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
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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![Suppose \( A \) is a \( 2 \times 2 \) matrix that has a repeated eigenvalue \( \lambda \) with linearly independent eigenvectors \( \vec{v}_1 \) and \( \vec{v}_2 \). Note that since
\[
A(\alpha \vec{v}_1 + \beta \vec{v}_2) = A\alpha \vec{v}_1 + A\beta \vec{v}_2 = \alpha A\vec{v}_1 + \beta A\vec{v}_2 = \alpha \lambda \vec{v}_1 + \beta \lambda \vec{v}_2 = \lambda (\alpha \vec{v}_1 + \beta \vec{v}_2)
\]
the vector \( \alpha \vec{v}_1 + \beta \vec{v}_2 \) is also an eigenvector for \( \lambda \). Since \( \vec{v}_1 \) and \( \vec{v}_2 \) are linearly independent, we can scale them using \( \alpha \) and \( \beta \) and add the scaled vectors together to get any nonzero vector \( \vec{v}_3 \) in the plane. (Think of \( \vec{v}_1 \) and \( \vec{v}_2 \) as arrows in the plane and how you could scale and combine them as in the figure below.)
*Diagram Description:*
The diagram illustrates vectors \( \vec{v}_1 \), \( \vec{v}_2 \), and \( \vec{v}_3 = \alpha \vec{v}_1 + \beta \vec{v}_2 \). Vector \( \vec{v}_1 \) is shown as an arrow in one direction, \( \vec{v}_2 \) is shown in another direction, and the combination \( \vec{v}_3 \) is shown as the resultant vector summing both scaled vectors.
So in fact every nonzero vector \( \vec{v}_3 \) is an eigenvector for \( A \)! In particular, the vectors \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) are eigenvectors for the eigenvalue \( \lambda \). This tells](/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 \( 2 \times 2 \) matrix that has a repeated eigenvalue \( \lambda \) with linearly independent eigenvectors \( \vec{v}_1 \) and \( \vec{v}_2 \). Note that since
\[
A(\alpha \vec{v}_1 + \beta \vec{v}_2) = A\alpha \vec{v}_1 + A\beta \vec{v}_2 = \alpha A\vec{v}_1 + \beta A\vec{v}_2 = \alpha \lambda \vec{v}_1 + \beta \lambda \vec{v}_2 = \lambda (\alpha \vec{v}_1 + \beta \vec{v}_2)
\]
the vector \( \alpha \vec{v}_1 + \beta \vec{v}_2 \) is also an eigenvector for \( \lambda \). Since \( \vec{v}_1 \) and \( \vec{v}_2 \) are linearly independent, we can scale them using \( \alpha \) and \( \beta \) and add the scaled vectors together to get any nonzero vector \( \vec{v}_3 \) in the plane. (Think of \( \vec{v}_1 \) and \( \vec{v}_2 \) as arrows in the plane and how you could scale and combine them as in the figure below.)
*Diagram Description:*
The diagram illustrates vectors \( \vec{v}_1 \), \( \vec{v}_2 \), and \( \vec{v}_3 = \alpha \vec{v}_1 + \beta \vec{v}_2 \). Vector \( \vec{v}_1 \) is shown as an arrow in one direction, \( \vec{v}_2 \) is shown in another direction, and the combination \( \vec{v}_3 \) is shown as the resultant vector summing both scaled vectors.
So in fact every nonzero vector \( \vec{v}_3 \) is an eigenvector for \( A \)! In particular, the vectors \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) are eigenvectors for the eigenvalue \( \lambda \). This tells
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