Homework 4
Ding Zhao
24-677 Special Topics: Linear Control Systems
Due: Sept 25, 2019, 08:30 am. Submit within deadline.
• At the beginning of each question you will find the key words for the knowledge that the exercise will help you to practice.
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• Note that we do not have programming questions in Homework 4.
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Exercise 1. eigenvalues (practice solving the eigenvalues manually instead of using Python, the same for the entire homework 4)
2 0 0 2 −2 3 ForA1=0 2 0A2=1 1 1
0 0 1 1 3 −1
Find the eigenvalues of A1 and A2, such that the eigenvalues are in ascending order
(λ1 ≤λ2 ≤λ3)
Solution:
1. For A1, eigenvalues are λ1 = 1,λ2 = 2, λ3 = 2.
2. For A2, eigenvalues are λ1 = −2,λ2 = 1, λ3 = 3.
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Exercise 2. Singular values.
2 1 −1 0 1
ForA1=42A2= 2 −10
1. Find the eigenvalues of A1T A1. 2. Find the eigenvalues of A1A1T . 3. Find the singular values of A1 4. Find the singular values of A2
Write eigenvalues in ascending order.
Solution:
T 20 10
1. A1 A1 = 10 5 .
Trace tr(A1) = 20+5 = 25 = λ1 +λ2 and , det(A1) = 20×5−10×10 = 0 = λ1 ×λ2
soλ1 =0,λ2 =25.
2. The eigenvalues for A1T A1 and its transpose A1A1T are the same. Therefore, for
A1A1T ,still λ1 = 0, λ2 = 25.
3. WecanuseeitherA1TA1 orA1A1T: σ1 =√λ2 =√25=5,σ2 =√λ1 =√0=0
T −1 0 1−1 2 2 −2 4.A2A2= 2−100−1=−25
10 ∆(λ)=(λ−2)(λ−5)−4=λ2 −7λ+6=(λ−1)(λ−6)
the eigenvalues of A2T A2 are 6 and 1, thus the singular values of √
A2are 6,1
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Exercise 3. Characteristic polynomial
−α1 −α2 −α3 −α4 A=1000 0 1 0 0
0010 A matrix in this form is called a companion form matrix
1. Calculate the characteristic polynomial in terms of αi, i = 1, 2, 3, 4.
2. Derive the normalized eigenvector v of A in terms of its eigenvalues λ, s.t. ∥v∥2 = 1.
(As the solution is analytical, we will check it manually. No AutoGrader for this question.)
Solution:
The characteristic polynomial is
λ4 +α1λ3 +α2λ2 +α3λ+α4 = 0
now let us assume that we have an eigenvector v for eigenvalue λ as,
v1 v = v2
v 3 v4
for (A − λI)v = 0 we can get a set of linear equations
(−α1 −λ)v1 −α2v2 −α3v3 −α4v4 = 0 v1 − λv2 = 0
v2 − λv3 = 0
v3 − λv4 = 0
which we can see that if we use
it satisfies the set of linear equations.
(−α1 −λi)v1 −α2v2 −α3v3 −α4v4 = 0
Substituting v in the above equation, the characteristic polynomial becomes: 4
λ3 v = aλ2
λ 1 1
λ4 +α1λ3 +α2λ2 +α3λ+α4 = 0 The corresponding eigen vector for λ
for ∥v∥2 = 1,
λ3 v = aλ2
λ 1 1
a2(1+λ2 +λ4 +λ6) = 1 1
∴a= (1+λ2 +λ4 +λ6) therefore the corresponding eigenvector is
λ3 1 λ2
v= (1+λ2 +λ4 +λ6)λ1 1
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Exercise 4. Jordan form, decomposition
Find the Jordan-form of the following matrices.
148 0 1 0 A1=0 2 0 A2=0 0 1
0 0 3 −2 −4 −3
10−1 043 A3=0 1 0 A4=0 20 16
0 0 2 0 −25 −20
(Write the Jordan form such that eigenvalues should be in ascending order of their absolute
values. The absolute value of a complex number is defined as |a + bi| = Solution:
Follow the procedure descried in the class:
1 0 0 A1 is Type-I matrix, therefore the Jordan form is 0 2 0.
003 −10 0
A2 is Type-I, Jordan form is 0 −1−i 0 . 0 0 −1+i
1 0 0 A3 is Type-II1, Jordan form is 0 1 0.
002 0 1 0
A4 is Type-II2, Jordan form is 0 0 1. 000
√
a2 + b2)
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Exercise 5. Function of matrices Given
110 A=0 0 1
001
Find A10,A103, and eAt (when submitting to the Autograder, substitute t with 1).
Solution:
To calculate A10:
The characteristic polynomial of A is ∆(λ) = det(λI − A) = λ(λ − 1)2. Let h(λ) = β0 + β1λ + β2λ2.
Substitute the eigenvalues.
For λ = 0, 010 = β0.
For λ = 1,f(1) = h(1) :
110 = β0 + β1 + β2.
We also need to use the derivative: f′(1) = h′(1) :
10 · 19 = β1 + 2β2.
Thenwehaveβ0 =0,β1 =−8,β2 =9
Now, as A10 = −8A + 9A2
110 111 119 =−80 0 1+90 0 1=0 0 1
001 001 001
To calculate A103:
0103 = β0
β0 =01103 =β0 +β1 +β2 103·1102 =β1 +2β2
β2 = 102
⇒ β1 = −101
A103 = −101A + 102A2
To compute eAt :
et =β0 +β1 +β2 ⇒β1 =2et −tet −2tet =β1 +2β2 β2 = tet − et + 1
1 1 1
001 001 001
1 1 0
=−1010 0 1+1020 0 1=0 0 1
1 1 102
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eAt =β0I+β1A+β2A2
100 110 111
=0 1 0+(2et−et−2)0 0 1+(2et−et+1)0 0 1 001 001 001
et et−1 tet−et+1 =0 1 et−1
0 0 et When t = 1,
ee−11 eAt = 0 1 e − 1
00e
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Exercise 6. Diagonalization Diagonlize the following matrix.
Such that A = MΛM−1 values.)
200 A=121
−1 0 1
(Write the diagonal matrix Λ such that its eigenalues are in ascending order of their absolute
Solution:
Steps to diagonalize the matrix A:
Step 1: Find the eigenvalues of A
2−λ00
det(A − λI) = det 1 2 − λ 1 = (2 − λ)2(1 − λ) = 0
−1 0 1−λ
Eigenvalues of A : λ = 1 and λ = 2
Step 2. Find three linearly independent eigenvectors of A.
By solving (A − λI)x = 0, for each value of λ, we obtain the following: Basis for λ = 1 :
Basis for λ = 2 :
0 v1 = −1
1 0
v2 = 1 0
−1 v3= 0
1 Step 3: Construct M from the vectors in step 2.
0 0 −1 M=−11 0
101 9
Step 4 : Construct Λ from the corresponding eigenvalues.
100 Λ=0 2 0
002
Such that A = MΛMT Λ is the diagonalized Matrix. For checking purpose only:
ΛM = MΛ
200 00−1 0 0−2 ΛM= 121−11 0=−12 0
−101 101 102
00−1100 0 0−2 MΛ=−11 0020=−12 0
101002102
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