Eigenvectors with a repeated eigenvalue

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.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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