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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2. is answered e.g. in math.stackexchange.com/questions/92837/… –  Julian Kuelshammer Oct 20 '12 at 13:21
I was wondering about one adjugate matrix property. I know that If $A$ is symmetric so is $adj(A)$. I was thinking to myself, is the opposite direction is also TRUE? If $adj(A)$ is symmetric then $A$ is symmetric ? Thank you. –  SyndicatorBBB Oct 29 '12 at 8:48
If $A$ is invertible, yes. –  user26857 Oct 29 '12 at 10:07

1) ${rank}(AB)\geq {rank}(A)+{rank}(B)-n$, for a proof, see e.g. http://ysharifi.wordpress.com/2010/09/09/rank-of-the-product-of-two-matrices/
2) $A\cdot adj(A)=det(A)\cdot I_n$.
@Guy: $adj(det(A)\cdot I)=det(A)^{n-1}$ can be calculated by hand just by the definition –  Julian Kuelshammer Oct 20 '12 at 14:22