How do you find this eigenvector?

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The discussion focuses on finding eigenvectors for the 2x2 matrix A = [0, 1; -2, -2] using the eigenvalues λ = -1 + i and λ = -1 - i. Participants emphasize substituting the eigenvalues into the matrix equation λI - A and solving the resulting linear equations to find the eigenvectors. The method involves setting up the equation (λI - A)v = 0 and solving for the vector v, which includes a free parameter. This approach leads to the determination of eigenvectors corresponding to each eigenvalue.

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msell2
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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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