Mathematics 54W Professor A. OgusSpring 2005Final Solution Set—May 18, 2005Work each problem on a separate sheet of paper. Be sure to put your name,your section number, and your GSI’s name on each sheet of paper. Also,at the top of the page, in the center, write the problem number, and besure to put the pages in order. Write clearly—explanations (with completesentences when appropriate) will help us understand what you are doing.Math 49 students taking the linear algebra portion should work problems1–5; Math 49 students taking the differential equation should work problems6–10.1. Let A :=1 −2 1 1 −2−1 2 −1 0 11 −2 1 −1 0.(2 points each)(a) Find a matrix B which is in reduced row echelon form and whichis row equivalent to A.B =1 −2 1 0 −10 0 0 1 −10 0 0 0 0(b) Find a basis for the null space of A.10011,−10100,21000.(c) Find a basis for the column space of A from among the columnsof A.1−11,10−1.(d) Find all X such that AX =2−10.X =10010+ X0,where X0is any element of the null space of A.(e) Find at least one X such that ATAX = ATB, where B =3−21.Hint: you do not need to calculate ATA to do this.Recall that the above equation says that AX is the orthogonalprojection B0of B on the column space W of A. Luckily we haveabove an orthogonal basis w1, w2for W , and soB0=(B|w1)(w1|w1)w1+(B|w2)(w2|w2)w2= 21−11+10−1Hence X =20010plus any element of the null space of A will do.2. Write the definition of each of the following concepts. Use completesentences and be as precise as you can.(2 points each)(a) The inverse of a matrix.Let A be an n × n matrix. Then the inverse of A is a matrix Bsuch that BA = AB = I; such a matrix is unique if it exists. Itcan be proved that either of these conditions implies the other,but only for n × n matrices.(b) A linearly independent sequence in a vector space V .A sequence (v1, v2, . . . vn) is linearly independent if for every se-quence of numbers (c1, c2, . . . cn) such that c1v1+ ···cnvn= 0,each ci= 0.(c) The dimension of a vector space. State the theorem which makesthis definition meaningful.The dimension of V is the number of elements in a basis for V .To know this makes sense, we need to use the theorems that say(1) V has a basis and (2) any two bases have the same numberof elements.(d) The orthogonal projection of a vector in Rnonto a linear subspaceW .The orthogonal projection of v onto W is the vector w in W suchthat v − w ∈ W⊥. Equivalently, it is the vector in W whichminimizes ||v − w||.(e) An eigenvector of a linear operator T : V → V .An eigenvector of V is a vector v such that T v = λv for somescalar λ.23. Let V be the space of vectors in R4such that x1+ x2+ x3+ x4= 0.and let W be the set of vectors in V such that x1= x4.(a) (3 pts). Find an orthogonal basis (v1, v2, v3) for V with v1=(0, 1, −1, 0) and such that (v1, v2) is a basis for W .v1= (0, 1, −1, 0), v2= (1, −1, −1, 1), v3= (1, 0, 0, −1).(b) (4 pts). Find the orthogonal projection of v := (2, 1, 3, −6) on W.πW(v) =(v|v1)(v1|v1)v1+(v|v2)(v2|v2)v2= −v1− 2v2= (−2, 1, 3, −2)(c) 3 pts). Find the distance from v to W .This is||v − πW(v)|| = ||(4, 0, 0, −4)|| = 4√24. Suppose that A is a matrix with 4 rows and 8 columns, and supposethat the rows of A span a three dimensional subspace of R8. Answerthe following questions, explaining you reasoning.(a) (2 pts.) What is the dimension of the space spanned by thecolumns of A?The column space of a matrix has the same dimension as the rowspace, so the dimension is 3.(b) (2 pts.) What is the dimension of the null space of A?The rank plus the nullity is n = 8, so the answer is 8 − 3 = 5.(c) (2 pts.) What is the dimension of the null space of ATA?The null space of ATA is the same as the null space of A (seebelow) so the answer is again 5.(d) (4 pts.) Prove that for any m × n real matrix A, the ranks ofATA and of A are the sameATA is an n × n matrix and A is an m × n matrix, so by therank-nullity formula, it is enough to prove that the dimensions ofthe null spaces are the same. In fact we prove the null spaces arethe same. Evidently AX = 0 implies ATAX = 0. Conversely, ifATAX = 0, then (ATAX|X) = 0, hence (AX|AX) = 0, hence||AX|| = 0, hence AX = 0.35. Let A =6 −14 2.(a) (2 pts.) Find the eigenvalues of A.These are the roots of the equation λ2− 8λ + 16. Thus λ = 4.(b) (3 pts.) We know that there exist matrices S and T such thatA = ST S−1, where T is upper triangular. Find T.T can be4 a0 4, for any nonzero value of a. (An answer withany a 6= 0 is acceptable here.)(c) (3 pts.) Now compute etA, as a function of t.We have A = 4I + N, where N =2 −14 −2and N2= 0. HenceetA= e4tI+tN= e4tetN= e4t(I + tN )= e4t1 + 2t −t4t 1 − 2t(d) (2 pts.) Find a matrix B with positive eigenvalues such thatB2= A.If B = 2I + M with M2= 0, then B2= 4I + 4M , so this equalsA if and only if M = (1/4)N. ThusB =2.5 −.251 1.546. Consider the system of differential equations:f0= gg0= 2f − g(a) (2 pts.) Write this system as a vector-valued differential equation.X0=0 12 −1X(b) (3 pts.) Find a fundamental solution set for the equation in part(a).The characteristic polynomial of this matrix A isλ2+ λ − 2 = (λ − 1)(λ + 2).Eig−2(A) = NS2 12 1= span−12Eig1(A) = NS−1 12 −2= span11.Hence a fundamental solution set is given byet11, e−2t−12(c) (2 pts.) Compute the Wronskian of your solution set.detet−e−2tet2e−2t= 3e−t.(d) (3 pts.) Find a pair of functions f, g satisfying the original systemand such that f(0) = −1 and g(0) = 5.We must havefg= aet11+ be−2t−12.So we need to find a, b such that15= a11+ b−12Thus a = 1 and b = 2, so f (t) = et− 2e−2tand g(t) = et+ 4e−2t57. For each of the following matrices, sketch and describe the trajec-tories of the solutions to the differential equation Y0(t) = AY (t). Inparticular, exhibit any asymptotes and/or invariant lines, draw arrowsindicating the direction of the flow along the solution, and say whetherthe origin is a stable or unstable node, saddle point, etc.(a) (3 pts.) A =2 00 1This is an