Suppose A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors and . Note that since A(avi + 352) = Aavi + Αβύ, = αλΰι + βαΰ, = αλύ, + βλΰg = λ(αύ1 + B52) the vector au + Bu₂ is also an eigenvector for A. Since ₁ and 2 are linearly independent we can scale them using a and 8 and add the scaled vectors together to get any nonzero vector 3 in the plane. (Think of ₁ and 2 as arrows in the plane and how you could scale and combine them as in the figure below.) BEZ V3 = avi + βV So in fact every nonzero vector 3 is an eigenvector for A! In particular, the vectors H eigenvectors for the eigenvalue A. This tells us that if A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors then A has to have the following form: 91 and H are
Suppose A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors and . Note that since A(avi + 352) = Aavi + Αβύ, = αλΰι + βαΰ, = αλύ, + βλΰg = λ(αύ1 + B52) the vector au + Bu₂ is also an eigenvector for A. Since ₁ and 2 are linearly independent we can scale them using a and 8 and add the scaled vectors together to get any nonzero vector 3 in the plane. (Think of ₁ and 2 as arrows in the plane and how you could scale and combine them as in the figure below.) BEZ V3 = avi + βV So in fact every nonzero vector 3 is an eigenvector for A! In particular, the vectors H eigenvectors for the eigenvalue A. This tells us that if A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors then A has to have the following form: 91 and H are
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.7: The Inverse Of A Matrix
Problem 31E
Related questions
Question
![Suppose A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors
₁ and 2. Note that since
A(αΰn + Büy) = Aav + Αβύ, = «Αύ, + BAΰg = αλής + βλύ, = \(αΰ1 + βύη)
the vector au + Bu is also an eigenvector for A. Since ₁ and 2 are linearly independent we can
scale them using a and 8 and add the scaled vectors together to get any nonzero vector 3 in the
plane. (Think of u and ₂ as arrows in the plane and how you could scale and combine them as in the
figure below.)
(You can write lambda for A.)
BUZ
▶ Hint
V₂
So in fact every nonzero vector 3 is an eigenvector for A! In particular, the vectors
H and H
eigenvectors for the eigenvalue A. This tells us that if A is a 2 x 2 matrix that has a repeated
eigenvalue A with linearly independent eigenvectors then A has to have the following form:
V3 = αν + βυ
are](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4100c518-1a8f-4898-88a2-d0f1a7669694%2F94c8e5fa-5e0a-400a-b0f7-b44cf6ba285b%2Fun5zakwj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors
₁ and 2. Note that since
A(αΰn + Büy) = Aav + Αβύ, = «Αύ, + BAΰg = αλής + βλύ, = \(αΰ1 + βύη)
the vector au + Bu is also an eigenvector for A. Since ₁ and 2 are linearly independent we can
scale them using a and 8 and add the scaled vectors together to get any nonzero vector 3 in the
plane. (Think of u and ₂ as arrows in the plane and how you could scale and combine them as in the
figure below.)
(You can write lambda for A.)
BUZ
▶ Hint
V₂
So in fact every nonzero vector 3 is an eigenvector for A! In particular, the vectors
H and H
eigenvectors for the eigenvalue A. This tells us that if A is a 2 x 2 matrix that has a repeated
eigenvalue A with linearly independent eigenvectors then A has to have the following form:
V3 = αν + βυ
are
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