Review for the final exam• Bring a number 2 pencil to the exam!• Room assignments for Friday, Dec 12, 7-10pm:◦ if your last name starts with A-J: 114 DKH◦ if your last name starts with K-Z: Foellinger AuditoriumWhat we have learned• Basic notions of linear algebra:Make sure that you can say, precisely, what each notion means.◦ vector spaces and subspaces◦ linear independence◦ basis and dim ension◦ linear tr ansformations (and matrices representing them)◦ orthogonal vectors, spaces, matrices, and projections◦ the four fundamental subspaces associated with a matrix• Technical skills:◦ matrix multiplication◦ Gaussia n eli mination and solving lin e ar s ystems◦ LU decompositio n◦ computing matrix inverses◦ Gram–Schmidt and QR decomposition◦ finding least squ ares solutions◦ determine (orthonormal) bases of spaces and thei r orthogonal complements◦ computing determinants◦ eigenvalues, eigen vectors and diagonali z a t ion• Applications◦ finite differenc e s (not on the exam)◦ directed graph s (both null spaces of edge-node inciden ce matrix)◦ Fourier series◦ linear regression (least squares lines)◦ difference equations (steady state, PageRank)◦ systems o f diffe r e ntial equati onsAr min [email protected] problemsExample 1. Are the following vector spaces?(a) The set of all functions f: R → R with f′′(0) = 7.No. Miss ing the zero function.(b) The set of all ort hogonal 2 × 2 matrices.No. Miss ing the zero matrix.(c) The set of all vectorsx1x2such that x1+ x2= 0.Yes. This isNul([1 1]).(d) The set of all eigenvectors of a matrix A.No. Take, for instance,A =1 00 2.(e) The set of solutions y(x) of y′′+ 7y′− y = 0.Yes. For instance, ify1′′+ 7y1′− y1= 0 and y2′′+ 7y2′− y2= 0,then(y1+ y2)′′+ 7(y1+ y2)′− (y1+ y2) = 0.Example 2. Decide if each criterion is tru e or false.Ann × n matrix A is invertible if and on ly if(a) the columns of A are independent.True.(b) 0 is not an eigenvalue of A.True.(c) A has no zero column.False. Take, for inst ance,A =1 21 2.(d) A has no free variables.True.(e) A is row equival e nt to I.True.(f) A has n independent eigenvectors.False. Take, for inst ance,A =0 00 1.Ar min [email protected] 3. Decide if each criterion is tru e or false.The linear systemAx = b is consistent if and only if(a) an echelon form of [A b] has no row of the form [00 β]with β0.True.(b) b is in Col(AT).False. (b is in Col(A) would be true.)(c) A has no free variables.False. Take, for inst ance,11x =10.(d) b is orthogonal to Nul(AT).True.Example 4. What is1 1 1 42 2 2 80 3 3 12 −7 0 5?Solution. The determinant is 0 because the matrix is not invertible (sec ond row is amultiple of the first).Ar min [email protected] 5. Solve the initial value problemy′=0 −2−4 2y, y(0) =11.Solution. The solution to y′= Ay, y(0) = y0is y(t ) = eA ty0.• Diagonalize A =0 −2−4 2:◦−λ −2−4 2 − λ= λ2− 2λ − 8, so the e igenvalues are −2, 4◦ λ = 4 has eigenspace Nul−4 −2−4 −2= spann1−2o◦ λ = −2 has eigenspace Nul2 −2−4 4= spann11o◦ Hence, A = PDP−1with P =1 1−2 1and D =4 00 −2.• Compute the solution y = eA ty0:y = PeD tP−1y0=1 1−2 1"e4t00 e−2t#131 −12 111=131 1−2 1"e4t00 e−2t#03=131 1−2 1"03e−2t#="e−2te−2t#Ar min [email protected] 6. Determi ne a basis for Nu l(A) and Nul(AT), where A is th e edge-nodeincidence matrix of the directed graph below.12345678912345678Solution.•Basis for Nul(A) from connected subgraphs.For each connected subgraph, get a basis vectorx tha t assig ns 1 to all nodes in that subgraph,an d0 to al l other nodes.Basis forNul(A):111100000,000011110,000000001• Basis for Nul(AT) from (independent) loops.For each (independen t) loop, get a basis vec tory that assigns 1 and −1 (depending on direction)to the edges in that loop, and0 to al l other edges.Basis forNul(AT):1−1100000,00001−1−11Ar min [email protected] 7. Let A =1 0 11 1 0−2 3 1.(a) Determine the LU decomposition of A.(b) What is det (A)?Solution.(a)1 0 11 1 0−2 3 1>R2→R2− R1R3→R3+2R11 0 10 1 −10 3 3L =11 1−2 ∗ 1>R3→R3−3R11 0 10 1 −10 0 6= U L =11 1−2 3 1The LU decomposition is A =1 0 01 1 0−2 3 11 0 10 1 −10 0 6.In the exam: check!!!(b) det (A) = det (L)de t(U) = 1 · (1 · 1 · 6) = 6Example 8. Suppose that, with respect to the bases01,1−1of R2, and110,100,001of R3,the linear transformation T : R3→ R2is represented by1 0 23 −1 1.(a) What exactly does the matrix encode?(b) What is T 112!?Solution.(a)T 110!= 101+ 31−1=3−2T 100!= 001− 11−1=−11T 001!= 201+ 11−1=11(b) T 112!= T 110!+ 2T 001!=3−2+ 211=50Ar min
View Full Document
Unlocking...