Infinite matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:52:20Z http://mathoverflow.net/feeds/question/1886 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1886/infinite-matrices Infinite matrices Gabe Cunningham 2009-10-22T15:40:21Z 2013-02-06T12:30:25Z <p>Suppose we have an infinite matrix A = (a<sub>ij</sub>) (i, j positive integers). What is the "right" definition of determinant of such a matrix? (Or does such a notion even exist?) Of course, I don't necessarily expect every such matrix to have a determinant -- presumably there are questions of convergence -- but what should the quantity be? The problem I have is that there are several ways of looking at the determinant of a finite square matrix, and it's not clear to me what the "essence" of the determinant is.</p> http://mathoverflow.net/questions/1886/infinite-matrices/1888#1888 Answer by Qiaochu Yuan for Infinite matrices Qiaochu Yuan 2009-10-22T15:56:17Z 2009-10-22T16:02:04Z <p>There are a lot of subtleties that you need to watch out for. First of all, "infinite matrices" aren't well-defined as linear transformations without additional hypotheses. A typical case in combinatorics is that the matrix is triangular and you're only interested in how it acts on a space of formal power series; the t-adic topology is what gives you convergence here. A typical case in analysis is that you're describing a bounded linear operator between separable Hilbert spaces, and then there is the notion of orthonormal basis to work with. In any case you need a topology on the underlying vector space to make sense of infinite sums.</p> <p>If you define the determinant of a matrix to be the product of its eigenvalues, then you run into immediate trouble: "infinite matrices" don't necessarily have any, even over an algebraically closed field. And in the nicest case, e.g. compact self-adjoint, the eigenvalues tend to zero and their product is zero. I also believe one can show that there is no nontrivial continuous homomorphism GL(H) -> C for H a Hilbert space. Finally, if you think of the determinant in terms of exterior powers, then it's not hard to see that for an infinite-dimensional space H, however you want to define the exterior powers of H they should always be infinite-dimensional. </p> <p>Having said all that, there is a notion of <a href="http://www.google.com/search?q=regularized+determinant&amp;ie=utf-8&amp;oe=utf-8&amp;aq=t&amp;rls=org.mozilla:en-US:official&amp;client=firefox-a" rel="nofollow">regularized determinant</a> in the literature, but I'm afraid I couldn't tell you anything about it.</p> http://mathoverflow.net/questions/1886/infinite-matrices/2052#2052 Answer by Andrew Stacey for Infinite matrices Andrew Stacey 2009-10-23T07:48:25Z 2009-10-23T07:48:25Z <p>There is a class of linear operators that have a determinant. They are, for some strange reason, known as "operators with a determinant".</p> <p>For Banach spaces, the essential details go along these lines. Fix a Banach space, X, and consider the <strong>finite rank</strong> linear operators. That means that T: X &rarr; X is such that Im(T) is finite dimensional. Such operators have a well-defined trace, tr(T). Using this trace we can define a norm on the subspace of finite-rank operators. If our operator were diagonalisable, we would define it as the sum of the absolute values of the eigenvalues (of which only finitely many are non-zero, of course). This norm is finer than the operator norm. We then take the closure in the space of all operators of the space of finite-rank operators with respect to this trace norm. These operators are called <strong>trace class</strong> operators. For such, there is a well-defined notion of a trace.</p> <p>(Incidentally, these operators form a two-sided ideal in the space of all operators and are actually the dual of the space of all operators via the pairing (S,T) &rarr; tr(ST).)</p> <p>Now trace and determinant are very closely linked via the forumula e<sup>tr T</sup> = det e<sup>T</sup>. This means that we can use our trace class operators to define a new class of "operators with a determinant". The key property should be that the exponential of a trace class operator should have a determinant. This suggests looking at the family of operators which differ from the identity by a trace class operator. Within this, we can look at the group of units, that is invertible operators.</p> <p>So an "operator with a determinant" is an invertible operator that differs from the identity by one of trace class.</p> <p>For more details, I recommend the book "Trace ideals and their applications" by Barry Simon (MR541149) and the article "On the homotopy type of certain groups of operators" by Richard Palais (MR0175130).</p> <p>But defining the determinant of an arbitrary operator is, of course, impossible. One can always figure out a renormalisation for a <em>particular</em> operator but there just ain't gonna be a system that works for everything: obviously det(I) = 1 but then det(2I) = ?</p> <p>(I should also say that I picked Banach spaces for ease of exposition. One can generalise this to locally convex topological spaces, but that involves handling nuclear materials so caution is advised.)</p>