linear algebra - Find NullSpace of a very sparse symbolic matrix

Below is a matrix diagram, produced in Mathematica. In this case it's a $956\times 950$ rectangular matrix. The white parts are all zero.

sa = Import["https://pastebin.com/raw/fiErKrhU", "Package"];
MatrixPlot[sa]

I'm wondering if there is a way to efficiently compute the null space of this matrix. From a different calculation entirely (using a Molien series), I know in advance there should be 6 linearly independent vectors in this null space, and I already know one of them.

The NullSpace routine takes too long to be feasible. I am hoping that there is a better way. I know that using NullSpace[N[m]] will return the answer rather quickly, but I am hoping to be able to do this symbolically.

Any help would be appreciated.

Update

Added SparseArray data for this matrix. It was too large for this message so I put it on pastebin.

https://pastebin.com/raw/fiErKrhU

Answer

Turns out this can be done with exact methods and a good option setting. And a dose of patience. I won't copy the matrix itself. In my session I named it mat as below.

AbsoluteTiming[
 nullspace = NullSpace[mat, Method -> "OneStepRowReduction"];]

(* Out[18]= {1839.141297, Null} *)

Check size and correctness:

Length[nullspace]


(* Out[19]= 6 *)

LeafCount[nullspace]

(* Out[20]= 50803 *)

Max[Abs[mat.Transpose[N[nullspace]]]]

(* Out[21]= 1.00182351685*10^-10 *)

I also have tried the method here but with no luck thus far. Oh well (I blame the author...)

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],

Blog

Search This Blog