Complex Eigenvalues and Their Vectors

[jsb19227]

I was doing some system practice problems, and I ran into some issues with the eigenvectors I was getting and the eigenvectors my book was getting. for the system
A = ((3 -6),(3 5)), I got the eigenvalue to be 4 + sqrt(17)i, and when I set up my system for the eigenvectors, I got

V₁(1+sqrt(17)i) + 6V₂ = 0
-3V₁ + (sqrt(17)i - 1)V₂ = 0

Whenever I solved it, I got V₁ = sqrt(17)i - 1 and V₂ = 3, but my book got V₁ = 6 and V₂ = -1 - sqrt(17)i

I plugged both back into the system and both resulted in zeros, so for eigenvalues, are there multiple vectors that work or is there only one, and if there is only one, how do you know which one is the right one?

Also, I put the system into the symbolabs solver and they got the same answer I did.

 

[Romsek]

Your post is confusing.

The matrix A as you've written it has two unique eigenvalues, \(\displaystyle \text{$\lambda_1 = 4+i\sqrt{17}$ and $\lambda_2=4-i\sqrt{17}$}\)

The corresponding vectors are \(\displaystyle \text{$v_1 = \left\{\dfrac{-1+i\sqrt{17}}{3},1\right\}$ and $v_2 = \left\{\dfrac{-1-i\sqrt{17}}{3},1\right\}$}\)

 

[jsb19227]

Your post is confusing.

The matrix A as you've written it has two unique eigenvalues, \(\displaystyle \text{$\lambda_1 = 4+i\sqrt{17}$ and $\lambda_2=4-i\sqrt{17}$}\)

The corresponding vectors are \(\displaystyle \text{$v_1 = \left\{\dfrac{-1+i\sqrt{17}}{3},1\right\}$ and $v_2 = \left\{\dfrac{-1-i\sqrt{17}}{3},1\right\}$}\)

Sorry for the confusion, I was referring to just the 4 + sqrt(17)i one, my main question was that my textbook used the second vector I listed for their solution. When I plugged their vector into the system I showed, it stilled equaled 0. So is their vector wrong or are both vectors correct?

 

[Cubist]

I plugged both back into the system and both resulted in zeros, so for eigenvalues, are there multiple vectors that work or is there only one?

Think about the statement that you made, "I plugged both back into the system and both resulted in zeros". If I say that your calculations were correct, then this probably answers your question

Why is this so? Think about how you obtained the eigenvector. There was probably a choice that you made for one of the values? What would happen if you choose a different value?

Why was there a choice? Look at the original equation Mv = λv. Would this allow v to be different magnitudes (but while it points in a certain direction determined by the combination of M and λ?

NOTE: For an eigenvector the only magnitude to avoid is 0 (since it is a trivial solution to Mv = λv). But I think λ can be 0 and that's OK.