Can I choose any number for the variable when finding eigenvectors in matrices?

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When finding eigenvectors in matrices I choose something for some x-es.

Like sometimes x3 or x4 is chosen to be s or t or 2s etc...

What I´d like to ask about is, does it not matter what the number is? Can I chose whatever I want to?

If the matrix has 3 eigenvalues and after gauss jordan I get:
100
014
000

I chose x3 to be 1.
But could I just as well chose it to be 5 or six?
 
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sciencegirl1 said:
When finding eigenvectors in matrices I choose something for some x-es.

Like sometimes x3 or x4 is chosen to be s or t or 2s etc...

What I´d like to ask about is, does it not matter what the number is? Can I chose whatever I want to?

If the matrix has 3 eigenvalues and after gauss jordan I get:
100
014
000

I chose x3 to be 1.
But could I just as well chose it to be 5 or six?
I have no idea what "x3 or x4" you are talking about. It is true that if you can get an eigenvector in the form <ax, bx, cx> where x is a variable you can take x to be any (non-zero) number you want. That is because any multiple of an eigenvector is also an eigenvector.
 
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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