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Math Help - Canonical form of linear operators- upper triangular equivalence

  1. #1
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    Canonical form of linear operators- upper triangular equivalence

    Show that all upper triangular matrices with diagonal entries \alpha and arbitrary other elements are equivalent if a_{12},a_{23},...,a_{n-1,n} are nonzero
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Consider

    A=\begin{bmatrix} \alpha  & a_{12} & \ldots & a_{1n}\\ 0 &\alpha  & \ldots & a_{2n} \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & \alpha \end{bmatrix}\in \mathbb{R}^{n\times n}

    with a_{12}\cdot a_{23}\cdot \ldots \cdot a_{n-1,n}\neq 0 , then

    \lambda=\alpha is the only eigenvalue of A (multiplicity n).

    Besides,

    \dim \ker(A-\alpha I)=n-r(A-\alpha I)=n-(n-1)=1.

    This means that the canonical form of Jordan J for A has only one block:

    J=\begin{bmatrix} \alpha  & 1 & \ldots & 0\\ 0 &\alpha  & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & \alpha \end{bmatrix}

    As a consequence, all matrices A are equivalent to J (even more, similar to J).

    Regards.

    Fernando Revilla
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    I add the following to my previous post:

    (i) If \alpha\neq 0 then, r(A)=n and all matrices A are equivalent to I_n (by a well known theorem).

    (ii) If \alpha=0 then, r(A)=n-1 and all matrices A are equivalent to

    \begin{bmatrix}{I_{n-1}}&{0}\\{0}&{0}\end{bmatrix} (by a well known theorem).

    So, we can avoid similarity and only use the concept of equivalence.

    Regards.

    Fernando Revilla
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