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For the matrix, $A$ with a size of $n\times 200$, does the inverse of $AA^T$ always exist? It is clear that if $n>200,$ $AA^T$ could not be inverse by the argument of the rank of $AA^T$. But how about the $n=200$ or $n<200$?

Arctic Char
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Hermi
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  • $AA^T$ is invertibile if and only if the rows of $A$ are linearly indpendent – Ben Grossmann Sep 27 '20 at 12:47
  • @BenGrossmann Yes, I know this statement. But how to use that in my question. – Hermi Sep 27 '20 at 12:48
  • @TheSilverDoe Please see this question https://math.stackexchange.com/questions/3297333/does-there-exist-the-inverse-of-xxt?noredirect=1&lq=1 – Hermi Sep 27 '20 at 12:51
  • @TheSilverDoe And my question is about the $n\times 200$ matrix but NOT a linearly independent matrix. Please check your "duplicate" proof in that answer... – Hermi Sep 27 '20 at 12:52
  • @BenGrossmann But we do not know "it has linearly independent rows"... – Hermi Sep 27 '20 at 12:53
  • If you don't even read the link I provided you, I cannot help you more. – TheSilverDoe Sep 27 '20 at 12:54
  • @Hermi Exactly. So of course, $AA^T$ will not always have an inverse, even in the case that $n \leq 200$ – Ben Grossmann Sep 27 '20 at 13:08
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    @Hermi It is difficult to understand how you could know this statement and fail to see any connection between this statement and your question; perhaps you are confused about what "if and only if" means. Also, it is not clear what you mean by "my question is about the $n \times 200$ matrix but not a linearly independent matrix." – Ben Grossmann Sep 27 '20 at 13:14

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The answer to the question as stated is no. For example, if we consider the matrix $A = (a_{ij})_{i,j=1}^{n,200}$ whose entries satisfy $$ a_{ij} = \begin{cases} d_i & i=j\\ 0 & i \neq j, \end{cases} $$ then we find that $AA^T$ will have an inverse when each of the values $d_1,\dots,d_n$ is non-zero and will fail to have an inverse otherwise.

Ben Grossmann
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