Simultaneous similarity of pairs of matrices

My question is: Given a pair of $n\times n$ matrices $(A,B)$ and the fact that $A$ is similar to $A^{\prime}$ and $B$ is similar to $B^{\prime}$, i.e., there exists invertible $n\times n$ matrices $P$ and $Q$ such that $PAP^{-1}=A^{\prime}$, $QBQ^{-1}=B^{\prime}$, is there a way to deduce if there exists an invertible $n \times n$ matrix $U$ such that $UAU^{-1}=A^{\prime}$, $UBU^{-1}=B^{\prime}$?

• My instinct is that you may very well not even have such a $U$ but I could be totally wrong. Matrix similarity is just another name for conjugation (in the group theoretic sense of the word), and afaik it can happen that you have two pairs of elements which are conjugate pairs, but whose conjugating elements are not the same (that is, the $U$'s are different for $A$ and $B$, and there is no shared $U$ that works for both). The reverse construction however will work, given $A$, $B$, you can choose a $U$ and then create the conjugates. – Justin Benfield Mar 12 '16 at 7:21
• I don't think $U$ exists for generic choices of $(A,B)$ and $(A^{\prime},B^{\prime})$. I'm just wondering if there are any sufficient conditions that the pairs of matrices satisfy which ensure the existence of such $U$. – Zitao Wang Mar 12 '16 at 7:44
• An obvious sufficient condition would be $A=B$ and $A'=B'$. But apart from that I know of no such conditions (admittedly I know relatively little linear algebra). – Justin Benfield Mar 12 '16 at 7:47
• The simplest condition I can think of is when $A$ and $B$ are diagonalizable, and diagonalize to $A^{\prime}$ and $B^{\prime}$ respectively. Then $A$ commutes with $B$ ensures that such $U$ exists. – Zitao Wang Mar 12 '16 at 7:48

The problem you mentioned is a known problem of simultaneous similarity of pairs of matrices over a field and is called "wild" as it is basically reasonably considered hopeless in terms of finding a solution. It has been discovered a long time ago by Gelfand and Ponomaryov that the problem of simultaneous similarity of pairs of matrices contains the problem of simultaneous similarity of n-tuples of matrices for arbitrary given n. The problem has only been solved for specific (pretty low) dimensions. Overall, it is one of the typical problem types in modern algebra to determine whether a given classification problem (for instance, classification of group and algebra representations, linear groups, representations of posets, and more) to determine whether it's wild (i.e. contains the problem mentioned in your post) or tame (i.e. not wild). So, for instance, a group $G$ is called wild over a commutative ring $R$ if the problem of classification of all $R$-representations of $G$ is wild. So, for instance, abelian $(2, 2)$-group is tame over $F_2 = Z/2Z$, but $(2, 2, 2)$ is wild over the same field.