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Math Help - Eigenvalues and eigen vectors

  1. #1
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    Eigenvalues and eigen vectors

    The question is, for which x is the equation A_x= lambda_x valid? And then I am a given a square matrix of say (9 5)(first row) (6 7) (second row.)Then you are given a lambda value of say 4 which apparently is the eigenvalue and x_1 and x_2. I don't understand the question properly. I can calculate for the eigenvectors since I have the eigenvalue by subtracting them from the diagonal matrix so I will have 9-4 and 7-4 and so on and then I can find the value of x to get the eigenvector. But I don't understand the question properly. I can't multiply the vector by the matrix since it is not a square matrix so, how do you go about it? Thanks all in advance.
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  2. #2
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    Hi

    It seems not very clear to you ...
    Read this and post if you have additional questions

    http://www.mathhelpforum.com/math-he...envectors.html
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  3. #3
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    Thank you gag. I understand the equation and can get the eigenvectors and we are already given the eigenvalue. What I don't understand is just the question which is: for which equation is the A_x= lambda_x valid? We already have lambda which is the eigenvalue, we can get x which is the eigenvector and we have A which is the matrix. This equation is also the same as the det (A - lambdaI)=0. I can calculate for the eigenvectors of both, we already have the eigenvalue so what does for which x is A_x= lambda_x valid mean then?
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  4. #4
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    For each eigenvalue \lambda, you can find an eigenvector X

    Thus you have for instance AX_1 = \lambda_1\:X_1

    All the vectors Y_1 = k\:X_1 are such that AY_1 = \lambda_1\:Y_1 because AY_1 = A\: k\:X_1 = k\:AX_1 = k\:\lambda_1\:X_1 = \lambda_1\:k\:X_1 = \lambda_1\:Y_1

    In other words all the vectors parallel to X1 are solutions
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  5. #5
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    Quote Originally Posted by running-gag View Post
    For each eigenvalue \lambda, you can find an eigenvector X

    Thus you have for instance AX_1 = \lambda_1\:X_1

    All the vectors Y_1 = k\:X_1 are such that AY_1 = \lambda_1\:Y_1 because AY_1 = A\: k\:X_1 = k\:AX_1 = k\:\lambda_1\:X_1 = \lambda_1\:k\:X_1 = \lambda_1\:Y_1

    In other words all the vectors parallel to X1 are solutions
    Thank you. It is just the question and not solving the equation. I know how to get the eigenvalues and eigenvectors but I just didn't get the question: for which x does the equation A_X=lambda_X valid? What does that mean? The matrices I have all have eigen vectors and we are already given their eigen values so, what does for which x (eigenvektor) is the equation valid mean? I think maybe I will try to seek clarification from the teacher but I appreciate your help. Thanks.
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  6. #6
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    Quote Originally Posted by Keep View Post
    Thank you. It is just the question and not solving the equation. I know how to get the eigenvalues and eigenvectors but I just didn't get the question: for which x does the equation A_X=lambda_X valid? What does that mean? The matrices I have all have eigen vectors and we are already given their eigen values so, what does for which x (eigenvektor) is the equation valid mean? I think maybe I will try to seek clarification from the teacher but I appreciate your help. Thanks.
    Assuming that by A_X= lambda_X, you mean AX= \lambda X, then asking "for what X is this equation valid" is just asking "what values of x satisfy this equation". If \lambda is NOT an eigenvalue, only x= 0 satisfies that equation. If \lambda is an eigenvalue the all eigenvectors, as well as 0, satisfy the equation.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Assuming that by A_X= lambda_X, you mean AX= \lambda X, then asking "for what X is this equation valid" is just asking "what values of x satisfy this equation". If \lambda is NOT an eigenvalue, only x= 0 satisfies that equation. If \lambda is an eigenvalue the all eigenvectors, as well as 0, satisfy the equation.
    Oh OK, I think I get it now. Looks easier than I thought. Thanks!
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