Math 110 Professor K. A. RibetMidterm Exam September 26, 2002Please put away all books, calculators, digital toys, cell phones, pagers, PDAs, and otherelectronic devices. You may refer to a single 2-sided sheet of notes. Please write your nameon each sheet of paper that you turn in. Don’t trust staples to keep your papers together.The symbol “R” denotes the field of real numbers. In this exam, “0” was used to denotethe vector space {0} consisting of the single element 0.These solutions were written by Ken Ribet. Sorry if they’re a little terse. If you havea question about the grading of your problem, see Ken Ribet for problems 1–2 and TomCoates for 3–4.1. (9 points) Let T : R4→ R3be the linear transformation whose matrix with respect tothe standard bases is A =−1 −1 7 51 0 −5 −30 −1 2 2. (In the book’s notation, T = LA.)Find bases for (i) the null space and (ii) the range of T .The null space consists of quadruples (x, y, z, w) satisfying three equations of which thefirst two are −x −y −7z +5w = 0 and x−5z −3w = 0. It turns out that the third equationis the sum of the first two, so we can forget it. (If you don’t notice this circumstance, you’llstill get the right answer.) Replace the first equation by the sum of the first two, leavingthe second alone. We get the two equations−y + 2z + 2w = 0x − 5z − 3w = 0.The interpretation is that z and w can be chosen freely, and then x and y are determinedby z and w. If we take z = 1, w = 0, we get the solution (5, 2, 1, 0). With the reversechoice, we get the solution (3, 2, 0, 1). These form a basis for the null space.Once we know that the null space has dimension 2, we deduce that the range has dimen-sion 2 as well. In fact, it consists of the space of triples (a, b, c) with c = a + b. A basiswould be the set containing (1, 0, 1) and (0, 1, 1). Of course, there are other correct an-swers: bases aren’t unique! Let me stress that R(T ) lies in 3-space. If your answer consistsof vectors in R4, you’ve messed up.Note from Ribet: An answer that just has a bunch of numbers with no explanation as towhat is going on is very unlikely to receive full credit. You need to tell the reader (me, inthis case) what you are doing.2. (9 points) Let V be a vector space over a field F . Suppose that v1, . . . , vnare ele-ments of V and that w1, . . . , wn, wn+1lie in the span of { v1, . . . , vn}. Show that the set{ w1, · · · , wn+1} is linearly dependent.Let W be the span of { v1, . . . , vn}. Then W is generated by n elements, so its dimensiond is at most n (for example, by Theorem 1.9 on page 42). If the vectors wiwere linearlyindependent, the set { w1, · · · , wn+1} could be extended to a basis of W . This is impossiblebecause all bases of W have d elements.3. (10 points) Let W1and W2be subspaces of a finite-dimensional F -vector space V .Recall that W1× W2denotes the set of pairs (w1, w2) with w1∈ W1, w2∈ W2. Thisproduct comes equipped with a natural addition and scalar multiplication:(w1, w2) + (w01, w02) := (w1+ w01, w2+ w02), a(w1, w2) := (aw1, aw2).This addition and scalar multiplication make W1× W2into an F -vector space. (There wasno requirement or expectation that students verify this point.)(1) Check that the mapT : W1× W2→ V, (w1, w2) 7→ w1+ w2is a linear transformation.This is fairly routine. For example, T (a(w1, w2)) = T (aw1, aw2) = aw1+ aw2= a(w1+w2) = aT (w1, w2). A similar computation shows that T of a sum is the sum of the T ’s.(2) Prove that N(T ) = 0 if and only if W1∩ W2= 0.The space N(T ) consists of pairs (w1, w2) with w1+ w2= 0. Hence w2is completelydetermined by w1as its negative. The wrinkle is that w1has to be in W1while w2= −w1has to be in W2. Thus w1has to be in both W1and W2, i.e., in the intersection of the twospaces. The null space N (T ) is in 1-1 correspondence with W1∩W2, with (w1, w2) ∈ N (T )corresponding to w1∈ W1∩ W2and w ∈ W1∩ W2corresponding to (w, −w) ∈ N (T ). Inparticular, N(T ) = 0 if and only if W1∩ W2= 0.(3) Show that dim(W1∩ W2) ≥ dim W1+ dim W2− dim V .Consider T : W1× W2→ V . We havedim(W1× W2) = dim N (T ) + dim R(T ) = dim(W1∩ W2) + dim R(T ).Here, I have used the fact that the identification between N(T ) and W1∩ W2that wediscussed above is a linear identification—one that respects addition and scalar multiplica-tion. Hence it preserves dimension. For this problem, we have to use another fact, namelythat dim(W1× W2)) is the sum of the dimensions of W1and W2. This follows from theRibet’s 110 first midterm—page 2fact that we get a basis for the product by taking the union of β1× {0} and {0} × β2,where the βiare bases of the Wi(i = 1, 2). This fact givesdim(W1× W2) = dim W1+ dim W2= dim(W1∩ W2) + dim R(T ) ≤ dim(W1∩ W2) + dim V.The desired inequality follows.4. (12 points) Label the following stateme nts as being true or false. For each statement,explain your answer(1) The span of the empty set is the empty vector space.There is no empty vector space! The span of the empty set is {0}.(2) If v is a vector in a vector space V that has more than two elements, then V is spannedby the set S = { w ∈ V | w 6= v }.The span of S contains S, so it contains all w different from v. Does it contain v as well?Yes, indeed: choose w ∈ V different from 0 and v; this choice is possible because V hasmore than two elements. Write v = (v − w) + w. The vectors v − w and w are both in S:neither is v. Hence v is in the span of S. Our conclusion is that the span of S contains allof V , so the assertion is true.(3) Suppose that T : V → W is a linear transformation between finite-dimensional R-vector spaces. If dim V > dim W and w lies in the range of T , then there are infinitelymany v ∈ V such that T (v) = w.If w lies in the range of T , then there is some v ∈ V such that T (v) = w. Fix this v, andnotice that T (v0) = w if and only if T (v0− v) = 0. Thus the set of v0mapping to w isthe set of v + x where x runs over N (T ). Thus there are infinitely many elements of Vthat map to w if …