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    Jordan form of an operator on a $6$-dimensional space with $T^6=0$ and $T^5\neq 0$, and similarity of any two such operators

    P-QJKMC

    Area
    Algebra
    Topics
    Canonical Forms, Jordan Canonical Form, Nilpotence
    Status
    draft

    Suppose \(V\) is \(6\)-dimensional and \(T\in\operatorname{End}(V)\) satisfies \[ T^6=0, \qquad T^5\ne0. \]

    1. Determine the Jordan canonical form of \(T\).
    2. Show that any two such operators are similar.
    Solution

    Because \(T\) is nilpotent, every Jordan block has eigenvalue \(0\). The condition \(T^6=0\) says every block has size at most \(6\), while \(T^5\ne0\) says at least one block has size at least \(6\). Hence there is a block of size exactly \(6\).

    Since \(V\) itself has dimension \(6\), that block is the whole Jordan form: \[ J(T)=J_6(0). \]

    If \(S\) is another operator satisfying the same hypotheses, then also \[ J(S)=J_6(0). \] Thus there exist invertible \(P,Q\) with \[ T=PJ_6(0)P^{-1}, \qquad S=QJ_6(0)Q^{-1}. \] Taking \[ A=QP^{-1} \] gives \[ ATA^{-1}=S. \] Hence all such operators are similar.

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