Use the result of the Example above to compute Al0 (for A the matrix from the Example). Hint: consider (PBP1)100 What is Blo0
Use the result of the Example above to compute Al0 (for A the matrix from the Example). Hint: consider (PBP1)100 What is Blo0
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
please send handwritten solution for Q 6.1
![Read the following carefully, and then answer the questions that follow.
We say that two nxn matrices are cousins, if there is some invertible matrix P so that
A = PBP-1
"Notice that if A and B are cousins, then B = PAP = (PA(P). Thus, the definition is symmetric" (ie the order we say it
doesn't matterk if A and B are csasins then B and A are coesins.
One of the first things you might notice about the notion of cousins, it that it is clearly connected to
diagonalizability:
Theorem 1
A matrix A is diagonalizable if and only if there is a diagonal matrix D so that A and D are
cousins.
One might think from the definition that the main use of the definition of cousin matrices is to
reframe diagonalizability. However, it actually allows us to generalize it in useful ways - in the
following example and the first question below, we'll see matrices A and B which are cousins,
but where neither of which is diagonal, and for which techniques from our study of diagonalizable
matrices can be applied.
-7 -5]
Let A=
10
and let B= Ry/2 =
7
be the matrix of rotation by 37/2.
Then with P=
, if you compute PBP1, you will see that the end result is A.
Thus, A and B are cousins.
The next theorem shows us that cousins share a lot of important properties in common:
Theorem 2
If A and B are cousins, then:
1. det(A) = det(B), and
2. CA(x) = cg(x) (and hence A and B have the same eigenvalues.)
Proof: Let A = PBP for some invertible matrix P.
1. det(A) = det(PBP-1) = det(P) - det(B) - det(P') = det(B) (using that det (P) = 1/det(P-1).)
2. CA(x) = det(xl - A) = det (xl – PBP-1). But we can rewrite I as PIP-1 and substitute
this in; so, continuing we have: .= det(xPIPp-1- PBP-1).
But then you can check that XPIP – PBP- = P(xl – B)P, by multiplying out the
right-hand side. This allows us to continue with:
. = det (P(xl – B)P) = det(P) - det (xl – B) - det(P-1) = det(xl – B) = cs(x)
Continues on the next page.
MAT223 - Winter 2022
Midterm 2
6.1
Use the result of the Example above to compute Al00 (for A the matrix from the Example).
Hint: consider (PBP )100, What is Bl00,
6.2 (
cousins. Hint: part of Theorem 2 should be useful.
) True or False: If A and B are nx n matrices so that det(A) = det(B), then A and B are
63
• *) True or False: If A and B are cousins and v is an eigenvector of A, then v is an
eigenvector of B.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa39a0292-955d-43cc-aa67-690c8fee190d%2F7071cb32-a0eb-4362-8ae9-20322274db9f%2F9sgsep_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Read the following carefully, and then answer the questions that follow.
We say that two nxn matrices are cousins, if there is some invertible matrix P so that
A = PBP-1
"Notice that if A and B are cousins, then B = PAP = (PA(P). Thus, the definition is symmetric" (ie the order we say it
doesn't matterk if A and B are csasins then B and A are coesins.
One of the first things you might notice about the notion of cousins, it that it is clearly connected to
diagonalizability:
Theorem 1
A matrix A is diagonalizable if and only if there is a diagonal matrix D so that A and D are
cousins.
One might think from the definition that the main use of the definition of cousin matrices is to
reframe diagonalizability. However, it actually allows us to generalize it in useful ways - in the
following example and the first question below, we'll see matrices A and B which are cousins,
but where neither of which is diagonal, and for which techniques from our study of diagonalizable
matrices can be applied.
-7 -5]
Let A=
10
and let B= Ry/2 =
7
be the matrix of rotation by 37/2.
Then with P=
, if you compute PBP1, you will see that the end result is A.
Thus, A and B are cousins.
The next theorem shows us that cousins share a lot of important properties in common:
Theorem 2
If A and B are cousins, then:
1. det(A) = det(B), and
2. CA(x) = cg(x) (and hence A and B have the same eigenvalues.)
Proof: Let A = PBP for some invertible matrix P.
1. det(A) = det(PBP-1) = det(P) - det(B) - det(P') = det(B) (using that det (P) = 1/det(P-1).)
2. CA(x) = det(xl - A) = det (xl – PBP-1). But we can rewrite I as PIP-1 and substitute
this in; so, continuing we have: .= det(xPIPp-1- PBP-1).
But then you can check that XPIP – PBP- = P(xl – B)P, by multiplying out the
right-hand side. This allows us to continue with:
. = det (P(xl – B)P) = det(P) - det (xl – B) - det(P-1) = det(xl – B) = cs(x)
Continues on the next page.
MAT223 - Winter 2022
Midterm 2
6.1
Use the result of the Example above to compute Al00 (for A the matrix from the Example).
Hint: consider (PBP )100, What is Bl00,
6.2 (
cousins. Hint: part of Theorem 2 should be useful.
) True or False: If A and B are nx n matrices so that det(A) = det(B), then A and B are
63
• *) True or False: If A and B are cousins and v is an eigenvector of A, then v is an
eigenvector of B.
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