Math 531 Handout Winter 2012 Do the following. SECTION 7.1 EXERCISES (page 495) Problems 2 and 3 SECTION 7.2 EXERCISES (page 509) Problems 4 and 5 1. Let T 2 L(V ). A nonzero vector x in V is a generalized eigenvector of T corresponding to the eigenvalue λ if (T − λI)p(x) = 0 for some positive integer p. Note: if p is the smallest positive integer such that (T − λI)p(x) = 0, then (T − λI)p−1(x) is an eigenvector of T . 2. Let T 2 L(V ) and let x in V be a generalized eigenvector of T corresponding to the eigenvalue λ. If p is the smallest positive integer such that (T − λI)p(x) = 0, then the ordered set f(T − λI)p−1(x); (T − λI)p−2(x);:::; (T − λI)(x); xg is called a cycle of generalized eigenvectors of T corresponding to the eigenvalue λ. The length of cycle, f(T − λI)p−1(x); (T − λI)p−2(x);:::; (T − λI)(x); xg, is p. The vectors (T − λI)p−1(x), and x are called the initial vector, and the end vector, respectively, of the cycle. Note: (1) the initial vector of a cycle of generalized eigenvectors of a linear operator T is the only eigen- vector of T in the cycle. (2) If x is an eigenvector of T corresponding to the eigenvalue λ then the set fxg is a cycle of generalized eigenvectors of T, corresponding to the eigenvalue λ, of length 1. 3. The generalized eigenspace of T corresponding to the eigenvalue λ of T, denoted by Kλ, is the set p fx 2 V j(TλI) for some positive integer pg. Note Eλ ⊆ Kλ. 4. If λ is an eigenvalue of T with multiplicity m, then (a) dim(Kλ) = m. m (b) Kλ = N((T − λI) ). 5. (DOT Diagram) Let rj denote the number of dots in the jth row of the dot diagram of Ti, the restriction of T to Kλi . Then (a) r1=dim(V)-rank(T − λi) j−1 j (b) rj = rank((T − λi)) ) - rank((T − λi)) )(j > 1). 6. INTERPRETATION of a DOT diagram: ••• The block corresponding to the dot diagram •• consists of a cycle of length 3 (# of dots • in column 1), a cycle of length 2 (# of dots in column 2), and a cycle of length 1 (# of dots in column 1). 1. For each of the following matrices A, find a Jordan canonical form J and an invertible matrix Q such that J = Q−1AQ. 2 3 3 1 −2 3 − λ 1 −2 (a) A = 4 −1 0 5 5. jA − λIj = −1 0 − λ 5 = −(λ − 3)(λ − 2)2 = 0. −1 −1 4 −1 −1 4 − λ So λ = 2 and its multiplicity is 2 and λ = 3 . E3: 2 0 1 −2 3 2 1 0 1 3 −1 −3 5 0 1 −2 A − 3I = 4 5 −−−−!RREF 4 5. −1 −1 1 0 0 0 A basis for E3 is f(−1; 2; 1)g. E2: 2 1 1 −2 3 2 3 − 0 2 −3 3 2 1 0 1 3 −1 −2 5 0 1 − 0 0 0 1 −3 A − 2I = 4 5 = 4 5 −−−−!RREF 4 5. 2 2 −2 − λ −1 −1 2 0 0 0 A basis for E2 is f(1; −3; −1)g. Now dim(K2) = 2= multiplicity of 2 in jA − 2Ij. 2 3 2 1 1 3 2 1 −1 1 2 − 2 2 −4 −2 2 0 0 0 (A − 2I) = 4 5 −−−−!RREF 4 5. −2 −1 1 0 0 0 Thus, a basis for (A − 2I)2 is f(−1; 2; 0); (1; 0; 2)g. A basis for the generalized eigenspace for K2 is either a union of two 1-cycles or a 2-cycle. However, it cannot be two 1-cycles because each cycle contains one eigenvector but a basis for E2 contains one eigenvector, since dim(E2) = 1. Therefore, a basis for the generalized eigenspace for K2 is a 2-cycle. This is confirmed by the following dot diagram: • r = 4 - rank(A − 2I)=3-2=1 and r = rank(A − 2I)-rank((A − 2I)2)= 2-1=1 . In or- 1 2 • der to form a basis (2-cycle) for K2, we now look for a nonzero generalized eigenvector x such that (A − 2I)(x) 6= 0 but (A − 2I)2(x) 6= 0. From the basis f(−1; 2; 0); (1; 0; 2)g of (A − 2I)2 given above, we choose x = (−1; 2; 0) (although both of these work) and note that 2 1 1 −2 3 2 −1 3 2 1 3 4 −1 −2 5 5 4 2 5 = 4 −3 5 So, the required 2-cycle is f(1; −3; −1); (−1; 2; 0)g. −1 −1 −2 0 −1 The required Jordan canonical basis is β0 = f(−1; 2; 1); (1; −3; −1); (−1; 2; 0)g. Hence, Q = 2 −1 1 −1 3 2 3 0 0 3 4 2 −3 2 5 and J = 4 0 2 1 5. 1 2 0 0 0 2 2 −2 −1 1 3 Check: Q−1 = 4 −2 −1 0 5 and Q−1AQ = J. −1 0 −1 2 −1 −1 0 3 (b) A = 4 0 −1 −2 5. 0 0 −1 −1 − λ −1 0 jA − λIj = 0 −1 − λ −2 = −(1 + λ)3 = 0. 0 0 −1 − λ So λ = −1 and its multiplicity is 3 . E−1: 2 0 −1 0 3 2 0 1 0 3 0 0 −2 0 0 1 A + I = 4 5 −−−−!RREF 4 5. 0 0 0 0 0 0 A basis for E−1 is f(1; 0; 0)g. Now dim(K−1) = 3= multiplicity of -1 in A + I. 2 0 0 2 3 2 0 0 1 3 2 0 0 0 0 0 0 (A + I) = 4 5 −−−−!RREF 4 5. 0 0 0 0 0 0 2 Thus, a basis for (A + I) is f1; xg. A basis for the generalized eigenspace for K−1 is either a union of three 1-cycles, or a union of a 1-cycle and a 2-cycle, or a 3-cycle. However, it cannot be a union of three 1-cycles or a union of a 1-cycle and a 2-cycle because each of these cases requires two eigenvectors but but a basis for E−1 contains one eigenvector, since dim(E−1) = 1. Therefore, a basis for the generalized eigenspace for K−1 is a 3-cycle. This is confirmed by the following dot diagram: 2 2 r1= 3 - rank(A + I)=3-2=1, r2= rank(A + I)-rank((A + I) )= 2-1=1, and r3= rank(A + I) - • 3 rank((A + I) ) = 1-0=1. • . In order to form a basis (3-cycle) for K−1, we now look • for a nonzero generalized eigenvector v such that (A + I)2(v) 6= 0 and (A + I)3(v) = 0. 2 0 0 0 3 Note (A + I)3 = 4 0 0 0 5 has a basis f1; x; x2g. We choose v = x2 and note that 0 0 0 2 3 2 3 2 3 3 2 3 0 0 2 0 2 0 −1 0 0 (A + I)2v = 4 0 0 0 5 4 0 5 = 4 0 5 and (A + I)v = 0 0 −2 5 4 0 5 = 0 0 0 1 0 0 0 0 1 2 0 3 4 −2 5 So, the required 3-cycle is f2; −2x; x2g. The required Jordan canonical basis is 0 2 2 0 0 3 2 −1 0 0 3 β0 = f(2; 0; 0); (0; −2; 0); (0; 0; 1)g. Hence, Q = 4 0 −2 0 5 and J = 4 0 −1 1 5. 0 0 1 0 0 −1 2 1 3 2 0 0 −1 1 −1 Check: Q = 4 0 − 2 0 5 and Q AQ = J. 0 0 −1 2 3 2 −4 2 2 2 − λ −4 2 2 6 −2 0 1 3 7 −2 0 − λ 1 3 (c) A = 6 7. jA−λIj = = (λ−2)2(λ−4)2 = 0. 4 −2 −2 3 3 5 −2 −2 3 − λ 3 −2 −6 3 7 −2 −6 3 7 − λ So λ = 2 and its multiplicity is 2, λ = 4 and its multiplicity is 2. E : 2 2 2 3 0 −4 2 2 1 0 0 −1 1 1 6 −2 −2 1 3 6 0 1 − − 7 A − 2I = 6 RREF 6 2 2 7. 4 −2 −2 1 3 −−−−! 4 0 0 0 0 5 −2 −6 3 5 0 0 0 0 A basis for E2 is f(0; 1; 2; 0); (2; 1; 0; 2)g. E4: 2 −2 −4 2 2 3 2 1 0 0 0 3 6 −2 −4 1 3 7 6 0 1 0 −1 7 A − 4I = 6 7 = 6 7. 4 −2 −2 −1 3 5 4 0 0 1 −1 5 −2 −6 3 3 0 0 0 0 A basis for E4 is f(0; 1; 1; 1)g. Now dim(K4) = 2= multiplicity of 4 in jA − 4Ij. 2 4 8 −4 −4 3 2 1 0 1 −1 3 6 4 4 0 −4 7 6 0 1 −1 0 7 (A − 4I)2 = 6 7 RREF 6 7.