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    An endomorphism of $\RR^5$ with eigenvalue $0$ is neither injective nor surjective

    P-DJFL4

    Area
    Algebra
    Topics
    Eigenvalues and Eigenvectors, Linear Algebra, Rank and Nullity
    Status
    draft

    Let \(L:\RR^5\to\RR^5\) be linear. Show that if \(0\) is an eigenvalue of \(L\), then \(L\) is neither injective nor surjective.

    Solution

    1. The map \(L\) is not injective.

    Since \(0\) is an eigenvalue, there is a nonzero vector \(v\) such that \(L(v)=0\). Thus \(0\ne v\in\ker L\).

    2. The map \(L\) is not surjective.

    By rank-nullity, \[ 5=\dim\ker L+\dim\operatorname{im}L. \] Since \(\dim\ker L\ge1\), \[ \dim\operatorname{im}L\le4<5=\dim\RR^5. \] Hence \(\operatorname{im}L\ne\RR^5\).

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