# Eigenvectors with a repeated eigenvalue

1. Jun 6, 2010

### Snippy

1. The problem statement, all variables and given/known data
For the following linear system:
$$\frac{dx}{dt}$$ = -2x
$$\frac{dy}{dt}$$ = -2y

Obtain the general solution.

2. Relevant equations

3. The attempt at a solution
A= -2 0
0 -2

Using the determinant of A-$$\lambda$$I I got a repeated eigenvalue of -2. I am not sure how to get the eigenvectors since substituting -2 back into A-$$\lambda$$I 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 6, 2010

### Staff: Mentor

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.

3. Jun 6, 2010

Thanks :)

4. Jun 6, 2010

### vela

Staff Emeritus
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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