation we just need to show Tpax ` yq atpxq ` Tpyq for all x, y P V and a P R. This is somewhat of a trivial observation, but it will make checking if a function is a linear transformation a little less cumbersome. Example 2. Another simple transformation is the identity transformation I V : V Ñ V to be the function that takes every vector in V to itself: I V pxq x for all v P V. To see that this is a linear transformation, let x, y P V and let a P R. Then we have Tpax ` yq ax ` y atpxq ` Tpyq. Hence, T is a linear transformation. Example 3. We can consider the vectors in R 2 as points on the Cartesian plane (i.e. the x-y plane). Then reflection about x-axis is a linear transformation. Specically T : R 2 Ñ R 2 given by Tpx, yq px, yq. 2

3 To see that this is a linear transformation, let px, yq, px 1, y 1 q P R 2 and let a P R. Then we have Tpapx, yq ` px 1, y 1 qq Tpax ` x 1, ay ` y 1 q Hence, T is a linear transformation. pax ` x 1, ay ` y 1 q pax ` x 1, ay y 1 q pax, ayq ` px 1, y 1 q apx, yq ` px 1, y 1 q atpx, yq ` Tpx 1, y 1 q. Here are several more general examples. You should try to quickly verify that each is a linear transformation. Example 4. Let n ě 2 be an integer. Then we have a projection linear transformation T : R n Ñ R n 1 dened by Tpa 1,..., a n q pa 1,..., a n 1 q. Example 5. Let PpRq denote the vector space consisting of all polynomials with coefcients in R. Then differentiation gives us a linear transformation T : PpRq Ñ PpRq dened by Tpppxqq p 1 pxq. Example 6. You might remember from your calculus class that the integral of a sum of two functions is the sum of their integrals, and the integral of a constant times a function is the constant times the integral of the function. This is saying that integration is a linear transformation on the vector space of real-valued integrable functions. For example, for any two real numbers a ď b we have the linear transformation T a,b : PpRq Ñ R dened by T a,b pppxqq ż b a ppxq dx. Note that if a b this is the zero transformation between PpRq and R. Example 7. Let R 8 tpa 1, a 2, a 3,... q : a i P Ru. We dene the backward shift linear transformation T : R 8 Ñ R 8 by Tppa 1, a 2, a 3, a 4,... qq pa 2, a 3, a 4,... q 3

4 We also have the forward shift linear transformation T : R 8 Ñ R 8 dened by Tppa 1, a 2, a 3,... qq p0, a 1, a 2, a 3,... q. Example 8. Let n be a positive integer and let M n prq be the vector space of n ˆ n matrices with coefcients in R. Then transposition is a linear transformation, i.e. T : M n prq Ñ M n prq dened by TpMq M T. 2 matrix representation of a linear transformation It turns out that any linear transformation T : V Ñ W between vector spaces can be represented by a matrix. In this class we re primarily interested in the vector spaces R n, so we ll focus on this case. Suppose we have a linear transformation T : R n Ñ R m and let te 1,..., e n u denote the standard ordered basis for R n (i.e. e i is the column vector in R n with a 1 in the i th position and zeros elsewhere). Let v P R n with and note that This means that v a 1. a n v a 1. a n f fl, f fl a 1 e 1 ` ` a n e n. Tpvq Tpa 1 e 1 ` ` a n e n q Let s record what we ve just done. a 1 Tpe 1 q ` ` a n Tpe n q a 1 Tpe 1 q Tpe n qfl f. fl. a n 4

5 Denition 2.1. Let T : R n Ñ R m a linear transformation of vector spaces and let te 1,..., e n u denote the standard ordered basis for R n. Then Tpvq Av, where A is the m ˆ n matrix given by A Tpe 1 q Tpe 2 q Tpe n qfl. Let s look at some examples. Example 9. Consider the linear transformation T : R 2 Ñ R 3 dened by Tpx, yq px ` y, 0, x yq. The standard ordered basis for R 2 is tp1, 0q, p0, 1qu and we ahve Tp1, 0q p1, 0, 1q Tp0, 1q p1, 0, 1q, Hence the matrix representation of T is A Tp1, 0q Tp0, 1qfl fl. 1 1 Let s see how this works for an arbitrary vector on R 2. Consider p2, 3q P R 2. Then Tp2, 3q p5, 0, 1q. Let s see if we get the same vector when we use the matrix representation of T. We have 1 1 j fl 2 0 fl p5, 0, 1q It works! Notice that the matrix representation is unique. That is, there cannot be two different matrix representations for the same linear transformation. Furthermore, any linear transformation T : R n Ñ R m is completely determined 5

6 Theorem 2.2. For each set of n vectors from R m, say ty 1,..., y n u, there exists exactly one linear transformation T : R n Ñ R m satisfying Tpe i q y i for all i. Proof. In order for T to be linear and satisfy the condition that Tpe i q y i for all i, we know that T must dened it as follows: for any x P R n with x a 1 e 1 ` ` a n e n we have Tpxq Tpa 1 e 1 ` ` a n e n q a 1 Tpe 1 q ` ` a n Tpe n q a 1 y 1 ` ` a n y n. Suppose that T : R n Ñ R m is another linear transformation satisfying Tpe i q y i for all i. We want to show that Tpxq Tpxq for all vectors x P R n. In other words, we want to show that T and T are the same linear transformation. Let x P R n with Then we have x a 1 e 1 ` ` a n e n. Tpxq Tpa 1 e 1 ` ` a n e n q a 1 Tpe 1 q ` ` a n Tpe n q a 1 y 1 ` ` a n y n a 1 Tpe 1 q ` ` a n Tpe n q Tpxq. Since T and T agree for all vectors v P R n, we know that they are the same linear transformation. Hence, T is unique. 6