# 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)&=adj(C^{-1}BC)\\&=adj(BC)adj(C^{-1})\\&=adj(C)adj(B)adj(C^{-1})\\&=C^{-1}adj(B)C\end{align}So $adj(A)$ similar to $adj(B)$.
• That's obviously not a valid proof. For two reasons, even. a) How do you define $A^{-1}$, if $\det A = 0$? And if it is not 0, as I said, the proof is easy. b) I'm not talking about matrices being equal --- I'm talking about similar matrices (and I even defined the term in my question) Commented Apr 20, 2016 at 6:09
• I don't think you can actually prove that $\mathrm{adj}(MN)=\mathrm{adj}(N)\mathrm{adj}(M)$ ( * ) using the fact that $\mathrm{adj}(M)=\det(M)M^{-1}$, but that's okay; I seem to recall that it is valid. Anyway, yeah, now it's nice and correct, I can't beleive I didn't see that. I'll accept a bit later --- maybe ( * ) is not correct and someone will disprove it. Commented Apr 20, 2016 at 6:45