MATH 304Linear AlgebraLecture 23:Diagonalization.Review for Test 2.DiagonalizationLet L be 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.Theorem 1 If v1, v2, . . . , vkare eigenvectors of a linearoperator L associated with distinct eigenvalues λ1, λ2, . . . , λk,then v1, v2, . . . , vkare linearly independent.Theorem 2 Let λ1, λ2, . . . , λkbe distinct eigenvalues of alinear operator L. For any 1 ≤ i ≤ k let Sibe a basis for theeigenspace associated with the eigenvalue λi. Then the unionS1∪ S2∪ ··· ∪ Skis a linearly independent set.Corollary Let A be an n×n matrix such that thecharacteristic equation det(A − λI ) = 0 has n distinct realroots. Then(i) there exists a basis for Rnconsisting of eigenvectors of A;(ii) all eigenspaces of A are one-dimensional.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=x1y1, v2=x2y2is a basis forR2formed by eigenvectors of A, i.e., Avi= λiviforsome 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=10, v2=−11.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.Initial value problem for a system of linear ODEs:(dxdt= 4x + 3y,dydt= y,x(0) = 1, y(0) = 1.The system can be rewritten in vector form:dvdt= Av, where A =4 30 1, v =xy.Matrix A is diagonalizable: A = UBU−1, whereB =4 00 1, U =1 −10 1.Let w =w1w2be coordinates of the vector v relative to thebasis v1= (1, 0)T, v2= (−1, 1)Tof eigenvectors of A. Thenv = Uw =⇒ w = U−1v.It follows thatdwdt=ddt(U−1v) = U−1dvdt= U−1Av = U−1AUw.Hencedwdt= Bw ⇐⇒(dw1dt= 4w1,dw2dt= w2.General solution: w1(t) = c1e4t, w2(t) = c2et, where c1, c2∈ R.Initial condition:w(0) = U−1v(0) =1 −10 1−111=1 10 111=21.Thus w1(t) = 2e4t, w2(t) = et. Thenx(t)y(t)= Uw(t) =1 −10 12e4tet=2e4t−etet.There are two obstructions to diagonalization.They are illustrated by the 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 eigenvalues/eigenvectors.)Topics for Test 2Coordinates and linear transformations (Leon 3.5, 4.1–4.3)• Coordinates relative to a basis• Change of basis, transition matrix• Matrix transformations• Matrix of a linear mappingOrthogonality (Leon 5.1–5.6)• Inner products and norms• Orthogonal complement, orthogonal projection• Least squares problems• The Gram-Schmidt orthogonalization processEigenvalues and eigenvectors (Leon 6.1, 6.3)• Eigenvalues, eigenvectors, eigenspaces• Characteristic polynomial• DiagonalizationSample problems for Test 2Problem 1 (15 pts.) Let M2,2(R) denote the vector spaceof 2 × 2 matrices with real entries. Consider a linear operatorL : M2,2(R) → M2,2(R) given byLx yz w=x yz w1 23 4.Find the matrix of the operator L with respect to the basisE1=1 00 0, E2=0 10 0, E3=0 01 0, E4=0 00 1.Problem 2 (20 pts.) Find a linear polynomial which is thebest least squares fit to the following data:x −2 −1 0 1 2f (x) −3 −2 1 2 5Problem 3 (25 pts.) Let V be a subspace of R4spannedby the vectors x1= (1, 1, 1, 1) and x2= (1, 0, 3, 0).(i) F ind an orthonormal basis for V .(ii) F ind an orthonormal basis for the orthogonal complementV⊥.Problem 4 (30 pts.) Let A =1 2 01 1 10 2 1.(i) F ind all eigenvalues of the matrix A.(ii) For each eigenvalue of A, find an associated eigenvector.(iii) Is the mat rix A diagonalizable? Explain.(iv) Fi nd all eigenvalues of t he ma trix A2.Bonus Problem 5 (15 pts.) Let L : V → W be a linearmapping of a finite-dimensional vector space V to a vectorspace W . Show thatdim Range(L) + dim ker(L) = dim V .Problem 1. Let M2,2(R) denote the vector space of 2×2matrices with real entries. Consider a linear operatorL : M2,2(R) → M2,2(R) given byLx yz w=x yz w1 23 4.Find the matrix of the operator L with respect to the
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