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.
Best Answer
I thought of the following possibly simpler proof to the ones given above (inasmuch as it uses elementary linear algebra) which makes use of the Cauchy-Binet formula and importantly, works for non-invertible matrices too. In the following, $A^{ij}$ will indicate the matrix derived from some matrix $A$ after eliminating the $i$th row and the $j$th column, and $A^{i0}$ will indicate that only the $i$th row was eliminated (with all columns remaining). Similarly $A^{0j}$ will mean $A$ with the $j$th column eliminated.
For any $1\leq i,j \leq n$, $$\left(\operatorname{adj}\left(B\right)\operatorname{adj}\left(A\right)\right)_{ij}=\sum_{k=1}^{n}\left(\operatorname{adj}B\right)_{ik}\left(\operatorname{adj}A\right)_{kj}\\=\sum_k \left(-1\right)^{k+i}\det\left(B^{ki}\right)\left(-1\right)^{j+k}\det\left(A^{jk}\right)\\=\left(-1\right)^{i+j}\sum_k \det\left(A^{jk}\right)\det\left(B^{ki}\right)\\=\left(-1\right)^{i+j}\sum_k \det\left(A^{jk}B^{ki}\right)=\left(*\right)$$
But we notice that $\left(AB\right)^{ji}=A^{j0}B^{0i}$ and so by the Cauchy-Binet formula we have: $$\det\left(AB\right)^{ji}=\sum_k \det\left(A^{jk}B^{ki}\right)$$
which gives us: $$\left(*\right)=\left(-1\right)^{i+j}\det\left(AB\right)^{ji}=\left(\operatorname{adj}\left(AB\right)\right)_{ij}$$ and we are done.
I hope that this is correct and if so that it helps.