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    Multiplicativity of the operator determinant and invertibility criterion

    E-HK-54-6

    Area
    Algebra
    Topics
    Determinants, Invertibility, Linear Algebra
    Status
    draft

    Let \(T\) and \(U\) be linear operators on the finite-dimensional vector space \(V\). Prove

    1. \(\det(TU) = (\det T)(\det U)\);

    2. \(T\) is invertible if and only if \(\det T \neq 0\).

    Solution

    Choose an ordered basis of \(V\), and let \(A=[T]\) and \(B=[U]\) be the corresponding matrices. Then \[ [TU]=AB. \] By multiplicativity of the matrix determinant, \[ \det(TU)=\det(AB)=\det(A)\det(B)=(\det T)(\det U). \] This proves (a).

    For (b), \(T\) is invertible exactly when its matrix \(A\) is invertible. For a square matrix over a field, \[ A\text{ is invertible}\iff \det A\ne0. \] Since \(\det T\) is defined as \(\det A\), this is exactly \[ T\text{ is invertible}\iff \det T\ne0. \]

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