Are adjugates of similar matrices similar? Is it true in general, that if $A \sim B$, i.e. $\exists C: A =C^{-1}BC$, then $\mathrm{adj}(A) \sim \mathrm{adj}(B)$? It's quite obvious if the matrices in question are invertible (that is, $\det A \neq 0,\ \det B \neq 0$); if they are not, however, it's not as easy. I couldn't prove nor disprove this fact – could someone here do this?
We need the identities $adj(MN)=adj(N)adj(M)$ and $adj(M)=\det(M)M^{-1}$. See the wiki for a proof. See the answers in this site for another source. So we have \begin{align}adj(A)&amp;=adj(C^{-1}BC)\\&amp;=adj(BC)adj(C^{-1})\\&amp;=adj(C)adj(B)adj(C^{-1})\\&amp;=C^{-1}adj(B)C\end{align}So $adj(A)$ similar to $ adj(B)$.
We need the identities $adj(MN)=adj(N)adj(M)$ and $adj(M)=\det(M)M^{-1}$. See the wiki for a proof. See the answers in this site for another source. So we have \begin{align}adj(A)&amp;=adj(C^{-1}BC)\\&amp;=adj(BC)adj(C^{-1})\\&amp;=adj(C)adj(B)adj(C^{-1})\\&amp;=C^{-1}adj(B)C\end{align}So $adj(A)$ similar to $ adj(B)$.