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Math Help - Linear algebra: Jordan forms & eigenvectors

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    Senior Member bkarpuz's Avatar
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    Linear algebra: Jordan forms & eigenvectors

    Dear MHF members,

    I have the following problem.
    If the Jordan form of A is a single Jordan block of size n \times n,
    then how many independent eigenvectors does A have and what is the dimension of \ker (A-\lambda I)^2?

    Thanks for your help.
    bkarpuz
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    Quote Originally Posted by bkarpuz View Post
    Dear MHF members,

    I have the following problem.
    If the Jordan form of A is a single Jordan block of size n \times n,
    then how many independent eigenvectors does A have and what is the dimension of \ker (A-\lambda I)^2?

    Thanks for your help.
    bkarpuz


    As the number of Jordan blocks corr. to an eigenvalue equals the dimension of the eigenspace corr. to that eigenvalue, we have in this

    case that there's one unique eigenvalue for that matrix and its eigenspace is of dimension 1...

    Tonio
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by bkarpuz View Post
    and what is the dimension of \ker (A-\lambda I)^2?

    One way for this question (not the most elegant): if J is the Jordan form of A, then


    (J-\lambda I)^2=\begin{bmatrix} 0 & 1 & 0 &  & \ldots &0 & 0 & 0\\ 0 & 0 & 1 & &\ldots &0 & 0 & 0 \\0 & 0 & 0 & & \ldots & 0 & 0 & 0\\ \vdots&&&&&&&\vdots \\ 0 & 0 & 0 &  & \ldots & 0 & 1 & 0\\ 0 & 0 & 0 &  & \ldots & 0 & 0 & 1\\0 & 0 & 0 & &\ldots & 0 & 0 & 0\end{bmatrix}^2=\begin{bmatrix} 0 & 0 & 1 &  & \ldots &0 & 0 & 0\\ 0 & 0 & 0 & &\ldots &0 & 0 & 0 \\0 & 0 & 0 & & \ldots & 0 & 0 & 0\\ \vdots&&&&&&&\vdots \\ 0 & 0 & 0 &  & \ldots & 0 & 0 & 1\\ 0 & 0 & 0 &  & \ldots & 0 & 0 & 0\\0 & 0 & 0 & &\ldots & 0 & 0 & 0\end{bmatrix}

    As a consequence,

    \dim \ker (A-\lambda I)^2=\dim \ker (J-\lambda I)^2=n-\textrm{rank}(J-\lambda I)^2=n-(n-2)=2
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