{"id":6177,"date":"2026-02-05T17:08:00","date_gmt":"2026-02-05T17:08:00","guid":{"rendered":"https:\/\/sinatootoonian.com\/?p=6177"},"modified":"2026-02-06T15:54:23","modified_gmt":"2026-02-06T15:54:23","slug":"eigenvalue-density-via-the-stieltjes-transform","status":"publish","type":"post","link":"https:\/\/sinatootoonian.com\/index.php\/2026\/02\/05\/eigenvalue-density-via-the-stieltjes-transform\/","title":{"rendered":"Eigenvalue Density via the Stieltjes Transform"},"content":{"rendered":"\n

These are my notes on how to compute the eigenvalue density of a random matrix using the Stieltes transform. They’re based on Chapter 2 of “A First Course in Random Matrix Theory,<\/em>” and conversations with ChatGPT. I am new to this topic so these notes may contain errors.<\/p>\n\n\n\n

We’re interested in the distribution of eigenvalues of a symmetric $N \\times N$ random matrix $X_N$. We’ll ultimately be interested in the limit of infinitely large $N$, but we’ll start with the finite case.<\/p>\n\n\n\n

There are lots of ways we can encode the eigenvalues. One natural way is to encode them as a probability distribution. Since we (for now) only have a discrete set of eigenvalues, we can put a delta function at each eigenvalue, so $$ \\rho_N(\\lambda) = {1 \\over N} \\sum_{i=1}^N \\delta(\\lambda – \\lambda_i).$$ <\/p>\n\n\n\n

We will want to bring in tools from complex analysis, so another way to encode the eigenvalues is to create an “indicator” function which has poles at the eigenvalues: $$g_N(z) ={1 \\over N} \\sum_{i=1}^N {1 \\over z – \\lambda_i}.$$<\/p>\n\n\n\n

We can relate these two encodings by thinking of our indicator function as the expectation of $1\/(z – \\lambda)$ relative to the probability distribution we defined: $$ g_N(z) = \\int_{-\\infty}^\\infty { \\rho_N(\\lambda) \\over z – \\lambda} d\\lambda.$$ <\/p>\n\n\n\n

It’s then natural to take the $N \\to \\infty$ limit, and arrive at the Stieltjes transform <\/strong>of our density, $$ g(z)= \\int_{-\\infty}^\\infty { \\rho(\\lambda) \\over z – \\lambda} d\\lambda,$$ where our discrete distribution $\\rho_N$ has now been replaced by the density, $\\rho$, that we’re after.<\/p>\n\n\n\n

Two questions arise at this point:<\/p>\n\n\n\n

    \n
  1. The function $g(z)$ encodes the density, but we don’t know the density! So how do we compute $g$ without that knowledge?<\/li>\n\n\n\n
  2. If we somehow manage to compute $g(z)$ by some other means, how do we extraction the density from it?<\/li>\n<\/ol>\n\n\n\n

    To answer the first, we return to the finite dimensional case and notice that $g_N(z)$ is the trace of the resolvent<\/strong>:<\/p>\n\n\n\n

    $$g_N(z) = {1 \\over N} \\sum_{i=1}^N {1 \\over z – \\lambda_i} = {1 \\over N} \\tr((zI – X_N)^{-1}).$$ So, by analogy, $$g(z) \\propto \\tr((zI – X)^{-1}).$$ The general strategy will then be to compute the finite dimensional trace, and extend somehow to the infinite-dimensional case. The main point is that this doesn’t require direct knowledge of the density.<\/p>\n\n\n\n

    So now for our second question: Suppose we have $g(z)$, how do we extract the density?<\/p>\n\n\n\n

    This is where complex analysis comes in. Notice that our function $g(z)$ is undefined on<\/em> the real line in the ordinary sense, due to the $(z – \\lambda)$ in the denominator. However, it’s well-behaved (analytic i.e. complex differentiable – we haven’t shown this) everywhere else, in particular above and below the line. The real line in this case is called a “jump”, because the value of the function “jumps” across it.<\/p>\n\n\n\n

    We can’t evaluate our function on<\/em> the real line, but we can consider taking limit values of our function from below<\/em> the real line: $$g(x – 0 i) \\triangleq \\lim_{\\veps \\to 0^+} g(x – \\veps i),$$ where by $0^+$ I mean taking the limit from the right.<\/p>\n\n\n\n

    The Sokhotski\u2013Plemelj theorem<\/a> then relates this limit to the value of our density on the line. In particular, it says that, for any interval $[a,b]$ containing 0, $$ \\im\\left[ \\lim_{\\veps \\to 0^+} \\int_{a}^b {f(x) \\over x – \\veps i} dx \\right] = \\pi f(0).$$<\/p>\n\n\n\n

    To apply this to our problem, we first extend the integration interval to $[-\\infty, \\infty]$. We then have $$ g(x – 0 i) = \\lim_{\\veps \\to 0^+} \\int_{-\\infty}^\\infty {\\rho(t) \\over t – \\veps i – \\lambda} dt = \\lim_{\\veps \\to 0^+} \\int_{-\\infty}^\\infty {\\rho(t – \\lambda) \\over t – \\veps i} dt.$$<\/p>\n\n\n\n

    This is just the form needed to apply the theorem, so we get $$ \\im{g(x – 0 i)} = \\pi \\rho(x).$$<\/p>\n\n\n\n

    So, the recipe is:<\/p>\n\n\n\n

      \n
    1. Use the resolvent to determine $g(z)$.<\/li>\n\n\n\n
    2. Evaluate directional limits of $g(z)$ on the real line.<\/li>\n\n\n\n
    3. Use the S-P theorem to read off the density from those limits.<\/li>\n<\/ol>\n\n\n\n

      Miscellaneous Statements<\/h2>\n\n\n\n

      Below are some elaborations on miscellaneous statements from the book chapter.<\/p>\n\n\n\n

      They state the definition of the finite dimensional transform $$g_N(z) = \\int_{-\\infty}^\\infty {\\rho_N(\\lambda) \\over z – \\lambda} d\\lambda,$$ then say that’s well behaved at infinity, and give a series expansion. The expansion can be derived by using the series representation $$ {1 \\over z- \\lambda} = {1 \\over z}{1 \\over 1 – {\\lambda \\over z}} = {1 \\over z} \\sum_{k=0}^\\infty {\\lambda^k \\over z^k} = \\sum_{k=0}^\\infty {\\lambda^k \\over z^{k+1}}.$$ Then $$g_N(z) = \\int_{-\\infty}^\\infty \\rho_N(\\lambda)\\sum_{k=0}^\\infty {\\lambda^k \\over z^{k+1}} d\\lambda = \\sum_{k=0}^\\infty {1 \\over z^{k+1}} \\int \\rho_N(\\lambda) \\lambda^k d\\lambda = \\sum_{k=0}^\\infty {1 \\over z^{k+1}} {1 \\over N} \\tr(A^k).$$ <\/p>\n\n\n\n

      The importance of infinity is that $z$ will then be beyond all the eigenvalues, so there is no risk of $g_N(z)$ blowing up. <\/p>\n\n\n\n

      Later, they mention that “knowledge of $g(z)$ near infinity is equivalent to knowledge of all the moments of $A$. What they mean by this is in the moment expansion above.<\/p>\n\n\n\n

      $$\\blacksquare$$<\/p>\n\n\n\n

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

      Notes on how to compute the eigenvalue density of a random matrix using the Stieltes transform. based on Chapter 2 of “A First Course in Random Matrix Theory,” and conversations with ChatGPT. <\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,30],"tags":[],"class_list":["post-6177","post","type-post","status-publish","format-standard","hentry","category-blog","category-notes-blog"],"_links":{"self":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/6177","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=6177"}],"version-history":[{"count":47,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/6177\/revisions"}],"predecessor-version":[{"id":6235,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/6177\/revisions\/6235"}],"wp:attachment":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/media?parent=6177"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/categories?post=6177"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/tags?post=6177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}