# Proof: $\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A$ for $A \in \mathbb{R}^{n\times n}$

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?

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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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Alright, I thought about $A^{-1} = \frac{\operatorname{adj}(A)}{\operatorname{det}A}$ but forgot about $\operatorname{adj}(AB) = \operatorname{adj}A \cdot \operatorname{adj}B$. Thank you, now that I see it, it seems quite easy. Damn, missed out 10 points! – user21385 Dec 19 '11 at 22:57
@Ief2 You can combine that formula with $(A^{-1})^{-1}=A$, and it works. Of course you also need to discuss separately the case $\det(A)=0$. It works since $A= (A^{-1})^{-1} = \frac{adj(A^{-1})}{\det(A^{-1})}$, and $adj(A^{-1})= adj( \frac{(adj(A)}{det(A)}) = \frac{1}{\det(A)^{n-1}} adj(adj(A))$... But this is exactly the same proof, written in a more complicated way. – N. S. Dec 19 '11 at 23:03
@N.S. Thank you for pointing that out. I think my exam question stated that the $A$ was an invertible matrix. But if that was not the case, and $\operatorname{det}A = 0$, would I just fill in the $0$ and conclude the following?: $\operatorname{adj}(\operatorname{adj}A) = 0$ only if $A = 0$? I doubt it because, if $A$ consists of $2$'s only, its adjugate matrix will also be zero, as will be $A$'s determinant. – user21385 Dec 19 '11 at 23:15
@Ief2 If $A$ is not invertible, the relation to prove becomes $adj(adj(A))=0$, which is true. It is a little harder to prove, but can be probably be proven the following way: Step 1: If $rank(A) \leq n-2$ then $adj(A) =0_n$. Step 2: Since $A adj(A)=0$, if $n \geq 3$ then either $rank(A) \leq n-2$ or $rank(adj(A)) \leq n-2$. This would solve the problem in the case $n \geq 3$, the rest is simple... I am sure there is a much simpler solution... – N. S. Dec 20 '11 at 1:54
@N.S. Thank you for your comments. It clears things up. – user21385 Dec 20 '11 at 11:06

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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Ok, maybe this level of maths is a little bit to high for me. But you say that this equality holds over a cummutative ring. I believe you when you say that, but that's not only the case right? Generally square matrices aren't a commutative ring, right? – user21385 Dec 20 '11 at 17:22
Dear Ief2: There is a misunderstanding about this commutative ring stuff. You asked the question for a real matrix. But neither Davide's answer nor mine uses the assumption that the entries of the matrix are real. This shows that the statement holds for matrices with entries in any commutative ring. I agree that $n$ by $n$ matrices with entries in a commutative ring form a non commutative ring. - A slightly different way of phrasing the argument I gave is this... – Pierre-Yves Gaillard Dec 20 '11 at 17:40
... Let's work over the complex numbers. Your equality holds for diagonalizable matrices, which are dense, and both sides of the equality are continuous in $A$. – Pierre-Yves Gaillard Dec 20 '11 at 17:40
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 :). – user21385 Dec 20 '11 at 22:55