Help with the concept of eigenvectors

AI Thread Summary
The discussion focuses on understanding eigenvectors after finding eigenvalues for a 2x2 matrix. The user is confused about solving the resulting equations, which are identical, indicating redundancy. The solution involves expressing one variable in terms of the other, allowing for the determination of the eigenvector. By choosing a value for one variable, a specific eigenvector can be calculated, and any scalar multiple of this vector is also valid. The conversation concludes with the user expressing gratitude for the clarification.
kel
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Hi, I just need a little bit of help with the concept of eigenvectors.

I have a basic 2x2 matrix and have found the eigenvalues to be: 4 and 9

I have also tried going through the process of finding the eigenvectors that my lecturer has shown me, but I'm not sure where to go from there, I'm currently stuck with:

2x + 3y = 0 and
2x + 3y = 0 (yes they are both the same!)

anyway, could anyone tell me where I go from here? step by step please.

Thankyou
Kel
 
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I assume this is the system you get for one of those eigenvalues.
In that case, it's completely normal that the equations are the same (or multiples of each other) since you want its solution to be a vector: the eigenvector.

One of the equations is now redundant, solve the remaining equation by letting one of the two variables (x and y) equal a parameter (e.g. t).
 
You mean like this?

2x + 3y = 0,

3y = t

therefore, t = -2x
 
Well, you want x and y in function of t, not the other way arround.
But it goes like that, more or less: just do it as you would solve an equation with two unknowns.
 
2x + 3y = 0 means that y=-(2/3)x, a relationship between the components. Once you choose an x (say x=1), you completely determine an eigenvector. Note that if you choose something else for x (say x=2 [or more generally, using the suggestion made above, x=t]), then you determine a constant multiple of the original eigenvector, which is still an eigenvector. Note: If v is an eigenvector, then so is kv.
 
Ok, got it.

Thanks for the help
 
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