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Given that $A_{m \times n}$ has real entries, I want to prove that $\operatorname{trace}(A^TA) = 0$ if and only if $A = 0$. In other words, I want to show that the only way for the trace of $(A^TA)$ to be zero is if $A$ is a zero matrix, and that if $A$ is a zero matrix then $A^TA$ has a trace of zero.

Intuitively this makes sense to me. My idea is that in order to get zeros on the main diagonal of any product of matrices I'd need at least one of the matrices to have zeros on its main diagonal. In this case, because the product is between $A$ and its transpose, I figure it $A$ does indeed need to be a zero matrix. However, I'm having difficulty turning my intuition into an actual proof.

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    $\begingroup$ Trace equal to zero does not mean that there are zeroes on the diagonal. $\endgroup$
    – Arthur
    Commented Feb 18, 2015 at 16:41
  • $\begingroup$ Oh, right, the sum is zero. Not necessarily that the entire diagonal must be zeros. $\endgroup$
    – GaMbiTaaaa
    Commented Feb 18, 2015 at 16:43

4 Answers 4

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If $A = 0$, then $A^TA = 0$ and ${\rm trace}(A^TA) = 0$, good. For the other direction you must use the definition of matrix multiplication. $$(A^TA)_{ij} = \sum_k (A^T)_{ik}A_{kj} = \sum_kA_{ki}A_{kj}.$$ And: $${\rm trace}(A^TA) = \sum_i (A^TA)_{ii} = \sum_i\sum_k (A_{ki})^2.$$ If you have $$A_{11}^2+\ldots+A_{1n}^2+A_{21}^2+\ldots+A_{2n}^2+\ldots+A_{n1}^2+\ldots+A_{nn}^2 = 0,$$ what can you say about the $A_{ij}$?

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$A^TA$ is positive semi-definite, since it is symmetric ($(A^T A)^T = A^T (A^T)^T = A^T A$) and $x^T A^T A x = (A x)^T (A x) = || A x||^2 \geq 0$ (so its Rayleigh quotient is non-negative). So, $A^T A$ has non-negative eigenvalues. $tr(A^T A)$ is simply the sum of the eigenvalues of $A^T A$, and is thus zero if and only if all the eigenvalues of $A^T A$ are zero (since they are non-negative). Now, the eigenvalues of $A^T A$ are simply the squared singular values of $A$, so this is equivalent to all the singular values of $A$ being zero. Since the rank of $A$ is simply the number of non-zero singular values, this is equivalent to $A$ having rank zero, i.e. being the zero matrix.

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One line proof using SVD: $$ 0 = tr(AA^T)=tr(U\Lambda VV^T\Lambda U^T)=tr(\Lambda^2 U^TU)=tr(\Lambda^2) \implies \Lambda = 0 \implies A = 0. $$

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    $\begingroup$ what $Λ$ means? $\endgroup$
    – ESCM
    Commented Dec 23, 2020 at 15:10
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Consider the vector space $V=M_n(\Bbb R)$ of all real square matrices of size $n$ over the field $\Bbb R$. It can be easily verified that the map $\langle\cdot,\cdot\rangle:V\times V\to \Bbb R$ given by $\langle A,B \rangle=\text{trace}(B^T A)$ defines an inner product on $M_n(\Bbb R)$. From this point of view, it follows that

$$\text{trace}(A^T A)=\langle A,A \rangle=0\Longleftrightarrow A=0.$$

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    $\begingroup$ I think that the point of the exercise is exactly this step in proving that ${\rm trace}(A^TA)$ defines a norm.. so this wouldn't be helpful :P $\endgroup$
    – Ivo Terek
    Commented Feb 18, 2015 at 17:23

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