{"id":4588,"date":"2025-08-07T10:14:24","date_gmt":"2025-08-07T09:14:24","guid":{"rendered":"https:\/\/sinatootoonian.com\/?p=4588"},"modified":"2025-12-27T19:44:32","modified_gmt":"2025-12-27T19:44:32","slug":"a-free-connectivity-non-solution","status":"publish","type":"post","link":"https:\/\/sinatootoonian.com\/index.php\/2025\/08\/07\/a-free-connectivity-non-solution\/","title":{"rendered":"A Free Connectivity Non-Solution"},"content":{"rendered":"\n

In this post I explore one possible unconstrained connectivity solution that turns out to not work.<\/p>\n\n\n\n

As before, the loss function we’re optimizing is
$$ L(\\ZZ) = {1 \\over 2} \\|\\XX^T \\ZZ^T \\ZZ \\XX – \\CC\\|_F^2 + {\\lambda \\over 2 }\\|\\ZZ – \\II\\|_F^2.$$<\/p>\n\n\n\n

The gradient above is $$ \\nabla_\\ZZ L = \\ZZ (2 \\XX \\bE \\XX^T) + \\lambda (\\ZZ – \\II),$$ where $\\bE$ is the term in the first norm above. <\/p>\n\n\n\n

Setting this to zero and solving for $\\ZZ,$ we get
$$ \\ZZ = (2 \\XX \\bE \\XX^T\/\\lambda + \\II)^{-1} .$$ <\/p>\n\n\n\n

This is symmetric, so can be eigendecomposed as $$\\UU_Z \\SS_Z^{-1}\\UU_Z^T = \\beta \\XX (\\XX^T \\UU_Z \\SS_Z^2 \\UU_Z^T \\XX – \\CC)\\XX^T + \\II,$$ where $\\beta = 2\/\\lambda.$ Defining $$\\DD \\triangleq \\beta^{-1} (\\SS_Z^{-1} – \\II),$$ we get $$\\UU_Z \\DD \\UU_Z^T = \\XX (\\XX^T \\UU_Z \\SS_Z^2 \\UU_Z^T \\XX – \\CC)\\XX^T.$$ Multiplying on the left by $\\UU_Z^T$ and on the right by $\\UU_Z$, we get
$$ \\DD = \\UU_Z^T \\XX \\XX^T \\UU_Z \\SS_Z^2 \\UU_Z^T \\XX \\XX^T \\UU_Z -\\UU_Z^T \\XX \\CC \\XX^T \\UU_Z.$$ <\/p>\n\n\n\n

Defining the $$\\QQ \\triangleq \\UU_Z^T \\XX,$$ we can write this as $$ \\DD = \\QQ \\QQ^T \\SS_Z^2 \\QQ \\QQ^T- \\QQ \\CC \\QQ^T.$$<\/p>\n\n\n\n

By applying the right eigenvectors of $\\CC$ to $\\XX$ in the original loss, we can replace $\\CC$ by $\\SS_Y^2$. So we arrive at an interesting “quadratic” equation involving $\\QQ$ and diagonal terms, $$ \\DD = \\QQ \\QQ^T \\SS_Z^2 \\QQ \\QQ^T – \\QQ \\SS_Y^2 \\QQ^T.$$
Note:<\/em> We arrived at this also in Wrangling Quartics, IV<\/a>.<\/p>\n\n\n\n

Guessing a solution<\/h2>\n\n\n\n

Now suppose $\\QQ \\QQ^T = \\bLa^2$ is diagonal. Then we get, after rearrangement $$\\QQ \\CC \\QQ^T = \\bLa^2 \\SS_Z^2 \\bLa^2 – \\DD.$$ This implies that $\\QQ$ diagonalizes $\\CC = \\VV_C \\SS_C^2 \\VV_C^T$, so $$\\QQ = \\bLa \\VV_C^T.$$<\/p>\n\n\n\n

Assuming square matrices<\/h3>\n\n\n\n

The situation I have is one where $\\QQ$ is a tall matrix, since we have fewer odours than cells. But for now suppose we had as many cell as odours, so all the matrices were square. We have $$ \\QQ^T = \\XX^T \\UU_Z = \\VV_C \\bLa.$$
In otherwords, $\\XX^T$ transforms the columns of $\\UU_Z$ into the eigenvectors of $\\CC$! Is that the case? <\/p>\n\n\n\n

No – the $\\QQ \\QQ^T$ I find is not diagonal. And maybe it can’t be. That would imply $$ \\QQ = \\bLa \\VV_C^T = \\UU_Z^T \\XX \\implies \\XX = \\UU_Z \\bLa \\VV_C^T,$$ which is not the case in general – we can set $\\UU_Z$ and $\\bLa$, but the right eigenvectors of $\\XX$ won’t be $\\VV_C$ in general.

$$\\blacksquare$$<\/p>\n","protected":false},"excerpt":{"rendered":"

In this post I explore one possible unconstrained connectivity solution that turns out to not work.<\/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":[26,40],"class_list":["post-4588","post","type-post","status-publish","format-standard","hentry","category-blog","category-research","tag-connectivity","tag-work"],"acf":[],"_links":{"self":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/4588","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=4588"}],"version-history":[{"count":29,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/4588\/revisions"}],"predecessor-version":[{"id":5704,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/4588\/revisions\/5704"}],"wp:attachment":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/media?parent=4588"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/categories?post=4588"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/tags?post=4588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}