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    Trivial intersection of range and null space versus $T^2\alpha = 0 \Rightarrow T\alpha = 0$

    E-HK-31-13

    Area
    Algebra
    Topics
    Linear Algebra, Linear Transformations, Range and Null Space
    Status
    draft

    Let \(V\) be a vector space and \(T\) a linear transformation from \(V\) into \(V\). Prove that the following two statements about \(T\) are equivalent.

    1. The intersection of the range of \(T\) and the null space of \(T\) is the zero subspace of \(V\).

    2. If \(T(T\alpha) = 0\), then \(T\alpha = 0\).

    Solution

    Assume (a). If \(T(T\alpha)=0\), then \(T\alpha\in\ker T\). Also \(T\alpha\in\operatorname{range}T\) by definition. Hence \[ T\alpha\in\operatorname{range}T\cap\ker T=\{0\}, \] so \(T\alpha=0\). Thus (b) holds.

    Conversely, assume (b). Let \(v\in\operatorname{range}T\cap\ker T\). Since \(v\) lies in the range, \(v=T\alpha\) for some \(\alpha\). Since \(v\) lies in the nullspace, \[ T(T\alpha)=T(v)=0. \] By (b), \(T\alpha=0\), hence \(v=0\). Therefore \[ \operatorname{range}T\cap\ker T=\{0\}, \] which is (a).

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