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Let $A$ be a $m\times n$ matrix with real coefficients. It is proved, for example here Derive the Pseudo Inverse (Moore Penrose) of Rank 1 Matrix as a Scalar Multiple of Its Transpose that if $A$ is a rank one matrix, then the pseudoinverse $A^+$ is $\frac{1}{c}A^*$, where $c$ is the sum of the squares of the entries of $A$. Is the converse true?

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Yes. $A A^+$ is the orthogonal projection on the range $\text{Ran}(A)$. Its trace is the rank of $A$. But $\text{tr}(A A^*)$ is the sum of the squares of the entries of $A$. Thus if your condition holds, $$\text{rank}(A) =\text{tr}(A A^+)= \text{tr}(A A^*/c) = c/c = 1$$

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