In Exercises 1-12, determine whether the given matrix A is diagonalizable. If A is diagonalizable, calculate A5 using the method of Example 2. 3. A = 5. A = 7. A = 8. A = [ -3 2 -2 1 1 0 10 2 ] 3 -2 -4 8-7-16 -3 3 7 -1 -1 -4 -8-3-16 12 7 4. A = 4.7 Si 13 [6³] 01 6. A = [17] A= 01
In Exercises 1-12, determine whether the given matrix A is diagonalizable. If A is diagonalizable, calculate A5 using the method of Example 2. 3. A = 5. A = 7. A = 8. A = [ -3 2 -2 1 1 0 10 2 ] 3 -2 -4 8-7-16 -3 3 7 -1 -1 -4 -8-3-16 12 7 4. A = 4.7 Si 13 [6³] 01 6. A = [17] A= 01
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![In Exercises 1-12, determine whether the given matrix
A is diagonalizable. If A is diagonalizable, calculate A³
using the method of Example 2.
3. A =
5. A =
7. A =
8. A =
[
-3 2
-2 1
2]
3 -2 -4
8-7-16
-3 3 7
-1 -1 -4
-8-3-16
12
1 0
10 2
7
4. A =
4.7 Si
13
[8³]
01
6. A = [-17]
01](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7bf79e7e-db7e-4b5c-a300-38b09014f3cb%2F95408ed5-5f0f-4f16-bfd5-34aa39179de9%2F5bo428a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In Exercises 1-12, determine whether the given matrix
A is diagonalizable. If A is diagonalizable, calculate A³
using the method of Example 2.
3. A =
5. A =
7. A =
8. A =
[
-3 2
-2 1
2]
3 -2 -4
8-7-16
-3 3 7
-1 -1 -4
-8-3-16
12
1 0
10 2
7
4. A =
4.7 Si
13
[8³]
01
6. A = [-17]
01
![HE
EXAMPLE 1
Solution
Therefore, S-¹AS has the form
S-¹AS =
0 0 0
and we have shown that if A has n linearly independent eigenvectors, then A is similar
to a diagonal matrix.
λ₁ 0
0
22
0
0
Now suppose that C-¹AC = D, where C is nonsingular and D is a diagonal matrix.
Let us write C and D in column form as
C = [C1, C₂, ..., C₁] and D= [d₁e₁, d₂e2, ..., dnen].
From C-¹AC = D, we obtain AC = CD, and we write both of these in column form as
AC = [AC₁, AC2, ..., ACn]
CD= [d₁ Ce₁, d₂Ce2, ..., dn Cen].
But since Cek = Ck for k = 1, 2, ..., n, we see that AC = CD implies
ACk = dk Ck, k = 1, 2, ...,n.
Since C is nonsingular, the vectors C₁, C2, ...,Cn are linearly independent (and in
particular, no Ck is the zero vector). Thus the diagonal entries of D are the eigenvalues
of A, and the column vectors of C are a set of n linearly independent eigenvectors.
EXAMPLE 2
Note that the proof of Theorem 19 gives a procedure for diagonalizing an (n × n)
matrix A. That is, if A has n linearly independent eigenvectors u₁, U2, ..., un, then the
matrix S = [u₁, u₂, ..., u₂] will diagonalize A.
Solution
Show that A is diagonalizable by finding a matrix S such that S-¹AS = D:
5 -6
3 -4
Forming S = [u₁, u₂], we obtain
=
It is easy to verify that A has eigenvalues λ₁ = 2 and ₂ = -1 with corresponding
eigenvectors
A =
0
0
23
--8---8-
and u₂ =
0
0
0
:
an
5-6 -44]
-[21]
S =
2
Next, forming S-¹(AS), we obtain
S-¹ (AS) =
As a check on the calculations, we form S-¹AS. The matrix AS is given by
5
-6
2 1
AS = [3] [} } ] [3]
=
-4
1
=D;
4.7 Similarity Transformations and Diagonalization
−1 2
D¹0
S-1
4
[][][]
Use the result of Example 1 to calculate A¹0, where
5 -6
3 -4
- [ 20⁰
Hence A¹0 = SD¹0 S-¹ is given by
4 10.
A =
A¹
0
(-1) ¹⁰
=
As was noted in Eq. (2), D¹0 = S-¹A¹0 S. Therefore, A ¹⁰ SD10 S-1. Now by
Example 1,
>]-[1024]
2047 -2046
1023-1022
2 0
0 -1
= D.
329
Sometimes complex arithmetic is necessary to diagonalize a real matrix.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7bf79e7e-db7e-4b5c-a300-38b09014f3cb%2F95408ed5-5f0f-4f16-bfd5-34aa39179de9%2F3lqfuz8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:HE
EXAMPLE 1
Solution
Therefore, S-¹AS has the form
S-¹AS =
0 0 0
and we have shown that if A has n linearly independent eigenvectors, then A is similar
to a diagonal matrix.
λ₁ 0
0
22
0
0
Now suppose that C-¹AC = D, where C is nonsingular and D is a diagonal matrix.
Let us write C and D in column form as
C = [C1, C₂, ..., C₁] and D= [d₁e₁, d₂e2, ..., dnen].
From C-¹AC = D, we obtain AC = CD, and we write both of these in column form as
AC = [AC₁, AC2, ..., ACn]
CD= [d₁ Ce₁, d₂Ce2, ..., dn Cen].
But since Cek = Ck for k = 1, 2, ..., n, we see that AC = CD implies
ACk = dk Ck, k = 1, 2, ...,n.
Since C is nonsingular, the vectors C₁, C2, ...,Cn are linearly independent (and in
particular, no Ck is the zero vector). Thus the diagonal entries of D are the eigenvalues
of A, and the column vectors of C are a set of n linearly independent eigenvectors.
EXAMPLE 2
Note that the proof of Theorem 19 gives a procedure for diagonalizing an (n × n)
matrix A. That is, if A has n linearly independent eigenvectors u₁, U2, ..., un, then the
matrix S = [u₁, u₂, ..., u₂] will diagonalize A.
Solution
Show that A is diagonalizable by finding a matrix S such that S-¹AS = D:
5 -6
3 -4
Forming S = [u₁, u₂], we obtain
=
It is easy to verify that A has eigenvalues λ₁ = 2 and ₂ = -1 with corresponding
eigenvectors
A =
0
0
23
--8---8-
and u₂ =
0
0
0
:
an
5-6 -44]
-[21]
S =
2
Next, forming S-¹(AS), we obtain
S-¹ (AS) =
As a check on the calculations, we form S-¹AS. The matrix AS is given by
5
-6
2 1
AS = [3] [} } ] [3]
=
-4
1
=D;
4.7 Similarity Transformations and Diagonalization
−1 2
D¹0
S-1
4
[][][]
Use the result of Example 1 to calculate A¹0, where
5 -6
3 -4
- [ 20⁰
Hence A¹0 = SD¹0 S-¹ is given by
4 10.
A =
A¹
0
(-1) ¹⁰
=
As was noted in Eq. (2), D¹0 = S-¹A¹0 S. Therefore, A ¹⁰ SD10 S-1. Now by
Example 1,
>]-[1024]
2047 -2046
1023-1022
2 0
0 -1
= D.
329
Sometimes complex arithmetic is necessary to diagonalize a real matrix.
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