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I have a small question regarding the adjugate matrix.

Suppose we have a square ($n \times n$) singular matrix with a rank of $n-1$.

Now I have two questions I'm trying to investigate:

  1. Is it possible that the adjugate matrix rank won't be changed? That is, if we have a $n \times n$ matrix with a rank of $n-1$ the adjugate will have the same rank ($n-1$).

  2. I know that in this case (rank of A is $n-1$) that $\mathrm{adj}(\mathrm{adj}(A))$ is $0$. I don't understand why. Is there any relation between these two questions?

Thanks alot, Guy

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    If $A$ is invertible, yes.2012-10-29

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For the first question use the following two facts:

1) $\operatorname*{rank}(AB)\geq \operatorname*{rank}(A)+ \operatorname*{rank}(B)-n$, for a proof, see e.g. http://ysharifi.wordpress.com/2010/09/09/rank-of-the-product-of-two-matrices/

2) $A\cdot \operatorname* {adj}(A)=\det(A)\cdot I_n$.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6189/discussion-between-julian-kuelshammer-and-guy)2012-10-20
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If $A$ is an $n \times n$ matrix then one of the following mutually exclusive possibilities occurs:

  • $A$ is invertible and $\operatorname{adj}(A) = \det(A)A^{-1}$.
  • $\operatorname{rank} A = n-1$ and $\operatorname{rank}\operatorname{adj}(A) = 1$.
  • $\operatorname{rank} A \leq n-2$ and $\operatorname{adj}(A) = 0$.

The proof of the claim follows from the fact that the adjugate of $A$ can be identified with the matrix of the $(n-1)$st exterior power of the dual of the linear transformation with matrix $A$ (with respect to appropriate choices of bases).

Applying the claim twice, once to $A$ and once to its adjugate, shows that if $A$ is not invertible then the adjugate of the adjugate is $0$.