# Why is the adjugate matrix the null matrix? [duplicate]

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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.

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

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• 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"? – Omnomnomnom Oct 21 '18 at 18:20
• The result is explained/proved nicely in this post – Omnomnomnom Oct 21 '18 at 18:22
• @Omnomnomnom You also explained it nicely here yourself. – Dietrich Burde Oct 21 '18 at 18:42

## 1 Answer

If $$A$$ has rank$${} then all $$(n-1)\times(n-1)$$ minors are equal to zero, and so $${\mathrm adj}(A)$$ has rank$$~0$$.

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