“main”2007/2/16page 250iiiiiiii250 CHAPTER 4 Vector Spaces14. On R2, define the operation of addition by(x1,y1) + (x2,y2) = (x1x2,y1y2).Do axioms A5 and A6 in the definition of a vectorspace hold? Justify your answer.15. On M2(R), define the operation of addition byA + B = AB,and use the usual scalar multiplication operation. De-termine which axioms for a vector space are satisfiedby M2(R) with the above operations.16. On M2(R), define the operations of addition and mul-tiplication by a real number (⊕ and · , respectively) asfollows:A ⊕ B =−(A + B),k · A =−kA,where the operations on the right-hand sides of theseequations are the usual ones associated with M2(R).Determine which of the axioms for a vector space aresatisfied by M2(R) with the operations ⊕ and ·.For Problems 17–18, verify that the given set of objects to-gether with the usual operations of addition and scalar mul-tiplication is a complex vector space.17. C2.18. M2(C), the set of all 2 × 2 matrices with complexentries.19. Is C3a real vector space? Explain.20. Is R3a complex vector space? Explain.21. Prove part 3 of Theorem 4.2.6.22. Prove part 6 of Theorem 4.2.6.23. Prove that Pnis a vector space.4.3SubspacesLet us try to make contact between the abstract vector space idea and the solution of anapplied problem. Vector spaces generally arise as the sets containing the unknowns ina given problem. For example, if we are solving a differential equation, then the basicunknown is a function, and therefore any solution to the differential equation will be anelement of the vector space V of all functions defined on an appropriate interval. Con-sequently, the solution set of a differential equation is a subset of V . Similarly, considerthe system of linear equations Ax = b, where A is an m × n matrix with real elements.The basic unknown in this system, x, is a column n-vector, or equivalently a vector inRn. Consequently, the solution set to the system is a subset of the vector space Rn.Asthese examples illustrate, the solution set of an applied problem is generally a subsetof vectors from an appropriate vector space (schematically represented in Figure 4.3.1).The question we will need to answer in the future is whether this subset of vectors isa vector space in its own right. The following definition introduces the terminology wewill use:Vector space of unknownsSolution set of applied problem:Is S a vector space?SSVFigure 4.3.1: The solution set S of an applied problem is a subset of the vector space V ofunknowns in the problem.“main”2007/2/16page 251iiiiiiii4.3 Subspaces 251DEFINITION 4.3.1Let S be a nonempty subset of a vector space V .IfS is itself a vector space under thesame operations of addition and scalar multiplication as used in V , then we say thatS is a subspace of V .In establishing that a given subset S of vectors from a vector space V is a subspace ofV , it would appear as though we must check that each axiom in the vector space definitionis satisfied when we restrict our attention to vectors lying only in S. The first and mostimportant theorem of the section tells us that all we need do, in fact, is check the closureaxioms A1 and A2. If these are satisfied, then the remaining axioms necessarily hold inS. This is a very useful theorem that will be applied on several occasions throughout theremainder of the text.Theorem 4.3.2 Let S be a nonempty subset of a vector space V . Then S is a subspace of V if and onlyif S is closed under the operations of addition and scalar multiplication in V .ProofIf S is a subspace of V , then it is a vector space, and hence, it is certainly closedunder addition and scalar multiplication. Conversely, assume that S is closed under ad-dition and scalar multiplication. We must prove that Axioms A3–A10 of Definition 4.2.1hold when we restrict to vectors in S. Consider first the axioms A3, A4, and A7–A10.These are properties of the addition and scalar multiplication operations, hence since weuse the same operations in S as in V , these axioms are all inherited from V by the subsetS. Finally, we establish A5 and A6: Choose any vector1u in S. Since S is closed underscalar multiplication, both 0u and (−1)u are in S. But by Theorem 4.2.6, 0u = 0 and(−1)u =−u, hence 0 and −u are both in S. Therefore, A5 and A6 are satisfied.The idea behind Theorem 4.3.2 is that once we have a vector space V in place,then any nonempty subset S, equipped with the same addition and scalar multiplicationoperations, will inherit all of the axioms that involve those operations. The only possibleconcern we have for S is whether or not it satisfies the closure axioms A1 and A2. Ofcourse, we presumably had to carry out the full verification of A1–A10 for the vectorspace V in the first place, before gaining the shortcut of Theorem 4.3.2 for the subset S.In determining whether a subset S of a vector space V is a subspace of V , we mustkeep clear in our minds what the given vector space is and what conditions on the vectorsin V restrict them to lie in the subset S. This is most easily done by expressing S in setnotation as follows:S ={v ∈ V : conditions on v}.We illustrate with an example.Example 4.3.3Verify that the set of all real solutions to the following linear system is a subspace of R3:x1+ 2x2− x3= 0,2x1+ 5x2− 4x3= 0.Solution: The reduced row-echelon form of the augmented matrix of the system is10 3001−20,1This is possible since S is assumed to be nonempty.“main”2007/2/16page 252iiiiiiii252 CHAPTER 4 Vector Spacesso that the solution set of the system isS ={x ∈ R3: x = (−3r, 2r, r ), r ∈ R},which is a nonempty subset of R3. We now use Theorem 4.3.2 to verify that S is asubspace of R3:Ifx = (−3r, 2r, r ) and y = (−3s, 2s, s) are any two vectors in S, thenx + y = (−3r, 2r, r ) + (−3s, 2s, s) = (−3(r +s),2(r + s),r + s) = (−3t,2t,t),where t = r +s. Thus, x+y meets the required form for elements of S, and consequently,if we add two vectors in S, the result is another vector in S. Similarly, if we multiply anarbitrary vector x = (−3r, 2r, r ) in S by a real number k, the resulting vector iskx = k(−3r, 2r, r ) = (−3kr,2kr, kr) = (−3w, 2w, w),where w = kr. Hence, kx again has the proper form for membership in the subset S,and so S is closed under scalar multiplication. By Theorem 4.3.2, S is a subspace of R3.Note, of course, that our application of Theorem 4.3.2 hinges on our prior knowledgethat R3is a vector