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Math Help - Eigenvectors and Eigenvalues of a Linear Map

  1. #1
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    Eigenvectors and Eigenvalues of a Linear Map

    Suppose I had a linear map T:\mathbb{R}^2\rightarrow\mathbb{R}^2 such that T(x,y)=(10x+y,10y+x) and I want to find the eigenvalues and eigenvectors of T.

    I'm doing this without determinants. I have it that
    10x+y=\lambda{x}
    10y+x=\lambda{y}

    After rearranging terms and doing simple elimination I have
    -(10-\lambda)(10-\lambda)=0

    I believe the eigenvalue is then \lambda=10. Next to find the eigenvectors I would find null(T-10I)

    To find null(T-10I) I would end up having that
    10x+y-10x=0
    10y+x-10y=0

    and so I have x=y. I believe this means that all eigenvectors would be in the form (x,x) (i.e. (1,1),(2,2)...)

    Was everything I did correct?

    Thank you.
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  2. #2
    Member HappyJoe's Avatar
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    Check the step again, where you find \lambda.

    You are supposed to get that \lambda is 9 or 11 (10 is not an eigenvalue).
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  3. #3
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    Quote Originally Posted by Anthonny View Post
    Suppose I had a linear map T:\mathbb{R}^2\rightarrow\mathbb{R}^2 such that T(x,y)=(10x+y,10y+x) and I want to find the eigenvalues and eigenvectors of T.

    I'm doing this without determinants. I have it that
    10x+y=\lambda{x}
    10y+x=\lambda{y}

    After rearranging terms and doing simple elimination I have
    -(10-\lambda)(10-\lambda)=0
    No, you don't. Multiply the second equation by 10- \lambda and the subtract it from the first equation. I suspect you did that but forgot the x in the second equation. You get ((10-\lambda)^2-1)x= 0. In order that that equation not have x= 0 as its only solution we must have (10- \lambda)^2- 1= 0

    I believe the eigenvalue is then \lambda=10. Next to find the eigenvectors I would find null(T-10I)

    To find null(T-10I) I would end up having that
    10x+y-10x=0
    10y+x-10y=0

    and so I have x=y. I believe this means that all eigenvectors would be in the form (x,x) (i.e. (1,1),(2,2)...)

    Was everything I did correct?

    Thank you.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    No, you don't. Multiply the second equation by 10- \lambda and the subtract it from the first equation. I suspect you did that but forgot the x in the second equation. You get ((10-\lambda)^2-1)x= 0. In order that that equation not have x= 0 as its only solution we must have (10- \lambda)^2- 1= 0

    Thanks. I see what I did wrong, I did forget about the x.

    For the eigenvectors for \lambda=9 I would have to find null(T-9I) which would be null((10x+y,10y+x)-(9x,9y))=null(x+y,y+x).

    Setting (x+y,y+x)=(0,0) I would have x+y=0 and y+x=0 and so x=-y and y=-x. I believe the eigenvectors are then in the form (x,-x) i.e. (1,-1),(2,-2),...

    Is how I found the eigenvectors correct?
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  5. #5
    Member HappyJoe's Avatar
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    It is indeed correct that all eigenvectors corresponding to the eigenvector 9 are of the form (x,-x), but do notice that this also implies that stuff like (1/2,-1/2) and (pi,-pi) are eigenvectors - not just (1,-1), (2,-2), and so on.
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