MATH 304Linear AlgebraLecture 33:Bases of eigenvectors.Diagonalization.Eigenvalues and eigenvectors of an operatorDefinition. Let V be a vector space and L : V → Vbe a linear operator. A number λ is called aneigenvalue of the operator L ifL(v) = λv for anonzero vector v ∈ V . The vector v is called aneigenvector of L associated with the eigenvalue λ.(If V is a functional space then eigenvectors are alsocalled eigenfunctions.)If V = Rnthen the linear operator L is given byL(x) = Ax, where A is an n×n matrix.In this case, eigenvalues and eigenvectors of theoperator L are precisely eigenvalues andeigenvectors of the matrix A.Theorem If v1, v2, . . . , vkare eigenvectors of alinear operator L associated with distincteigenvalues λ1, λ2, . . . , λk, then v1, v2, . . . , vkarelinearly independent.Corollary Let A be an n×n matrix such that thecharacteristic equation det(A − λI ) = 0 has ndistinct real roots. Then Rnhas a basis consistingof eigenvectors of A.Proof: Let λ1, λ2, . . . , λnbe distinct real roots ofthe characteristic equation. Any λiis an eigenvalueof A, hence there is an associated eigenvector vi.By the theorem, vectors v1, v2, . . . , vnare linearlyindependent. Therefore they form a basis for Rn.Theorem If λ1, λ2, . . . , λkare distinct realnumbers, then the functions eλ1x, eλ2x, . . . , eλkxarelinearly independent.Proof: Consider a linear operatorD : C∞(R) → C∞(R) given by Df = f′.Then eλ1x, . . . , eλkxare eigenfunctions of Dassociated with distinct eigenvalues λ1, . . . , λk.Characteristic polynomial of an operatorLet L be a linear operator on a finite-dimensionalvector space V . Let u1, u2, . . . , unbe a basis for V .Let A be the matrix of L with respect to this basis.Definition. The characteristic polynomial of thematrix A is called the characteristic polynomialof the operator L.Then eigenvalues of L are roots of its characteristicpolynomial.Theorem. The characteristic polynomial of theoperator L is well defined. That is, it does notdepend on the choice of a basis.Theorem. The characteristic polynomial of theoperator L is well defined. That is, it does notdepend on the choice of a basis.Proof: Let B be the matrix of L with respect to adifferent basis v1, v2, . . . , vn. Then A = U BU−1,where U is the transition matrix from the basisv1, . . . , vnto u1, . . . , un. We have to show thatdet(A − λI ) = det(B − λI ) for all λ ∈ R. Weobtaindet(A − λI ) = det(UBU−1− λI )= detUBU−1− U(λI )U−1= detU(B − λI )U−1= det(U) det(B − λI ) det(U−1) = det(B − λI ).DiagonalizationLet L b e a linear operator on a finite-dimensional vector spaceV . Then the following conditions are equivalent:• the matrix of L with respect to some basis is diagonal;• there exists a basis for V formed by eigenvectors of L.The operator L is diagonalizable if it satisfies theseconditions.Let A be an n×n matrix. Then the following conditions areequivalent:• A is the matrix of a diagonalizable operator;• A is similar to a diagonal matrix, i.e., it is represented asA = UBU−1, where the matrix B is diagonal;• there exists a basis for Rnformed by eigenvectors of A.The matrix A is diagonalizable if it satisfies these conditions.Otherwise A is called defective.Example. A =2 11 2.• The matrix A has two eigenvalues: 1 and 3.• The eigenspace of A associated with theeigenvalue 1 is the line spanned by v1= (−1, 1).• The eigenspace of A associated with theeigenvalue 3 is the line spanned by v2= (1, 1).• Eigenvectors v1and v2form a basis for R2.Thus the matrix A is diagonalizable. Namely,A = UBU−1, whereB =1 00 3, U =−1 11 1.Example. A =1 1 −11 1 10 0 2.• The matrix A has two eigenvalues: 0 and 2.• The eigenspace corresponding to 0 is spanned byv1= (−1, 1, 0).• The eigenspace corresponding to 2 is spanned byv2= (1, 1, 0) and v3= (−1, 0, 1).• Eigenvectors v1, v2, v3form a basis for R3.Thus the matrix A is diagonalizable. Namely,A = UBU−1, whereB =0 0 00 2 00 0 2, U =−1 1 −11 1 00 0 1.Problem. Diagonalize the matrix A =4 30 1.We need to find a diagonal matrix B and aninvertible matrix U such that A = UBU−1.Suppose that v1= (x1, y1), v2= (x2, y2) is a basisfor R2formed by eigenvectors of A, i.e., Avi= λivifor some λi∈ R. Then we can takeB =λ100 λ2, U =x1x2y1y2.Note that U is the transition matrix from the basisv1, v2to the standard basis.Problem. Diagonalize the matrix A =4 30 1.Characteristic equation of A:4 − λ 30 1 − λ= 0.(4 − λ)(1 − λ) = 0 =⇒ λ1= 4, λ2= 1.Associated eigenvectors: v1= (1, 0), v2= (−1, 1).Thus A = UBU−1, whereB =4 00 1, U =1 −10 1.Problem. Let A =4 30 1. Find A5.We know that A = UBU−1, whereB =4 00 1, U =1 −10 1.Then A5= UBU−1UBU−1UBU−1UBU−1UBU−1= UB5U−1=1 −10 11024 00 11 10 1=1024 −10 11 10 1=1024 10230 1.Problem. Let A =4 30 1. Find a matrix Csuch that C2= A.We know that A = UBU−1, whereB =4 00 1, U =1 −10 1.Suppose that D2= B for some matrix D. Let C = UDU−1.Then C2= UDU−1UDU−1= UD2U−1= UBU−1= A.We can take D =√4 00√1=2 00 1.Then C =1 −10 12 00 11 10 1=2 10 1.There are two obstructions to existence of a basisconsisting of eigenvectors. They are illustrated bythe following examples.Example 1. A =1 10 1.det(A − λI ) = (λ − 1)2. Hence λ = 1 is the onlyeigenvalue. The associated eigenspace is the linet(1, 0).Example 2. A =0 −11 0.det(A − λI ) = λ2+ 1.=⇒ no real eigenvalues or eigenvectors(However there are complex
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