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    A nilpotent operator on an $n$-dimensional space satisfies $T^n = 0$

    E-HK-63-5

    Area
    Algebra
    Topics
    Annihilating Polynomials, Linear Algebra
    Status
    draft

    Let \(V\) be an \(n\)-dimensional vector space and let \(T\) be a linear operator on \(V\). Suppose that there exists some positive integer \(k\) so that \(T^k = 0\). Prove that \(T^n = 0\).

    Solution

    Let \(m_T\) be the minimal polynomial of \(T\). Since \(T^k=0\), the polynomial \(x^k\) annihilates \(T\), so \[ m_T(x)\mid x^k. \] Thus \(m_T=x^r\) for some \(r\ge1\). By the Cayley–Hamilton theorem, \(m_T\) divides the characteristic polynomial of \(T\), which has degree \(n\). Hence \(r\le n\). Therefore \[ T^n=T^{n-r}T^r=0. \]

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