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Thread: 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 $\displaystyle \alpha$ and arbitrary other elements are equivalent if $\displaystyle a_{12},a_{23},...,a_{n-1,n}$ are nonzero
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Consider

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

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

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

    Besides,

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

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

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

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

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

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

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

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

    Regards.

    Fernando Revilla
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