$\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$

Question

How can I prove that $\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$, if $A$ and $B$ are any two $n\times n$-matrices. Here, $\operatorname{adj} A$ means the adjugate of the matrix $A$.

I know how to prove it for non singular matrices, but I have no idea what to do in this case.

Answer 1

If we have an inner product space $\;V\;$ , then for all $\;v,u\in V\;$ :

$$\color{red}{\langle ABu,v\rangle}=\langle u,(AB)^*v\rangle$$

$$\langle u,B^*A^*v\rangle=\langle Bu,A^*v\rangle=\color{red}{\langle ABu,v\rangle}$$

Since the red terms in both lines are the same, then

$$\langle u,(AB)^*v\rangle=\langle u,B^*A^*v\rangle\implies \langle\,u\,,\,\left((AB)^*-(B^*A^*)\right)v\,\rangle=0\implies \ldots$$

Answer 2

In addition to DonAntonio's answer, if you want just something to do with matrices, you can go through this

We know that $A~ Adj A = |A|~ I $ ( how ?)

If $A$ and $B$ are matrices of the same order, then :

$=> (AB)~ adj (AB) = |A|~|B|~I$

$=> (AB)~ adj (AB) = |A|~I~|B|~I$

$=> (AB)~ adj (AB) = A~(adj~ A)~|B|~I$

$=> B~(adj~AB) = (adj A)~|B|~I $

$=> B~(adj~AB) = (adj A)~|B|~I $

Can you work out from here?

Answer 3

The easiest technique for dealing with the adjugate matrix is to consider the field of rational functions in $2n^2$ indeterminates $K=F(X,Y)$, where $X$ and $Y$ denote the sets of indeterminates $X_{ij}$ and $Y_{ij}$, for $1\le i,j\le n$. Here $F$ is the base field, in your case probably $\mathbb{R}$ or $\mathbb{C}$.

Then the matrices $X=[X_{ij}]$ and $Y=[Y_{ij}]$ with coefficients in $K$ are invertible, because they have nonzero determinant. By general rule, $$\def\adj{\operatorname{adj}} (\det X) X^{-1}=\adj X $$ and similarly for $Y$ and $XY$. Thus $$ \adj(YX)=\det(XY)\cdot(YX)^{-1}=(\det X\cdot\det Y)X^{-1}Y^{-1} $$ while $$ (\det X)X^{-1}\cdot(\det Y)Y^{-1}=\adj X\cdot\adj Y. $$ Comparing the two expressions we get $$ \adj X\cdot\adj Y=\adj(YX). $$ Now these matrices have coefficients in $F[X,Y]$, the ring of polynomials in the $2n^2$ indeterminates above. Substituting the coefficients of $A$ for $X_{ij}$ and those of $B$ for $Y_{ij}$ gives your claim: $$ \adj A\cdot \adj B = \adj(BA). $$