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Suppose square matrix $A$ with order-n, and $A^*$ is it's adjugate matrix, when $n>2$, prove: $\left(A^*\right)^*=\left|A\right|^{n-2}A$

Proof:

  • when $A$ is invertible, $A^{-1}=\frac{1}{|A|}A^*$, so $A^*=|A|A^{-1}$, and then

\begin{align} \left(A^*\right)^* = \left|\left(|A|A^{-1}\right)\right| \left(|A|A^{-1}\right)^{-1} = \left|A\right|^n\left|A\right|^{-1}A|A|^{-1} = \left|A\right|^{n-2}A \end{align}

  • how about the case $A$ is not invertible?

$|A|=0$, then seems we should show $\left(A^*\right)^*=0$

OR.
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integer
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  • The little star to denote the adjugate matrix seems to be a bad idea as it is used in waaaay another sense for matrices as well. Either use the more widely accepted Adj$,A;$ or simply use another letter, say $,B,$ . – DonAntonio Oct 06 '13 at 13:48
  • http://math.stackexchange.com/questions/92837/proof-mathrmadj-mathrmadja-mathrmdetan-2-cdot-a-for-a/92842#92842 – integer Oct 06 '13 at 15:03
  • @DonAntonio Fine, I found a similar question. – integer Oct 06 '13 at 15:04
  • Oh! This is a case where $0^0=1$. ;-) When $n=2$ the double adjugate of a noninvertible matrix $A$ is $A$. – egreg Oct 06 '13 at 16:41

2 Answers2

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You don't need to prove anything different for the singular case.

You can do your calculations in the field of rational functions in $n^2$ indeterminates $x_{ij}$ $(i,j=1,2,\dots,n)$ about the matrix $X$ having those indeterminates as coefficients, which is clearly invertible. Thus the relation $$ \operatorname{Adj}(\operatorname{Adj}(X))=(\det X)^{n-2}X $$ holds true with the same proof as above; but this expression lives in the ring of polynomials in those indeterminates, so substituting $a_{ij}$ for $x_{ij}$ gives a true expression.

Note that the relation holds also for $n=2$, for $$ \operatorname{Adj}(\operatorname{Adj}(A))=A $$ for $2\times2$ matrices, but only if we set $0^0=1$.

For $1\times1$ matrices, it makes sense to define $\operatorname{Adj}(A)=1$ and the relation is still true in the form $\det(A)\operatorname{Adj}(A)=A$.

egreg
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The relation that is always true is

$$\text{Adj}\,A\cdot A=|A|\cdot I\implies |\text{Adj}\,A||A|=|A|^n\,,\,\,\text{and from here that if $A$ isn't invertible} :$$

$$ \text{Adj}\,A\cdot A=0\cdot I$$

DonAntonio
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