Eigenvectors with a repeated eigenvalue

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Homework Help Overview

The discussion revolves around a linear system of differential equations characterized by a repeated eigenvalue. Participants are tasked with obtaining the general solution for the system defined by the equations dx/dt = -2x and dy/dt = -2y.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of a repeated eigenvalue and the challenges in finding eigenvectors when substituting the eigenvalue into the matrix results in a zero matrix. There is also a suggestion to consider the uncoupled nature of the system and the possibility of directly applying solutions to the differential equations.

Discussion Status

Some participants have provided guidance on the nature of the matrix being diagonal and the implications for eigenvectors. There is an acknowledgment that any vector can serve as an eigenvector in this context, and suggestions have been made regarding the selection of orthogonal vectors.

Contextual Notes

There is a noted uncertainty regarding the necessity of eigenvalues and eigenvectors for this particular system, given its uncoupled structure. Participants are navigating the implications of the repeated eigenvalue and the resulting zero matrix when applying the eigenvalue to the characteristic equation.

Snippy
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Homework Statement


For the following linear system:
\frac{dx}{dt} = -2x
\frac{dy}{dt} = -2y

Obtain the general solution.


Homework Equations





The Attempt at a Solution


A= -2 0
0 -2

Using the determinant of A-\lambdaI I got a repeated eigenvalue of -2. I am not sure how to get the eigenvectors since substituting -2 back into A-\lambdaI will give a matrix of zeroes. I checked with MATLAB and the eigenvectors should be (1,0) and (0,1) but I have no idea how to get them.
 
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Snippy said:

Homework Statement


For the following linear system:
\frac{dx}{dt} = -2x
\frac{dy}{dt} = -2y

Obtain the general solution.


Homework Equations





The Attempt at a Solution


A= -2 0
0 -2

Using the determinant of A-\lambdaI I got a repeated eigenvalue of -2. I am not sure how to get the eigenvectors since substituting -2 back into A-\lambdaI will give a matrix of zeroes. I checked with MATLAB and the eigenvectors should be (1,0) and (0,1) but I have no idea how to get them.
I don't see that there is any need for eigenvalues and eigenvectors at all. Your system of equations is uncoupled, which means that in the matrix equation x' = Ax, A is already diagonal.

If you have x'(t) = kx, the solution is x(t) = Aekt, right? Apply this idea to each of the differential equations in your system.
 
Thanks :)
 
Snippy said:
Using the determinant of A-\lambdaI I got a repeated eigenvalue of -2. I am not sure how to get the eigenvectors since substituting -2 back into A-\lambdaI will give a matrix of zeroes. I checked with MATLAB and the eigenvectors should be (1,0) and (0,1) but I have no idea how to get them.
As Mark44 noted, A is already diagonal, so you don't need to find eigenvectors so you can diagonalize A, but in case you're still wondering, when A-λI=0, any vector x will satisfy (A-λI)x=0, which means every vector x is an eigenvector of A. You can therefore choose any two vectors, like (1,0) and (0,1), as your eigenvectors. (Usually when you have this kind of freedom, it's a good idea to select vectors that are orthogonal to each other.)
 

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