'Weighted' diagonalization

Question

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?

Answer 1

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...)

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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).

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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).

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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.

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I hope this helps ^_^