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I had my exam of linear algebra today and one of the questions was this one.

Given $ A \in \mathbb{R}^{n \times n}$, prove that:

$\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A.$

Of course I was not able to prove this identity, otherwise I wouldn't post it here. But I'm still curious how one can prove this identity.

Could someone point me in the right direction?

3 Answers 3

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We use the identities
$\tag 1\operatorname{adj}(A)\cdot A=\det A \cdot I_n$ and $\tag 2\operatorname{adj}(AB)=\operatorname{adj}(B)\cdot \operatorname{adj}(A).$ We have by (1) $\operatorname{adj}(\operatorname{adj}(A)\cdot A)=(\det A)^{n-1}\cdot I_n$ and using (2) $\operatorname{adj}(A)\cdot \operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-1}I_n.$ Multiplying by $A$ we get $\det A \cdot I_n \cdot \operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-1}\cdot A.$ If $\det A\neq 0$, we get the wanted equality, otherwise it's clear if $n\geq 2$.

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    @N.S. Tha$n$k you for your comments. It clears things up.2011-12-20
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The equality holds over any commutative ring.

Short proof. It suffices to show that the equality holds for diagonal matrices, which is straightforward.

Slightly longer proof. Let $ (a_{ij})_{i,j=1}^n $ be indeterminates. It is enough to check that the equality holds for the matrix $ A\in M_n(\mathbb Q(a_{11},\dots,a_{nn})) $ whose $(i,j)$ entry is $a_{ij}$. But this clear since $A$ is semi-simple.

More details.

Why does it suffice to check the equality in this particular case?

Let $B$ be in $M_n(K)$, where $K$ is a commutative ring. The statement we must prove says that a certain matrix $F(B)$, depending on $B$, is zero. But each entry of $F(B)$ is a polynomial in the entries of $B$, and the coefficients of this polynomial are integers depending only on $n$.

Why is $A$ semi-simple?

Because the discriminant of its characteristic polynomial is nonzero.

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    Alright now I know what you meant when talking about the commutative ring. For the rest of you answer I'll certainly read it again when I'm a little bit more educated in math. I think I'm just not ready for it yet :).2011-12-20