MATH 304Linear AlgebraLecture 26:Review for the final exam.Topics for the final exam: Part IElementary linear algebra (Leon 1.1–1.4, 2.1–2.2)• Systems of linear equations: elementaryoperations, Gaussian elimination, back substitution.• Matrix of coefficients and augmented matrix.Elementary row operations, row echelon form andreduced row echelon form.• Matrix algebra. Inverse matrix.• Determinants: explicit formulas for 2×2 and3×3 matrices, row and column expansions,elementary row and column operations.Topics for the final exam: Part IIAbstract linear algebra (Leon 3.1–3.6, 4.1–4.3)• Vector spaces (vectors, matrices, polynomials, functionalspaces).• Subspaces. Nullspace, column space, and row space of amatrix.• Span, spanning set. Linear independence.• Bases and dimension.• Rank and nullity of a matrix.• Coordinates relative to a basis.• Change of basis, transit ion matrix.• Linear transformations.• Matrix transformations.• Matrix of a linear mapping.• Similarity of matrices.Topics for the final exam: Parts III–IVAdvanced linear algebra (Leon 5.1–5.7, 6.1–6.3)• Euclidean structure in Rn(length, angle, dot product)• Inner products and norms• Orthogonal complement• Least squares problems• The Gram-Schmidt orthogonalization process• Orthogonal polynomials• Eigenvalues, eigenvectors, eigenspaces• Characteristic polynomial• Bases of eigenvectors, diagonalization• Orthogonal matrices• Rotations in space• Matrix exponentialsBases of eigenvectorsLet A be an n×n matrix with real entries.• A has n distinct real eigenvalues =⇒ 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= AProblem. For each of the following matricesdetermine whether it allows(a) a basis of eigenvectors for Rn,(b) a basis of eigenvectors for Cn,(c) an orthonormal basis of eigenvectors for Rn.A =1 00 4(a),(b),(c): yesB =0 10 0(a),(b),(c): noProblem. For each of the following matricesdetermine whether it allows(a) a basis of eigenvectors for Rn,(b) a basis of eigenvectors for Cn,(c) an orthonormal basis of eigenvectors for Rn.C =2 31 4(a),(b): yes (c): noD =0 −11 0(b): yes (a),(c): noProblem 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 columns 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 −λExpand 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 diagonalizable in R4If A is diagonalizable in C4then A = UXU−1, where U is aninvertible matrix 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 02=−1 0 0 00 −1 0 00 −2 −1 02 0 0 −1.Since A26= −I , the matrix A is not diagonalizable in C4.Problem Consider a linear 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 range and kernel of L.(c) Find the eigenvalues of L.(d) Find the matrix of the operator L2010(L applied2010 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 +35ze2+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 range of the operator L is spanned by columnsof the matrix B. It follows that Range(L) is th eplane spanned by v1= (0, 1, 0) and v2= (4, 0, 3).The kernel of L is the nullspace of the matrix B,i.e., the solution set for the equation 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 kernel of L is the set of vectorsv ∈ R3such that L(v) = v0× v = 0.It follows that this 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 L2010is B2010.Since the matrix B has eigenvalues 0, i, and −i, it isdiagonalizable in C3. Namely, B = UDU−1, whereU is an invertible matrix with complex entries andD =0 0 00 i 00 0 −i.Then B2010= UD2010U−1. We have that D2010== diag0, i2010, (−i)2010= diag(0, −1, −1) = D2.HenceB2010= UD2U−1= B2=−0.64 0 −0.480 −1 0−0.48 0 −0.36.Problem. Let f1, f2, f3, . . . be the Fibonaccinumbers defined by f1= f2= 1, fn= fn−1+ fn−2for n ≥ 3. Find limn→∞fn+1fn.For any integer n ≥ 1,fn+2fn+1=1 11 0fn+1fn.That is, vn+1= Avn, where A=1 11 0, vn=fn+1fn.In particular, v2= Av1, v3= Av2= A2v1,v4= Av3= A3v1. In general, vn= An−1v1.Characteristic equation of the matrix A:1 − λ 11 −λ= 0 ⇐⇒ λ2− λ − 1 = 0.Eigenvalues: λ1=1+√52, λ2=1−√52.Let w1= (x1, y1) and w2= (x2, y2) beeigenvectors of A associated with the eigenvalues λ1and λ2. Then w1, w2is a
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