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: 4 −2 0 (a) A1 = 0 2 0, 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 1 1 0 4 0 0 S = 0 1 0 a D = 0 2 0 . 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. √ 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: 0,75 0,20 0,10 P = 0,05 0,60 0,10 . 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 ∈ R then the states 2 ∞ after each election are represented by the vectors x0, P x0,P x0,...,P x0, where ∞ k P denotes limk→∞ P (if it exists). The matrix P can be diagonalized as 2 1 0 0 3 −2 −1 −1 1 P = S 0 0,6 0 S , where S = 3 1 1 . 0 0 0,55 1 1 0 Thus, 1 0 0 2 2 2 1 P ∞ = S 0 0 0 S−1 = S S−1 = 1 1 1 . ∗1 1∗ 6 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 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 ∈ Ker(P − 1 · I3).
3 Problem 5. Compute the Jordan normal form and the (generalized) eigenvectors for the matrix 1 1 1 B = 0 1 0 . −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:
1 − λ 1 1 2 pB(λ) = 0 1 − λ 0 = (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) = span{(1, 0, 1)T }. 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) )\Ker(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 span{(2, −1, 1)T }. We construct the matrix S from the (generalized) eigenvectors as
1 0 2 S = 0 0 −1 . 1 1 1
The inverse of S equals 1 2 0 −1 S = −1 −1 1 . 0 −1 0 Finally, we can express B using the Jordan normal form as
1 1 1 1 0 2 2 1 0 1 2 0 −1 B = 0 1 0 = 0 0 −1 0 2 0 −1 −1 1 = 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., 1 1 1 B = 0 1 0 . −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:
2 1 0 2 1 0 4 4 0 2 J = 0 2 0 0 2 0 = 0 4 0 . 0 0 1 0 0 1 0 0 1
Thus,
1 0 2 4 4 0 1 2 0 0 2 4 2 2 −1 B = SJ S = 0 0 −1 0 4 0 −1 −1 1 = 0 1 0 . 1 1 1 0 0 1 0 −1 0 −4 −1 8
Similarly for J 3:
2 1 0 4 4 0 8 12 0 3 J = 0 2 0 0 4 0 = 0 8 0 . 0 0 1 0 0 1 0 0 1
Thus,
1 0 2 8 12 0 1 2 0 −4 2 12 3 3 −1 B = SJ S = 0 0 −1 0 8 0 −1 −1 1 = 0 1 0 . 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. For example, notice that only the contents of the Jordan cells is affected when taking a power of a matrix in Jordan normal form. Finally, we compute the eleventh power of J:
211 11 · 210 0 2048 11264 0 11 J = 0 211 0 = 0 2048 0 . 0 0 1 0 0 1
Thus,
1 0 2 2048 11264 0 1 2 0 11 11 −1 B = SJ S = 0 0 −1 0 2048 0 −1 −1 1 1 1 1 0 0 1 0 −1 0 −9216 −7170 11264 = 0 1 0 . −11264 −9217 13312
5 Problem 7. Construct a matrix A with eigenvalues λ1 = 2 and λ2 = 3 and the corresponding T T eigenvectors x1 = (1, 2) and x2 = (2, 5) . Solution: Since the eigenvectors are linearly independent, the matrix A is diagonalizable. Specifically, it can be decomposed as A = SDS−1, where the columns of S are the eigenvectors and D is a diagonal matrix with the corresponding eigenvalues on its diagonal. We get
2 0 1 2 D = ,S = . 0 3 2 5
It remains to compute the inverse of S:
5 −2 S−1 = −2 1
and we compte A as
1 2 2 0 5 −2 −2 2 A = SDS−1 = = . 2 5 0 3 −2 1 −10 7
6