Research level applications of "row rank = column rank"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T06:31:36Z http://mathoverflow.net/feeds/question/123194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123194/research-level-applications-of-row-rank-column-rank Research level applications of "row rank = column rank"? David Feldman 2013-02-28T05:19:07Z 2013-02-28T20:01:35Z <p>No less an authority than Gilbert Strang frames "row rank equals column rank" (and a couple of other facts) as "The Fundamental Theorem of Linear Algebra."</p> <p>I'd simply like to assemble (for teaching purposes) a list of research level applications of this basic fact.</p> <p>Applications can be theoretical or practical, and I would particularly appreciate learning what value this fact has in the minds of physicists.</p> <p>(It goes without say) please do not start a debate concerning the centrality of this fact (that's not what MO is for). My question merely seek insights or pointers to the literature that would support making a positive case for centrality. So if you use linear algebra all the time but never this fact, no need to chime in.</p> <p>"Row rank equals column rank" has the consequence for square matrices that $A$ singular makes $A^T$ singular; I'm sure this case comes up everywhere. Here I'm specifically looking for applications of the one-the-nose numerical equality of ranks.</p> <p>Feel free to offer a philosophical take on linear algebra that would support the centrality of "row rank equals column rank" even if that philosophy isn't grounded in specific results. (For example, does this simple statement offer a hidden paradigm for whole sophisticated theories.)</p> http://mathoverflow.net/questions/123194/research-level-applications-of-row-rank-column-rank/123205#123205 Answer by Duchamp Gérard H. E. for Research level applications of "row rank = column rank"? Duchamp Gérard H. E. 2013-02-28T10:29:26Z 2013-02-28T20:01:35Z <p>Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$; $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by $$(x,y)\to f(xy)$$<br> This holds even for infinite monoids when one considers the functions that have finite dimensional orbits by shifts [1] (for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ are of rank 2, $\exp$ has rank 1). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric. </p> <p>[1] For people who are familiar with these matters, this is the Sweedler's dual of $k[M]$ for the comultiplication of the monoid. </p> http://mathoverflow.net/questions/123194/research-level-applications-of-row-rank-column-rank/123212#123212 Answer by Benoît Kloeckner for Research level applications of "row rank = column rank"? Benoît Kloeckner 2013-02-28T13:00:20Z 2013-02-28T13:00:20Z <p>There is this proof of the De Bruijn-Erdös theorem: $p$ points in the plane, not all on the same line, at least $p$ lines go through at least two of the points.</p> <p>The linear algebraic proof goes like this: let $A$ be the incidence matrix of points versus lines (each row is labeled by a point, each column by a line going through at least two of the points, and the $ij$ coefficient is $1$ if the given point is on the given line, $0$ otherwise). Then it is easily seen that $det(AA^T)\neq0$. In particular the rank of $A$ is $p$, and since this is its column rank the number of columns must be at least $p$.</p> http://mathoverflow.net/questions/123194/research-level-applications-of-row-rank-column-rank/123222#123222 Answer by Noah Stein for Research level applications of "row rank = column rank"? Noah Stein 2013-02-28T15:10:14Z 2013-02-28T15:10:14Z <p>In some sense you can view the singular value decomposition as a sharpening of this theorem (for real and complex matrices, anyway). This, in turn, is useful all over the place.</p>