I’m trying to make some sense of
$$ {1 \over \la’} \left(S^2 \wt Z_{UU} S^2 – S \wt C_{VV} S\right) + I = \wt Z_{UU}^{-1}. \label{start}\tag{1}$$ Below I’m going to drop all the tildes and subscripts, for clarity. If we left multiply by $Z$ we get
$$ {1 \over \la’} Z(S^2 Z^2 S^2 – S C S) + Z = I,$$ while if we right-multiply by $Z$ we get $$ {1 \over \la’}(S^2 Z^2 S^2 – S C S) Z + Z = I.$$ Subtracting the latter from the former, we get a commutation relation
$$ Z (S^2 Z^2 S^2 – S C S) – (S^2 Z^2 S^2 – S C S) Z = 0.$$ Letting $H$ be the symmetric term in brackets, with eigendecomposition $H = U D U^T$, we get $$ Z U D U^T = U D U^T Z.$$ Left multiplying by $U^T$ and right by $U$, we get
$$ U^T Z U D = D U^T Z U,$$ i.e. $U^T Z U$ commutes with a diagonal matrix, which means it must be diagonal (assuming all the $D_{ii}$ are different). In otherwords, we’ve found out that $Z = U D’ U^T$, for some diagonal matrix $D’$. But we already knew that from $\Eqn{start},$ which tells us that $H = U(\la’ (\La^{-1}-1)) U^T$, where $Z = U \La U^T.$
The really remarkable thing is that at the solution, the eigenvectors of $Z$ are the same as those of $$H = S^2 Z^2 S^2 – S C S.$$ $Z^2$ clearly has the same eigenvectors as $Z$. But sandwiching it between two $S^2$ terms will shift the eigenvectors (and eigenvalues). But $Z$ is such that the subsequent subtraction of $S C S$ restores the eigenvectors.
Can we do any more with this? Let’s write things in terms of $U$ and $\La$.
$$ S^2 U \La^2 U^T S^2 – S C S = U \La’ U^T,$$ where $\La’ \equiv \la’ (\La^{-1} – 1).$ Left and right multiplying by $U^T$ and $U$, respectively, we get
$$ U^T S^2 U \La^2 U^T S^2 U – U^T S U C’ U^T S U = \La’,$$ where $C’ \equiv U^T C U.$ Letting $Q = U^T S U,$ we arrive at
$$ Q^2 \La^2 Q^2 – Q C’ Q = \La’.$$
This looks a lot like a quadratic equation, except that the variables sandwich the coefficients. Is there a way to make the analogy precise, so that we can apply a corresponding quadratic formula to solve the equation?
Leave a Reply