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    Vanishing of $\operatorname{trace}(A^t A)$ forces $A = 0$

    E-HK-37-5

    Area
    Algebra
    Topics
    Linear Algebra, Linear Functionals
    Status
    draft

    Let \(A\) be an \(m \times n\) matrix with real entries. Prove that \(A = 0\) if and only if \(\operatorname{trace}(A^t A) = 0\).

    Solution

    Let \(A=(a_{ij})\) be an \(m\times n\) real matrix. The diagonal entries of \(A^tA\) are \[ (A^tA)_{jj}=\sum_{i=1}^m a_{ij}^2. \] Therefore \[ \operatorname{tr}(A^tA) =\sum_{j=1}^n\sum_{i=1}^m a_{ij}^2. \] This is a sum of squares of real numbers. Hence it is zero if and only if every entry \(a_{ij}\) is zero, i.e. if and only if \(A=0\).

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