MATH 311-504Topics in Applied MathematicsLecture 3-14:Review for the final exam (continued).Topics for the final exam: Part I• n-dimensional vectors, dot product, crossproduct.• Elementary analytic geometry: lines and planes.• Systems of linear equations: elementaryoperations, echelo n and reduced form.• Matrix algebra, inverse matrices.• Determinants: explicit formulas for 2-by-2 and3-by-3 matrices, row and column expansions,elementary row and column o perations.Topics for the final exam: Part II• Vector spaces (vectors, matrices, polynomials,functional spaces).• Bases and dimension.• Linear mappings/transformations/operators.• Subspaces. Image and null-space of a linear map.• Matrix of a linear map relative to a basis.Change of coordinates.• Eigenvalues and eigenvectors. Characteristicpolynomial of a matri x. Bas es of eigenvectors(diagonal ization).Topics for the final exam: Part III• Norms. Inner products.• Orthogonal and orthonormal bases. T heGram-Schmidt orthogonalization process.• Orthogonal polynomials.• Orthonormal bases of eigenvectors. Symmetricmatrices.• Orthogonal matrices. Rotations in space.Bases of eigenvectorsLet A be an n×n m atrix with real entries.• A has n distinct real eigenva lues =⇒ a basis for Rnformed by eigenvectors of A• A has complex eigenvalues =⇒ no basis for Rnformed byeigenvectors of A• A has n distinct complex eigenvalues =⇒ a basis for Cnformed by eigenvectors of A• A has multiple eigenvalues =⇒ further information isneeded• an orthonormal basis for Rnformed by eigenvectors of A⇐⇒ A is symmetric: AT= A• an orthonormal basis for Cnformed by eigenvectors of A⇐⇒ A is normal: AAT= ATAProblem For each of the following matricesdetermine whether it allows(a) a basis of eigenvectors f or Rn,(b) a basis of eigenvectors f or Cn,(c) an orthonormal basis of eigenvectors for Rn,(d) an orthonormal basis of eigenvectors for Cn.A =1 00 4(a),(b),(c),(d): yesB =0 10 0(a),(b),(c),(d): noProblem For each of the following matricesdetermine whether it allows(a) a basis of eigenvectors f or Rn,(b) a basis of eigenvectors f or Cn,(c) an orthonormal basis of eigenvectors for Rn,(d) an orthonormal basis of eigenvectors for Cn.C =2 31 4(a),(b): yes (c),(d): noD =0 −11 0(b),(d): yes (a),(c): noProblem For each of the following matricesdetermine whether it allows(a) a basis of eigenvectors f or Rn,(b) a basis of eigenvectors f or Cn,(c) an orthonormal basis of eigenvectors for Rn,(d) an orthonormal basis of eigenvectors for Cn.(a),(b),(d): yes (c): no Impossible!(b): yes (a),(c),(d): no E =0 −21 0Problem Let V be the vector space spanned byfunctions f1(x) = x sin x, f2(x) = x cos x,f3(x) = sin x, and f4(x) = cos x.Consider the linear operator D : V → V ,D = d/dx.(a) Find the matrix A of the operator D relative tothe basis f1, f2, f3, f4.(b) Find the eigenvalues of A.(c) Is the matrix A diagonalizable in R4(in C4)?A is a 4×4 matrix whose co lumns are coordinates offunctions Dfi= f′irelative to the basis f1, f2, f3, f4.f′1(x) = (x sin x)′= x cos x + sin x = f2(x) + f3(x),f′2(x) = (x cos x)′= −x sin x + cos x= −f1(x) + f4(x),f′3(x) = (sin x)′= cos x = f4(x),f′4(x) = (cos x)′= − sin x = −f3(x).Thus A =0 −1 0 01 0 0 01 0 0 −10 1 1 0.Eigenvalues of A are roots of its characteristicpolynomialdet(A − λI ) =−λ −1 0 01 −λ 0 01 0 −λ −10 1 1 −λExpand the determinant by the 1st row:det(A − λI ) = −λ−λ 0 00 −λ −11 1 −λ− (−1)1 0 01 −λ −10 1 −λ= λ2(λ2+ 1) + (λ2+ 1) = (λ2+ 1)2.The eigenvalues are i and −i, both of multiplicity 2.Complex eigenvalues =⇒ A is not diagona l izable in R4If A is diago nalizable in C4then A = UXU−1, where U is aninvertible ma trix with complex entries andX =i 0 0 00 i 0 00 0 −i 00 0 0 −i.This would imply that A2= UX2U−1. But X2= −I so thatA2= U(−I )U−1= −I .A2=0 −1 0 01 0 0 01 0 0 −10 1 1 02=−1 0 0 00 −1 0 00 −2 −1 02 0 0 −1.Since A26= −I , the matrix A is not diag onalizable in C4.Problem Consider a l inear operator L : R3→ R3defined by L(v) = v0× v, wherev0= (3/5, 0, −4/5).(a) Find the matrix B of the operator L.(b) Find the image and null-space of L.(c) Find the eigenvalues of L.(d) Find the matrix of the operator L311(L applied311 times).L(v) = v0× v, v0= (3/5, 0, −4/5).Let v = (x, y, z) = xe1+ ye2+ ze3. ThenL(v) = v0× v =e1e2e33/5 0 −4/5x y z=45ye1−45x +35ze2+35ye3.In particular, L(e1) = −45e2, L(e2) =45e1+35e3,L(e3) = −35e2.Therefore B =0 4/5 0−4/5 0 −3/50 3/5 0.B =0 4/5 0−4/5 0 −3/50 3/5 0.The image of the operator L is spanned by columnsof the matrix B. It fol lows that Im L is the planespanned by v1= (0, 1, 0 ) and v2= (4, 0, 3 ).The null-space of L is the solution set for theequation Bx = 0.0 4/5 0−4/5 0 −3/50 3/5 0→1 0 3/40 1 00 0 0=⇒ x +34z = y = 0 =⇒ x = t(−3 /4, 0, 1).Alternatively, the null-space of L is the set ofvectors v ∈ R3such that L(v) = v0× v = 0.It follows that Null L is the line spanned byv0= (3/5, 0, −4/5).Characteristic polynomial of the matrix B:det(B − λI ) =−λ 4/5 0−4/5 −λ −3/50 3/5 −λ= −λ3−(3/5)2λ−(4/5)2λ = −λ3−λ = −λ(λ2+1).The eigenvalues are 0, i, and −i.The matrix of the operator L311is B311.Since the matrix B has eigenvalues 0, i, and −i, it isdiagonali zable in C3. Namely, B = UDU−1, whereU is an invertible matrix with complex entries andD =0 0 00 i 00 0 −i.Then B311= UD311U−1. We have that D311== diag0, i311, ( −i )311= diag(0, −i , i ) = −D.HenceB311= U(−D)U−1= −B =0 −4/5 04/5 0 3/50 −3/5