Vector Spaces

Part V Vector Spaces 385 Section 22 Vector Spaces Focus Questions By the end of this section, you should be able to give precise and thorough answers to the questions listed below. You may want to keep these questions in mind to focus your thoughts as you complete the section. What is a vector space? • What is a subspace of a vector space? • What is a linear combination of vectors in a vector space V ? • What is the span of a set of vectors in a vector space V ? • What special structure does the span of a set of vectors in a vector space V • have? Why is the vector space concept important? • Application: The Hat Puzzle In a New York Times article (April 10, 2001) “Why Mathematicians Now Care About Their Hat Color”, the following puzzle is posed. “Three players enter a room and a red or blue hat is placed on each person’s head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players’ hats but not his own. No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.” 387 388 Section 22. Vector Spaces The game can be played with more players, and the problem is to find a strategy for the group that maximizes its chance of winning. One strategy is for a designated player to make a random guess and for the others to pass. This gives a 50% chance of winning. However, there are much better strategies that provide a nearly 100% probability of winning as the number of players in- creases. One such strategy is based on Hamming codes and subspaces of a particular vector space to implement the most effective approach. Introduction We have previously seen that Rn and the set of fixed size matrices have a nice algebraic structure when endowed with the addition and scalar multiplication operations. In fact, as we will see, there are many other sets of elements that have the same kind of structure with natural addition and scalar multiplication operations. Due to this underlying similar structure, these sets are connected in some way and can all be studied jointly. Mathematicians look for these kinds of connections between seemingly dissimilar objects and, from a mathematical standpoint, it is convenient to study all of these similar structures at once by combining them into a larger collection. This motivates the idea of a vector space that we will investigate in this chapter. An example of a set that has a structure similar to vectors is a collection of polynomials. Let P1 be the collection of all polynomials of degree less than or equal to 1 with real coefficients. That is, P1 = a0 + a1t : a0; a1 R : f 2 g So, for example, the polynomials 2 + t, 5t, 7, and p12 πt are in P1, but pt is not in P1. − − Two polynomials a(t) = a0 + a1t and b(t) = b0 + b1t in P1 are equal if a0 = b0 and a1 = b1. We define addition of polynomials in P1 by adding the coefficients of the like degree terms. So if a(t) = a0 + a1t and b(t) = b0 + b1t, then the polynomial sum of a(t) and b(t) is a(t) + b(t) = (a0 + a1t) + (b0 + b1t) = (a0 + b0) + (a1 + b1)t: So, for example, (2 + 3t) + ( 1 + 5t) = (2 + ( 1)) + (3 + 5)t = 1 + 8t: − − We now consider the properties of the addition operation. For example, we can ask if polynomial addition is commutative. That is, if a(t) and b(t) are in P1, must it be the case that a(t) + b(t) = b(t) + a(t)? To show that addition is commutative in P1, we choose arbitrary polynomials a(t) = a0 + a1t and b(t) = b0 + b1t in P1. Then we have a(t) + b(t) = (a0 + b0) + (a1 + b1)t = (b0 + a0) + (b1 + a1)t = b(t) + a(t): Note that in the middle step, we used the definition of equality of polynomials since a0+b0 = b0+a0 and a1 + b1 = b1 + a1 due to the fact that addition of real numbers is commutative. So addition of elements in P1 is a commutative operation. Section 22. Vector Spaces 389 Preview Activity 22.1. (1) Now we investigate other properties of addition in P1. (a) To show addition is associative in P1, we need to verify that if a(t) = a0 + a1t, b(t) = b0 + b1t, and c(t) = c0 + c1t are in P1, it must be the case that (a(t) + b(t)) + c(t) = a(t) + (b(t) + c(t)): Either verify this property by using the definition of two polynomials being equal, or give a counterexample to show the equality fails in that case. (b) Find a polynomial z(t) P1 such that 2 a(t) + z(t) = a(t) for all a(t) P1. This polynomial is called the zero polynomial or the additive 2 identity polynomial in P1. (c) If a(t) = a0 + a1t is an element of P1, is there an element p(t) P1 such that 2 a(t) + p(t) = z(t); where z(t) is the additive identity polynomial you found above? If not, why not? If so, what polynomial is p(t)? Explain. (2) We can also define a multiplication of polynomials by scalars (real numbers). 1 (a) What element in P1 could be the scalar multiple 2 (2 + 3t)? (b) In general, if k is a scalar and a(t) = a0 + a1t is in P1, how do we define the scalar multiple ka(t) in P1? (c) If k is a scalar and a(t) = a0 + a1t and b(t) = b0 + b1t are elements in P1, is it true that k(a(t) + b(t)) = ka(t) + kb(t)? If no, explain why. If yes, verify your answer using the definition of two polynomials being equal. (d) If k and m are scalars and a(t) = a0 + a1t is an element in P1, is it true that (k + m)a(t) = ka(t) + ma(t)? If no, explain why. If yes, verify your answer. (e) If k and m are scalars and a(t) = a0 + a1t is an element in P1, is it true that (km)a(t) = k(ma(t))? If no, explain why. If yes, verify your answer. (f) If a(t) = a0 + a1t is an element of P1, is it true that 1a(t) = a(t)? If no, explain why. If yes, verify your answer. 390 Section 22. Vector Spaces Spaces with Similar Structure to Rn Mathematicians look for patterns and for similarities between mathematical objects. In doing so, mathematicians often consider larger collections of objects that are sorted according to their similarities and then study these collections rather than just the objects themselves. This perspective can be very powerful – whatever can be shown to be true about an arbitrary element in a collection will then be true for every specific element in the collection. In this section we study the larger collection of sets that share the algebraic structure of vectors in Rn. These sets are called vector spaces. In Preview Activity 22.1, we showed that the set P1 of polynomials of degree less than or equal to one with real coefficients, with the operations of addition and scalar multiplication defined by (a0 + a1t) + (b0 + b1t) = (a0 + b0) + (a1 + b1)t and k(a0 + a1t) = (ka0) + (ka1)t; has a structure similar to R2. By structure we mean how the elements in the set relate to each other under addition and multiplication by scalars. That is, if a(t) = a0 + a1t, b(t), and c(t) are elements of P1 and k and m are scalars, then (1) a(t) + b(t) is an element of P1, (2) a(t) + b(t) = b(t) + a(t), (3) (a(t) + b(t)) + c(t) = a(t) + (b(t) + c(t)), (4) there is a zero polynomial z(t) (namely, 0 + 0t) in P1 so that a(t) + z(t) = a(t), (5) there is an element a(t) in P1 (namely, ( a0) + ( a1)t) so that a(t) + ( a(t)) = z(t), − − − − (6) ka(t) is an element of P1, (7) (k + m)a(t) = ka(t) + ma(t), (8) k(a(t) + b(t)) = ka(t) + kb(t), (9) (km)a(t) = k(ma(t)), (10) 1a(t) = a(t). The properties we saw for polynomials in P1 stated above are the same as the properties for vector addition and multiplication by scalars in Rn, as well as matrix addition and multiplication n by scalars identified in Section 8. This indicates that polynomials in P1, vectors in R , and the set of m n matrices behave in much the same way as regards their addition and multiplication × by scalars. There is an even closer connection between linear polynomials and vectors in R2. An a0 2 element a(t) = a0 + a1t in P1 can be naturally associated with the vector in R . All the a1 results of polynomial addition and multiplication by scalars then translate to corresponding results of addition and multiplication by scalars of vectors in R2. So for all intents and purposes, as far as addition and multiplication by scalars is concerned, there is no difference between elements in P1 Section 22.

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