(b) Nonsingular matrir P such that D = P-1AP is diagonal. 1 Using the vectors in the bases for E and E4, we have P = 1 -2 How ?22 1/3 2/3 1/3 -1/3 hence P-1 and
(b) Nonsingular matrir P such that D = P-1AP is diagonal. 1 Using the vectors in the bases for E and E4, we have P = 1 -2 How ?22 1/3 2/3 1/3 -1/3 hence P-1 and
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
100%
I have labeled where the confusion of the solution is. I don't need an answer to the problem (also attached). I just need an explanation on how did we get this matrix that I've added a note in red? from where did the [ 2/3 1/3 ... ] came?? what was done in order to get it? Thank you :)
Transcribed Image Text:-[-
2
-1
Let A
3
(a) Find the eigenvalues and corresponding eigenvectors of A.
(b) Find a nonsingular matrix P such that D = P-' AP is diagonal.
![[
2 -1
1. Let A
-2
Observe first that
t – 2
1
Aa(t) = |
-
= (t – 2)(t – 3) – 2 = t2 – 5t + 6 – 2 = t? – 5t + 4 = (t – 4)(t – 1).
%3D
|
t – 3
(a) Eigenvalues and corresponding eigenvectors of A.
Since A4(t) = (t – 4)(t – 1), the eigenvalues areA= 1, 1 = 4.
-
-
[
1
with echelon
-1
The eigenspace for A = 1 is the null space of the matrix
2 -2
-1
1
form
A vector
is in E if and only if -x + y = 0, so y = x. Hence
0 0
{[:]
B, - {{:}
E1
| x
ER
and a basis for E1 is B1
2 1
The eigenspace for A= 4 is the null space of the matrix
with echelon form
1
A vector
is in E4 if and only if 2x + y =
0, so y = -2x. Hence
0 0
{[ }}
1
E4
|x ER
-2x
and a basis for E4 is B4
-2
(b) Nonsingular matrir P such that D = P-'AP is diagonal.
[
1
1
Using the vectors in the bases for E1 and E4, we have P
1 -2
How 222
2/3
1/3
hence P-1
and
1/3 -1/3
2/3
1/3
2
-1
1
D = P-'AP
1/3 -1/3 ]
1 -2
-2
3
1 0
0 4
2/3
1/3
1
4
= D.
1/3 -1/3
1 -8
(c) f(A), where f(t) = t – 4t3 + 3t – 7.
We have D = P-'AP, so A = PDP-l and f(A) = f(PDP-1) = Pf(D)P-1.
(See diagonal factorization, text page 298.)
Next,
4
1 0
0 4
1
1
1 0
0 4
1 0
+3
0 4
1 0
0 1
f(D)
-7
3
+
0 -256
1
-4
0 256
+
0 12
-7
-7 0
0 5
1- 4+3 - 7
256 – 256 + 12 – 7
and so
1/3
[
-7 0
0 5
2/3
f(A) = Pf(D)P-1
1 -2
1/3 -1/3
-7
5
2/3
1/3
-3
-4
= f(A).
-7 -10
1/3 -1/3
-8
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6214fda8-f992-4e88-b320-5195339361f5%2Ff50038f3-9752-4639-8df3-83f466e5d579%2F8ko6x5h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[
2 -1
1. Let A
-2
Observe first that
t – 2
1
Aa(t) = |
-
= (t – 2)(t – 3) – 2 = t2 – 5t + 6 – 2 = t? – 5t + 4 = (t – 4)(t – 1).
%3D
|
t – 3
(a) Eigenvalues and corresponding eigenvectors of A.
Since A4(t) = (t – 4)(t – 1), the eigenvalues areA= 1, 1 = 4.
-
-
[
1
with echelon
-1
The eigenspace for A = 1 is the null space of the matrix
2 -2
-1
1
form
A vector
is in E if and only if -x + y = 0, so y = x. Hence
0 0
{[:]
B, - {{:}
E1
| x
ER
and a basis for E1 is B1
2 1
The eigenspace for A= 4 is the null space of the matrix
with echelon form
1
A vector
is in E4 if and only if 2x + y =
0, so y = -2x. Hence
0 0
{[ }}
1
E4
|x ER
-2x
and a basis for E4 is B4
-2
(b) Nonsingular matrir P such that D = P-'AP is diagonal.
[
1
1
Using the vectors in the bases for E1 and E4, we have P
1 -2
How 222
2/3
1/3
hence P-1
and
1/3 -1/3
2/3
1/3
2
-1
1
D = P-'AP
1/3 -1/3 ]
1 -2
-2
3
1 0
0 4
2/3
1/3
1
4
= D.
1/3 -1/3
1 -8
(c) f(A), where f(t) = t – 4t3 + 3t – 7.
We have D = P-'AP, so A = PDP-l and f(A) = f(PDP-1) = Pf(D)P-1.
(See diagonal factorization, text page 298.)
Next,
4
1 0
0 4
1
1
1 0
0 4
1 0
+3
0 4
1 0
0 1
f(D)
-7
3
+
0 -256
1
-4
0 256
+
0 12
-7
-7 0
0 5
1- 4+3 - 7
256 – 256 + 12 – 7
and so
1/3
[
-7 0
0 5
2/3
f(A) = Pf(D)P-1
1 -2
1/3 -1/3
-7
5
2/3
1/3
-3
-4
= f(A).
-7 -10
1/3 -1/3
-8
1
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
