NMAI058 – Linear algebra 2 Tutorial 5 Diagonalization and Jordan normal form Date: March 29, 2021 TA: Pavel Hubáček Problem 1. Decide and justify whether the following matrices are diagonalizable: 04 −2 01 (a) A1 = @0 2 0A, 6 −5 1 0 1 (b) A = . 2 −2 2 Solution: (a) A matrix is diagonalizable if it can be expressed as A = SDS−1 for some regular matrix S and a diagonal matrix D. Equivalently, A must have n linearly independent eigenvectors and we test this equivalent characteri- zation. We compute the eigenvalues of A1. The characteristic polynomila of A1 is pA1 (λ) = (4 − λ)(2 − λ)(1 − λ): Thus, the eigenvalues are λ1 = 4; λ2 = 2 a λ3 = 1. The corresponding eigenvectors are found by solving the system (A1 − λI) = 0, where we sub- T stitute the above eigenvalues for λ. We get eigenvectors x1 = c · (1; 0; 2) , T T x2 = c · (1; 1; 1) , and x3 = c · (0; 0; 1) . The eigenvectors are linearly inde- pendent (they have to be, since each corresponds to a distinct eigenvalue). −1 Thus, the matrix A1 is diagonalizable and we can express it as A1 = SDS , where 01 1 01 04 0 01 S = @0 1 0A a D = @0 2 0A : 2 1 1 0 0 1 (b) We proceed analogously to matrix A1. The only (insignificant) difference is that we get complex eigenvalues. The characteristic polynomila of A2 2 is pA2 (λ) = λ − 2λ + 2. The roots of pA2 are 1 + i and 1 − i and the corresponding eigenvectors are (1; 1 + i)T and (1; 1 − i)T . Thus, the matrix −1 A2 is diagonalizable and we can express it as A2 = SDS , where 1 1 1 + i 0 S = a D = : 1 + i 1 − i 0 1 − i 1 0 1 Problem 2. Prove that the matrix B = ( 0 0 ) is not diagonalizable. Solution: The matrix B has 0 as eigenvalue with algebraic multiplicity two. Thus, if it were diagonalizable then it would had been similar to the zero matrix. Equivalently, for some regular S, 0 6= B = S0S−1 = 0; which is a contradiction. p Problem 3. For the diagonalizable matrix C below, compute C3 and B = C (i.e., a matrix B such that B2 = C). −11 30 C = −10 24 Solution: Let C = SDS−1 for a diagonal matrix D. Note that Ck = SDS−1k = SDkS−1: 1 −1 1 −1 −1 1 Similarly, SD 2 S · SD 2 S = SDS for the diagonal matrix D 2 containing on the diagonal the square roots of the corresponding diagonal elements in D, 1 2 p i.e., Di;i = Di;i. 3 3 −1 1 −1 The third power C can be computed as SD S and the square root as SD 2 S . We find the decomposition SDS−1 of C: −11 30 3 2 9 0 −1 2 C = = = SDS−1; −10 24 2 1 0 4 2 −3 which gives 3 2 729 0 −1 2 −1931 3990 SD3S−1 = = = C3; 2 1 0 64 2 −3 −1330 2724 1 −1 3 2 3 0 −1 2 −1 6 1 SD 2 S = = = C 2 : 2 1 0 2 2 −3 −2 6 Problem 4. There are three political parties in the city of Celestopolis: Anarchists (A), Bu- reaucrats (B), and Corruptionists (C). It was observed that the electorate be- haves according to the following principles: 75 % of the voters of party A votes always A but 5 % of their voters switch to party B and 20 % to party C. Among those who previously voted for party B, 20 % switch to party A and another 20 % switch to party C. Finally, 80 % of those who previously voted C will vote the same and the rest is evenly split among the voters of A and B. How will the distribution of the representatives on the city council evolve over time? 2 Solution: The behavior of the electorate is summarized in the diagram below: 0,20 0,75 A 0,05 B 0,60 0,10 0,20 0,20 0,10 C 0,80 The vertices of the graph represent the possible states (i.e., the parties), the ed- ges capture the transitions between the states (i.e., becoming a voter of another party), and the weights on the edges the respective probabilities of each tran- sition. In order to characterize the election process, we construct the transition matrix P , where pij is the probability of the transition from state j to state i: 00;75 0;20 0;101 P = @0;05 0;60 0;10A : 0;20 0;20 0;80 Thus, the elements of the first column of P correspond to the probabilities of transitioning from state A to state A (75 % votes again the Anarchists), to state B(5 % will vote Bureaucrats), and to state C (20 % will vote Corruptionists). 3 If the original state of the city council is given by a vector x0 2 R then the states 2 1 after each election are represented by the vectors x0; P x0;P x0;:::;P x0, where 1 k P denotes limk!1 P (if it exists). The matrix P can be diagonalized as 0 1 0 2 1 1 0 0 3 −2 −1 −1 1 P = S @0 0;6 0 A S ; where S = @ 3 1 1 A : 0 0 0;55 1 1 0 Thus, 01 0 01 02 2 21 1 P 1 = S 0 0 0 S−1 = S S−1 = 1 1 1 : @ A ∗1 1∗ 6 @ A 0 0 0 3 3 3 Let e = (1; 1; 1)T be the all-one vector. The limit state corresponds to the vector 1 1 T T T T P x0 = 6 (2e x0; e x0; 3e x0) and the electoral support will eventually have the ratio 2 : 1 : 3. Note that P is a positive matrix and, thus, the limit state is captured by the eigenvector v = S∗1 corresponding to the eigenvalue 1. Therefore, we could find the limit state directly by finding a nonzero vector v 2 Ker(P − 1 · I3). 3 Problem 5. Compute the Jordan normal form and the (generalized) eigenvectors for the matrix 0 1 1 11 B = @ 0 1 0A : −1 0 3 Solution: We need to find a regular matrix S and a matrix J in the Jordan normal form such that B = SJS−1. We proceed similarly to when trying to diagonalize the matrix. The characteristic polynomial of B is: 01 − λ 1 1 1 2 pB(λ) = @ 0 1 − λ 0 A = (1 − λ)(2 − λ) : −1 0 3 − λ The eigenvalues are λ1 = 2 and λ2 = 1. The kernel of (B − 2I) has dimension one. Specifically, Ker(B − 2I) = spanf(1; 0; 1)T g. The eigenvalue λ = 2 has algebraic multiplicity two (as a root of the characteristic polynomial) but its geometric multiplicity is one (it has only a single corresponding linearly inde- pendent eigenvector). Thus, matrix B does not have three linearly independent eigenvectors. Next, we proceed with finding the generalized eigenvectors corresponding to λ1. We compute Ker ((B − 2I)2) = span (1; 0; 0)T ; (0; 0; 1)T and pick a generalized 2 T eigenvector x2 from Ker((B−2I) )nKer(B−2I), e.g., x2 = (0; 0; 1) . We compute the eigenvector x1 as T x1 = (B − 2I)x2 = (1; 0; 1) : The system (A − 1I)x = 0 has solution set spanf(2; −1; 1)T g. We construct the matrix S from the (generalized) eigenvectors as 01 0 2 1 S = @0 0 −1A : 1 1 1 The inverse of S equals 0 1 2 01 −1 S = @−1 −1 1A : 0 −1 0 Finally, we can express B using the Jordan normal form as 0 1 1 11 01 0 2 1 02 1 01 0 1 2 01 −1 B = @ 0 1 0A = @0 0 −1A @0 2 0A @−1 −1 1A = SJS : −1 0 3 1 1 1 0 0 1 0 −1 0 Problem 6. Compute the second, third, and eleventh power of the matrix B from the previous problem, i.e., 0 1 1 11 B = @ 0 1 0A : −1 0 3 4 Solution: Using the Jordan normal form, we can proceed similarly to diagonalizable ma- trices. We get B2 = (SJS−1)(SJS−1) = SJ 2S−1 and B3 = SJ 3S−1. It remains to find the second and third power of the matrix J. First, we compute the second power of J: 02 1 01 02 1 01 04 4 01 2 J = @0 2 0A @0 2 0A = @0 4 0A : 0 0 1 0 0 1 0 0 1 Thus, 01 0 2 1 04 4 01 0 1 2 01 0 0 2 41 2 2 −1 B = SJ S = @0 0 −1A @0 4 0A @−1 −1 1A = @ 0 1 0A : 1 1 1 0 0 1 0 −1 0 −4 −1 8 Similarly for J 3: 02 1 01 04 4 01 08 12 01 3 J = @0 2 0A @0 4 0A = @0 8 0A : 0 0 1 0 0 1 0 0 1 Thus, 01 0 2 1 08 12 01 0 1 2 01 0 −4 2 121 3 3 −1 B = SJ S = @0 0 −1A @0 8 0A @−1 −1 1A = @ 0 1 0 A : 1 1 1 0 0 1 0 −1 0 −12 −5 20 Even though it is slightly more involved than for diagonalizable matrices, compu- tation of powers using Jordan normal form is less complex then a direct compu- tation.