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 $Z = U \\La U^T.$<\/p>\n\n\n\n
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.<\/p>\n\n\n\n
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
$$ Q^2 \\La^2 Q^2 – Q C’ Q = \\La’.$$<\/p>\n\n\n\n
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?<\/p>\n","protected":false},"excerpt":{"rendered":"
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 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1,148],"tags":[40],"class_list":["post-1503","post","type-post","status-publish","format-standard","hentry","category-blog","category-research","tag-work"],"acf":[],"_links":{"self":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/1503","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/comments?post=1503"}],"version-history":[{"count":20,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/1503\/revisions"}],"predecessor-version":[{"id":1523,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/1503\/revisions\/1523"}],"wp:attachment":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/media?parent=1503"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/categories?post=1503"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/tags?post=1503"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}