In this problem, if you give decimal answers then give at least three digits of accuracy beyond the decimal. The matrix has the following complex eigenvalues (give your answer as a comma separated list of complex numbers; use T for 1 and feel free to use a computer to solve the relevant quadratic equation): A = 1.65+1.548386257i, 1.65-1.548386257i Since A has non-real eigenvalues, it is not diagonalizable, but we can find a matrix C = section 5.5 of Lay): C= P = A cos(6) sin(8) -sin(8) cos(8) 2 -1.4 1.8 1.3 ]] and an invertible matrix P such that A = PCP-1 (I want you to cook up C, P as in a Further, we may factor Cas C = XY, where X is a matrix that scales by the positive real number radians, with →π/2 ≤ 0 ≤T/2) is and Y is rotation matrix whose counter-clockwise rotation angle (in This means that if we let B be the basis for R² consisting of the columns of P, then the B-matrix of A is C. So if we're willing to change our basis for R², then the linear transformation x →→ Ax really is just rotation followed by scaling!
In this problem, if you give decimal answers then give at least three digits of accuracy beyond the decimal. The matrix has the following complex eigenvalues (give your answer as a comma separated list of complex numbers; use T for 1 and feel free to use a computer to solve the relevant quadratic equation): A = 1.65+1.548386257i, 1.65-1.548386257i Since A has non-real eigenvalues, it is not diagonalizable, but we can find a matrix C = section 5.5 of Lay): C= P = A cos(6) sin(8) -sin(8) cos(8) 2 -1.4 1.8 1.3 ]] and an invertible matrix P such that A = PCP-1 (I want you to cook up C, P as in a Further, we may factor Cas C = XY, where X is a matrix that scales by the positive real number radians, with →π/2 ≤ 0 ≤T/2) is and Y is rotation matrix whose counter-clockwise rotation angle (in This means that if we let B be the basis for R² consisting of the columns of P, then the B-matrix of A is C. So if we're willing to change our basis for R², then the linear transformation x →→ Ax really is just rotation followed by scaling!
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![In this problem, if you give decimal answers then give at least three digits of accuracy beyond the decimal.
The matrix
has the following complex eigenvalues (give your answer as a comma separated list of complex numbers; use "i" for ✓-1 and feel free to use a computer to solve the relevant quadratic
equation):
λ = 1.65+1.548386257i, 1.65-1.548386257i
Since A has non-real eigenvalues, it is not diagonalizable, but we can find a matrix C =
section 5.5 of Lay):
C =
P =
cos(6)
A =
sin(8)
-sin(8)
cos(8)
[113]
1.8
Further, we may factor Cas C = XY, where X is a matrix that scales by the positive real number
radians, with —π/2 ≤0 ≤ π/2) is
and an invertible matrix P such that A = PCP-¹ (I want you to cook up C, P as in
and Y is rotation matrix whose counter-clockwise rotation angle (in
This means that if we let B be the basis for R² consisting of the columns of P, then the B-matrix of A is C. So if we're willing to change our basis for IR², then the linear transformation
x Ax really is just rotation followed by scaling!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3886e624-3ad7-4547-aeab-f36b3d1dd04a%2Ff049b82b-da2c-432a-8c03-8ea81e24310b%2Fdw7ejms_processed.png&w=3840&q=75)
Transcribed Image Text:In this problem, if you give decimal answers then give at least three digits of accuracy beyond the decimal.
The matrix
has the following complex eigenvalues (give your answer as a comma separated list of complex numbers; use "i" for ✓-1 and feel free to use a computer to solve the relevant quadratic
equation):
λ = 1.65+1.548386257i, 1.65-1.548386257i
Since A has non-real eigenvalues, it is not diagonalizable, but we can find a matrix C =
section 5.5 of Lay):
C =
P =
cos(6)
A =
sin(8)
-sin(8)
cos(8)
[113]
1.8
Further, we may factor Cas C = XY, where X is a matrix that scales by the positive real number
radians, with —π/2 ≤0 ≤ π/2) is
and an invertible matrix P such that A = PCP-¹ (I want you to cook up C, P as in
and Y is rotation matrix whose counter-clockwise rotation angle (in
This means that if we let B be the basis for R² consisting of the columns of P, then the B-matrix of A is C. So if we're willing to change our basis for IR², then the linear transformation
x Ax really is just rotation followed by scaling!
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