Suppose A is a 2 x 2 matrix that has a repeated eigenvalue A with linearly independent eigenvectors₁ and 2. Note that sinceA(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 canscale them using a and 8 and add the scaled vectors together to get any nonzero vector 3 in theplane. (Think of u and ₂ as arrows in the plane and how you could scale and combine them as in thefigure below.)(You can write lambda for A.)BUZ▶ HintV₂So in fact every nonzero vector 3 is an eigenvector for A! In particular, the vectorsH and Heigenvectors for the eigenvalue A. This tells us that if A is a 2 x 2 matrix that has a repeatedeigenvalue A with linearly independent eigenvectors then A has to have the following form:V3 = αν + βυare

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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

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