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    $AB - BA = I$ is impossible for complex matrices

    E-HK-35-5

    Area
    Algebra
    Topics
    Linear Algebra, Linear Functionals
    Status
    draft

    If \(A\) and \(B\) are \(n \times n\) complex matrices, show that \(AB - BA = I\) is impossible.

    Solution

    1. Suppose for contradiction that \(AB - BA = I\).

    assume the contrary.

    2. Take the trace of both sides: \(\operatorname{tr}(AB - BA) = \operatorname{tr}(I) = n\).

    the trace of the identity is \(n\).

    3. But \(\operatorname{tr}(AB - BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = 0\).

    \(\operatorname{tr}(AB) = \operatorname{tr}(BA)\) (the trace is cyclic).

    4. Contradiction: \(0 = n \neq 0\).

    2 and 3.

    5. Hence \(AB - BA = I\) is impossible.

    <1>1–<1>4.

    6. Q.E.D.

    5.

    Solution

    Suppose, for contradiction, that \[ AB-BA=I_n. \] Taking traces and using \(\operatorname{tr}(AB)=\operatorname{tr}(BA)\) gives \[ 0=\operatorname{tr}(AB)-\operatorname{tr}(BA) =\operatorname{tr}(I_n)=n. \] Over \(\CC\) the scalar \(n\) is nonzero. This contradiction proves that no such complex matrices \(A,B\) exist.

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