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Consider a matrix $A \in \mathbb{R}^{d \times m}$ such that $m \geq d$ and denote its columns i.e $A_{:, i}$ by $a_i$. Let $AA^T$ is invertible.

Now, consider the sum $S(A) = \sum_{r=1}^m a_r a^T_r$ which is an $d \times d$ matrix.

Note that each $a_r a_r^T$ is a rank 1 matrix. Note that when m = d, it follows from another question.

Is rank(S(A)) = d ?

Example: A = \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix}

S(A) = \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}

S in this example is full rank.

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    $\begingroup$ It's not clear what you're asking. It seems to me that $S(A) = AA^T$. So, if $AA^T$ is invertible, of course $S(A)$ is invertible since it is the same matrix. $\endgroup$ – Omnomnomnom Apr 5 at 2:35
  • $\begingroup$ Thanks. I did not see that. $\endgroup$ – listener Apr 6 at 1:31
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as @Omnomnomnom pointed out: Looks like $S(A) = AA^T$.

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  • $\begingroup$ Since the answer is from OP himself/herself, I think it's a valid answer. $\endgroup$ – AspiringMathematician Apr 6 at 4:04

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