'Weighted' diagonalization For given matrices $A$, $B$ and $\Lambda$. $\Lambda = \text{Diag}(\lambda_1,\cdots, \lambda_n)$ is a diagonal matrix without identical elements $\lambda_i \neq\lambda_j \ \forall i\neq j$. I want to find $Q$ such that $$ AQ\Lambda Q^{-1} + BQ\Lambda^{-1} Q^{-1} $$ is diagonal. Is this diagonalization unique?
 A: I was going to leave this as a comment, but as it was getting kind of long I decided to post it as an answer instead. That said, it is slightly incomplete, as I am not well versed in solving quadratics over matrix rings (or indeed over any noncommutative rings...)

That said, I suspect there are no solutions in general... 
Lets set $X = Q \Lambda Q^{-1}$ (notice $Q \Lambda Q^{-1} = (Q \Lambda^{-1} Q^{-1})^{-1}$).
Your question, then, is to solve $AX + BX^{-1} = D$ for $D$ diagonal. 
We can rearrange this into $AX^2 - DX + B = 0$, and a solution to your problem will exist if and only if 


*

*For some choice of $D$, a solution to this quadratic $X$ exists

*such that $X$ is conjugate to $\Lambda$ 
(that is, $X = Q \Lambda Q^{-1}$ for some $Q$)


While I admit I'm not an expert in solving matrix quadratics, it seems like the stars would have to align for a solution to your problem to exist in general. If you want to look into solving these kinds of quadratics to find solutions, I would caution you before your journey. It seems others have found this subject quite hard (cf. this question).

That said, if we assume everything in sight commutes, and further that $A$ is invertible, then we can solve this equation using the quadratic theorem.
(Note, however, that this is a very strong assumption to put on $A$, $B$, and $\Lambda$.)
In this setting, $AX^2 - DX + B = 0$ has solutions if and only if $\sqrt{D^2 - 4AB}$ exists (if and only if $(D^2 - 4AB)$ is positive semi-definite). We can set 
$$Q \Lambda Q^{-1} = X = \left ( D \pm \sqrt{D^2 - 4AB} \right ) (2A)^{-1}$$
Given specific matrices $A, B, \Lambda$, we could use this form in order to find those $D$ diagonal which work: 
We just symbolically evaluate 
$\left ( D \pm \sqrt{D^2 - 4AB} \right ) (2A)^{-1}$ to get a matrix $M$
whose entries are polynomials in $d_i$, the diagonal entries of $D$.
We then (symbolically) diagonalize this matrix, to write $M = Q \widetilde{M} Q^{-1}$
with $\widetilde{M}$ diagonal. (Again, the entries of $Q$ and $\widetilde{M}$ will be polynomials in the $d_i$.
Finally, we set $\widetilde{M} = \Lambda$ and solve for the $d_i$ (if a solution exists).

Edit: I couldn't help myself and computed an example. Note any decent computer algebra software can do these computations for you:
e.g.
Say 
$A = \begin{pmatrix} 1 & 1\\ 0 & 2 \end{pmatrix}$,
$B = \begin{pmatrix} 3 & 1\\ 0 & 4 \end{pmatrix}$,
$\Lambda = \begin{pmatrix} 4 & 0\\ 0 & 5 \end{pmatrix}$.
Then let's set $D = \begin{pmatrix} x & 0\\ 0 & y \end{pmatrix}$.
We find (proof by wolframalpha)
$$
M = (D + \sqrt{D^2 - 4AB})(2A)^{-1} = 
\begin{pmatrix} 
\frac{x + \sqrt{x^2 - 12}}{2} & \frac{-\sqrt{5}i}{2}\\
0 & \frac{y + \sqrt{y^2 - 32}}{4}
\end{pmatrix}
$$
Diagonalizing (shown at the bottom of the same wolframalpha page),
we see
$M = Q\widetilde{M}Q^{-1}$ with $Q$ too ugly for me to want to type up, and
$$
\widetilde{M} =
\begin{pmatrix}
\frac{x + \sqrt{x^2 - 12}}{2} & 0\\
0 & \frac{y + \sqrt{y^2 - 32}}{4}
\end{pmatrix}
$$
Thus, if $\Lambda = \widetilde{M}$, we see that 
$x = \frac{19}{4}$ and $y = \frac{54}{5}$ solve the problem.
That is, we can find a $Q$ (using the definition that was too ugly for me to type up) such that
$$
A Q \Lambda Q^{-1} + B Q \Lambda^{-1} Q^{-1} = 
\begin{pmatrix}
\frac{19}{4} & 0\\
0 & \frac{54}{5}
\end{pmatrix}
$$
In this case, $D$ seems to be unique (unless I'm missing something obvious). That said, I would still be surprised to see that $D$ is always unique.
Also (it bears repeating), this is only possible if $A$ and $B$ commute, and $A$ is invertible. There may be a similar approach in the noncommutative case, but I am unaware of the required techniques.

I hope this helps ^_^
