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This question already has an answer here:

I struggle to understand why if $A\in M_n(\mathbb{C}) $, $\det A=0$ and $rank A \le n-2$, then $A^*=O_n$. Could you please tell me why this claim holds? My textbook offers no proof for this.

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marked as duplicate by Namaste, Dietrich Burde, Nosrati, Lord Shark the Unknown, Key Flex Oct 21 '18 at 19:21

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  • $\begingroup$ Have you seen the result that $A$ has rank at most $n-2$ if and only if all $n-1$ minors are non-zero? That is, have you seen anything about "determinantal rank"? $\endgroup$ – Omnomnomnom Oct 21 '18 at 18:20
  • $\begingroup$ The result is explained/proved nicely in this post $\endgroup$ – Omnomnomnom Oct 21 '18 at 18:22
  • $\begingroup$ @Omnomnomnom You also explained it nicely here yourself. $\endgroup$ – Dietrich Burde Oct 21 '18 at 18:42
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If $A$ has rank${}<n-1$ then all $(n-1)\times(n-1)$ minors are equal to zero, and so ${\mathrm adj}(A)$ has rank$~0$.

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  • $\begingroup$ Could you give me some further explainations? How come $adj(A)$ has rank 0? $\endgroup$ – user69503 Oct 21 '18 at 18:39
  • $\begingroup$ See the answers here. $\endgroup$ – Dietrich Burde Oct 21 '18 at 18:42

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