Adjugate Matrix properties

Question

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:

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 ($n-1$).
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

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 08:48

score 4 · Answer 1 · answered Nov 21 '14 at 16:14

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 $0$.

score 2 · Accepted Answer · edited Nov 17 '19 at 12:31

2

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

edited Nov 17 '19 at 12:31

davyjones

659

answered Oct 20 '12 at 13:28

Julian Kuelshammer

9,750

@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
@Guy: There are not that many products of matrices around where you can apply 1). Maths is a lot about trying it out by oneself – Julian Kuelshammer Oct 20 '12 at 14:23
Okay I proved the adj(det(A)*I) identity. Thank you. I'll keep trying the to prove the second one. It seems much harder. never used the first fact you gave me. – SyndicatorBBB Oct 20 '12 at 14:39
Well I'll show you my way if you can just point out what is missing. First of all let's assume rankA = n-1 then:
n-1 = rankA >= rank(A*adjA) >= rankA + rank(adjA) - n

--> n-1 >= rankA + rank(adjA) - n

--> n-1 >= n-1 + rank(adjA) - n

--> n >= rank(adjA)

So not only that I didn't get what I want to prove (rank(adjA)) <= n-2 but I also got a wrong prove. rank(adjA) cannot be n. It shows it very bad here see my response below.
– SyndicatorBBB Oct 20 '12 at 14:54
@Guy: You have to use 2) in the first line. – Julian Kuelshammer Oct 20 '12 at 15:04
What is $det(A)$ if the rank of $A$ is $n-1$? – Julian Kuelshammer Oct 20 '12 at 15:05
Oh I understand you, thanks. working on it. – SyndicatorBBB Oct 20 '12 at 15:10
Okay so det(A) = 0 now : rank(A * adj(A)) = 0 so rank(A * adjA) <= rank(adjA) <= n-1 and also rank(A * adjA) >= rank(A) + rank(adjA) - n --> rank (A * adjA) >= 1. So now I know rank(adjA) >= 1 and rank(adjA) <= n-1 – SyndicatorBBB Oct 20 '12 at 15:15
let us continue this discussion in chat – Julian Kuelshammer Oct 20 '12 at 15:18

Adjugate Matrix properties

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