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$\newcommand{\adj}{\operatorname{adj}}$Let $A\in \mathbb{M}_n$ ($n \geq\ 2$) be a regular matrix and $\adj(A)$ its adjoint.
Prove that if A is regular then $\adj(\adj(A)) = (\det A)^{n-2} A$
(where $adj(adj(A))$ is the adjoint of $\adj(A)$).

For now, I know, or think I do, that $A^{-1} = \adj(A)/\det(A)$
from where we can see that $\adj(A) = A^{-1} \det(A)$

And that's all I know

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If $A$ is invertible its adjoint and inverse are related by $$\operatorname{adj}(A) = \det (A) A^{-1}$$ Thus $$\operatorname{adj}(\operatorname{adj}(A)) = \det(\det (A) A^{-1})(\det A^{-1})^{-1} = \det(A)^{n-1} \det(A)^{-1}A = \det (A)^{n-2}A$$ where we have just used the normal rules like $\det(\alpha A) = \alpha^n \det A$ and $\det(A^{-1}) = \det(A)^{-1}$

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  • $\begingroup$ Thank you, that helped a lot. I figured out a way how to prove this. $\endgroup$ – BowTie Jan 23 '14 at 20:37
  • $\begingroup$ @RanSch Sorry for my incorrect judgement, but I think instead of proposing an edit, it'd be better to leave a comment here and let the OP corrects it himself. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 2 '17 at 20:22
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Does "regular" mean "invertible"?

If A is invertible, then ${\rm adj}(A)={\rm det}(A) A^{-1}.$

${\rm adj} ({\rm adj}A )={\rm adj} (({\rm det} A)A^{-1})={\rm det}(({\rm det} A)A^{-1}).(({\rm det} A)A^{-1})^{-1}$

Note that

  • ${\rm det}(({\rm det} A)A^{-1}) =({\rm det} A)^n ({\rm det} A^{-1})=({\rm det} A)^{n-1}.$

  • $(({\rm det} A)A^{-1})^{-1}=({\rm det} A)^{-1}A.$

So ${\rm adj} ({\rm adj}A ) = ({\rm det} A)^{n-1}. ({\rm det} A)^{-1}A =({\rm det} A)^{n-2}A.$

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Hints: from basic properties of determinants (which we denote by $\;|A|\;$)

** For any scalar $\;a\in F\;,\;\;|aA|=a^n |A|\;$

** $$A^{-1}=\frac{\text{Adj} A}{| A|}\iff \text{Adj}A=|A|A^{-1}$$

So now

$$\text{Adj}\left(\text{Adj}(A)\right)=\text{Adj}\left(|A|A^{-1}\right)=\left||A|A^{-1}\left(|A|A^{-1}\right)^{-1}\right|=$$

$$=||A|A^{-1}|\cdot|(|A|A^{-1})^{-1}|=|A|^n|A|^{-1}\cdot|A|^{-1}|A|\;\ldots$$

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  • $\begingroup$ Thank you for your help too :D $\endgroup$ – BowTie Jan 23 '14 at 20:38

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