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Math Help - eigenvalue and eigenspaces

  1. #1
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    eigenvalue and eigenspaces

    So I've to find the eigenvalues and a basis for each corresponding eigenspaces.

    Matrix A=[(2,0,0), (1,-1,0), (-1,3,3)]
    After finding the determinant, I got(λ-2) (λ+1) (λ-3)=0. So λ = 2, -1, 3 are the eigenvalues.

    [(λ-2, 0,0),(-1, λ+1, 0),(1,-3,λ-3)]

    For λ=2; -v_2+v_3=0
    3v_2-3v_3=0
    -v_3=0

    Let v_3 = t, then v_2=-t, but what is v_1? Is it 0? So the span is [0, -1, 1]?

    For λ=-1; -3v_1-v_2+v_3=0
    -3v_3=0
    -4v_3=0

    Let v_3=t, then what is v_2 and v_1? Are they both 0? So the span is [0,0,1]?

    For λ=3; v_1-v_2+v_3=0
    4v_2-3v_3=0

    Let v_3=t, then v_2=(3/4)t and v_1=(-1/4)t. So the span is [(-1/4), (3/4), 1]? I think I got this right.
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Taurus3 View Post
    For λ=2; -v_2+v_3=0
    3v_2-3v_3=0
    -v_3=0

    I don't know what are you doing. For \lambda=2 the corresponding eigenspace is:

    \ker (A-2I) \equiv\begin{Bmatrix}0=0\\x_1-3x_2=0\\-x_1+3x_2+x_3=0\end{matrix}

    and a basis is:

    B_2=\{(3,1,0)\}



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    are you sure?
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Taurus3 View Post
    are you sure?
    More than that. Check:


    A \begin{bmatrix}{3}\\{1}\\{0}\end{bmatrix}=2\begin{  bmatrix}{3}\\{1}\\{0}\end{bmatrix}



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    I think you mistook the format of my matrix. My bad. (2,0,0) is one column, (1,-1,0) is the second one, and (-1,3,3) the third one. Now can you check?
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    Quote Originally Posted by Taurus3 View Post
    So I've to find the eigenvalues and a basis for each corresponding eigenspaces.

    Matrix A=[(2,0,0), (1,-1,0), (-1,3,3)]
    After finding the determinant, I got(λ-2) (λ+1) (λ-3)=0. So λ = 2, -1, 3 are the eigenvalues.
    Yes, for any upper or lower triangular matrices, the eigenvalues are the number on the main diagonal.

    [(λ-2, 0,0),(-1, λ+1, 0),(1,-3,λ-3)]
    Okay. I tend to think of eigenvector [itex]\lambda[/itex] and eigenvector v as satifying Av= \lambda v. You are writing it is (\lambda I- A)v= 0 which is the same thing.

    For λ=2; -v_2+v_3=0
    3v_2-3v_3=0
    -v_3=0[/tex]
    What happened to v_1? The equations, with \lambda= 2 are
    (0)v_1= 0
    -v_1+ 3v_2= 0
    v_1- 3v_2- v_3

    The first equation does NOT say " v_1= 0", it is true for any v_1.
    The second equation says v_1= 3v_2 and the third that v_3= v_1- 3v_2= 3v_2- 3v_2= 0
    Taking v_2= 1, an eigenvector, corresponding to eivenvalue 2, is [3, 1, 0] and the eigenspace is spanned by [3, 1, 0].

    (Added: If those were to be columns rather then rows, then your equations are
     v_2- v_3= 0
    -3v_2+ 3v_3= 0
    3v_3= 2v_3= 0
    The third equation gives v_3= 0, both first and second equations give v_2= 0 but there is no v_1 in any of the equations so it can be anything: [t, 0, 0]= t[1, 0, 0]. The eigenspace is spanned by [1, 0, 0])
    Of course, it is easy to check whether or not a given vector is an eigenvector. Did you do that?

    Let v_3 = t, then v_2=-t, but what is v_1? Is it 0? So the span is [0, -1, 1]?

    For λ=-1; -3v_1-v_2+v_3=0
    -3v_3=0
    -4v_3=0

    Let v_3=t, then what is v_2 and v_1? Are they both 0? So the span is [0,0,1]?

    For λ=3; v_1-v_2+v_3=0
    4v_2-3v_3=0

    Let v_3=t, then v_2=(3/4)t and v_1=(-1/4)t. So the span is [(-1/4), (3/4), 1]? I think I got this right.
    Last edited by HallsofIvy; February 6th 2011 at 04:45 AM.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Taurus3 View Post
    I think you mistook the format of my matrix. My bad. (2,0,0) is one column, (1,-1,0) is the second one, and (-1,3,3) the third one. Now can you check?
    To avoid cofusion, I suggest you next time to write in \LaTeX code. Alternatively, use (a,b,...)^T for columns.


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    for λ= -1, would the span be [0, 0, 0]?

    -3v_1-v_2+v_3=0
    -3v_3=0
    -4v_3=0

    Thanks.
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Taurus3 View Post
    for λ= -1, would the span be [0, 0, 0]?

    Impossible. If E is an eigenspace then E\neq \{0\} .


    -3v_1-v_2+v_3=0
    -3v_3=0
    -4v_3=0

    (v_1,v_2,v_3)^T=(\alpha,-3\alpha ,0)^T=\alpha(1.-3,0)^T\quad (\alpha\in \mathbb{R})


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