How do you find this eigenvector?

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To find the eigenvectors for the given 2x2 matrix, start by substituting one of the eigenvalues, λ = -1 + i, into the matrix equation λI - A. This results in a new matrix, which you then multiply by a vector (x, y) and set the equation equal to zero. The resulting linear equations will be dependent, allowing you to solve for one variable in terms of the other, typically leading to a free parameter. By choosing a value for x, you can determine y and thus obtain the eigenvector. Following this method for both eigenvalues will yield the corresponding eigenvectors.
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[0 1]
[-2 -2] This is the 2x2 matrix.

[λ -1]
[2 λ+2] This is the matrix that equals λI - A.

Here are the eigenvalues I found:
λ = -1 + i, -1 - i

I am really confused at what to do next to find the eigenvectors. I keep looking online for help but I still can't figure it out...

Thanks!
 
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msell2 said:
[0 1]
[-2 -2] This is the 2x2 matrix.

[λ -1]
[2 λ+2] This is the matrix that equals λI - A.

Here are the eigenvalues I found:
λ = -1 + i, -1 - i

I am really confused at what to do next to find the eigenvectors. I keep looking online for help but I still can't figure it out...

Thanks!

Now put one of your eigenvalues into λI - A and try to find a vector v such that (λI - A)v=0. Then v would be an eigenvector of λ. It's just a set of linear equations to solve with a free parameter.
 
msell2 said:
[0 1]
[-2 -2] This is the 2x2 matrix.

[λ -1]
[2 λ+2] This is the matrix that equals λI - A.

Here are the eigenvalues I found:
λ = -1 + i, -1 - i

I am really confused at what to do next to find the eigenvectors. I keep looking online for help but I still can't figure it out...

Thanks!

For ##\lambda## equal an eigenvalue, your two equations are dependent, so you only need to solve one and you have one free variable. The first one is the same as ##\lambda x - y =0## or ##y = \lambda x##. So let ##x = 1## and what do you get for ##y##? Do that for each value of ##\lambda## and you will have two [x,y] eigenvectors.
 
Dick said:
Now put one of your eigenvalues into λI - A and try to find a vector v such that (λI - A)v=0. Then v would be an eigenvector of λ. It's just a set of linear equations to solve with a free parameter.
I get what I need to do in theory, I just don't actually know how to do it. How do you find v such that (λI-A)v=0?
 
msell2 said:
I get what I need to do in theory, I just don't actually know how to do it. How do you find v such that (λI-A)v=0?

Just try it! What is the matrix (λI-A) when λ=-1+i? Write that matrix times a vector (x,y), set it equal to zero and try to find a solution for x and y.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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