Math 110 Final Exam with AnswersNAME (1 pt):TA (1 pt):Name of Neighbor to your left (1 pt):Name of Neighbor to your right (1 pt):Instructions: You are allowed to use 1 sheet (8 1/2 by 11 inches, both sides) of notes.Otherwise this is a closed book, closed notes, closed calculator, closed computer, closed PDA,closed cellphone, closed mp3 player, closed network, open brain exam.You get one point each for filling in the 4 lines at the top of this page. All other questionsare worth 10 points.Fill in the questions at the top of the page. Then stop and wait until we tell you toturn the page and start the rest of the exam. Do not start reading the rest of theexam until we tell you to start.After you start, read all the questions on the exam before you answer any of them, so youdo the ones you find easier first.Write all your answers on this exam. If you need scratch paper, ask for it, write your nameon each sheet, and attach it when you turn it in (we have a stapler).12345678Total1Question 1. (10 points) (version 1) Let P (R) be the vector space of all real polynomials.Define the linear map T : P (R) → P (R)byT (f)=f, the second derivative.Part 1: Show that T is onto but not one-to-one.Answer: If g(x)=di=0gixi,thenT (f)=g where f(x)=f0+ f1x +di=0gixi+2(i+1)(i+2)(soT is onto) and f0and f1are arbitrary (so T is not one-to-one).Part 2: Describe all eigenvectors of T , i.e. nonzero polynomials such that f(x)=λf(x)for some λ.Answer: Since the degree of fislessthanthedegreeoff, fcannot be a nonzero multipleof f .Thereforeλ =0and f(x)=0is satisfied by all linear polynomials f(x)=f0+ f1x.Question 1. (10 points) (version 2) Let P (C) be the vector space of all complex polyno-mials. Define the linear map S : P (C) → P (C)byS(g)=g, the second derivative.Part 1: Show that S is onto but not one-to-one.Answer: If f(x)=di=0fixi,thenS(g)=f where g(x)=g0+ g1x +di=0fixi+2(i+1)(i+2)(soS is onto) and g0and g1are arbitrary (so S is not one-to-one).Part 2: Describe all eigenvectors of S, i.e. nonzero polynomials such that g(x)=λg(x)for some λ.Answer: Since the degree of gislessthanthedegreeofg, gcannot be a nonzero multipleof g.Thereforeλ =0and g(x)=0is satisfied by all linear polynomials g(x)=g0+ g1x.2Question 2. (10 points) (version 1) Let A be an m-by-n complex matrix, and let B bean n-by-m complex matrix. Show that Im+ A · B is invertible if and only if In+ B · A isinvertible.Answer: Solution 1: Suppose Im+ A · B is invertible, and (In+ B · A) · v =0; we need toshow v =0.MultiplybyA to getA · (In+ B · A) · v = A · v + A · B · A · v =(Im+ A · B) · (A · v)so A · v =0since Im+ A · B is invertible. But then v = −B · A · v =0as desired. ThusIm+ A · B invertible implies In+ B · A invertible. The converse follows b y the same argument.Solution 2: From the practice final, we know A · B and B · A have the identical nonzeroeigenvalues. Therefore, −1 is an eigenvalue of A · B if and only if it an eigenvalue of B · A,implying 0 is an eigenvalue of Im+ A · B if and only if it i s an eigenvalue of In+ B · A,sothat Im+ A · B is singular if and only if In+ B · A is singular.Question 2. (10 points) (version 2) Let X be an m-by-n real matrix, and let Y be ann-by-m complex matrix. Show that Im+X ·Y is invertible if and only if In+Y ·X is invertible.Answer:Solution1:SupposeIm+ X · Y is invertible, and (In+ Y · X) · v =0; we need toshow v =0.MultiplybyX to getX · (In+ Y · X) · v = X · v + X · Y · X · v =(Im+ X · Y ) · (X · v)so X · v =0since Im+ X · Y is invertible. But then v = −Y · X · v =0as desired. ThusIm+X ·Y invertible implies In+Y ·X invertible. The converse follows by the same argument.Solution 2: From the practice final, we know X · Y and Y · X have the identical nonzeroeigenvalues. Therefore, −1 is an eigenvalue of X · Y if and only if it an eigenvalue of Y · X,implying 0 is an eigenvalue of Im+ X · Y if and only if i t is an eigenvalue of In+ Y · X,sothat Im+ X · Y is singular if and only if In+ Y · X is singular.3Question 3. (10 points) (version 1) Let A = PR· L · U · PCbe an LU decompositionof the m-by-n real matrix A of rank r>0. Thus PRand PCare permutation matrices, Lis m-by-r and unit lower triangular, and U is r-by-n and upper triangular with Uiinonzero.Show how to express an LU decomposition of Atusing simple modifications of the parts ofthis LU decomposition of A.Answer: Write U = D ·ˆU,whereD is r-by-r and diagonal with Dii= Uii,soˆU is unitupper triangular. Then A = PR· L · U · PC= PR· L · D ·ˆU · PCso At= PtC·ˆUt· (D · Lt) · PtR.This is an LU decomposition of At.Question 3. (10 points) (version 2) Let B = PR· L · U · PCbe an LU decomposition ofthe m-by-n complex matrix B of rank r>0. Thus PRand PCare permutation matrices, Lis m-by-r and unit lower triangular, and U is r-by-n and upper triangular with Uiinonzero.Show how to express an LU decomposition of B∗using simple modifications of the parts ofthis LU decomposition of B.Answer: Write U = D ·ˆU,whereD is r-by-r and diagonal with Dii= Uii,soˆU is unitupper triangular. Then B = PR· L · U · PC= PR· L · D ·ˆU · PCso B∗= P∗C·ˆU∗· (¯D · L∗) · P∗R.This is an LU decomposition of B∗.4Question 4. (10 points) (version 1) Let T : V → V be a linear operator. Supposethat T (vi)=λivifor i =1, ...m,andallλiare distinct. If W is an invariant subspace of Tand includes the vectormi=1ai· vi, where all the ai=0,thenprovethatW contains vifori =1, ..., m.Answer: Let w1=mi=1ai· vi.Sincew1∈ W ,soarew2= T · w1through wn= Tn−1· w1since W is invariant. Let˜V =[v1, ..., vm] and˜W =[w1, ..., wm].LetAij= ai· λj−1ibe m-by-m. Then we can express the dependence of all the wion all the viby˜W =˜V · A.NowA =diag(a1,a2, ..., an) · B,whereBij= λj−1i.ThusB is the m by m Vandermonde matrixwith distinct λi, and so nonsingular by homework question 4.3.22(c). Thus A is the productof nonsingular matrices and also nonsingular. Thus˜W · A−1=˜V , so all the columns of˜V ,namely the vi, are linear combinations of the wi,andsowithinW as desired.Question 4. (10 points) (version 2) Let S : X → X be a linear operator. Supposethat S(xi)=λixifor i =1, ...m,andallλiare distinct. If Y is an invariant subspace of Sand includes the vectormi=1bi· xi, where all the bi=0,thenprovethatY contains xifori =1, ..., m.Answer: Let y1=mi=1bi· xi.Sincey1∈