# Unable to find the Null space of a matrix R

R = {{-639880.9531714534 +
79.19371296239888 c^2, -453339.5840789839 +
55.56821108386519 c^2, 0, 0,
160.0595864532318}, {-453339.5840789839 +
55.56821108386519 c^2, -4.986933805092816*^6 +
74.24759881256949 c^2, 0, 0, -0.00447586740976628}, {0,
0, -639877.8131714534 +
79.19371296239888 c^2, -453339.5840789839 +
55.56821108386519 c^2, 160.0595864532318}, {0,
0, -453339.5840789839 +
55.56821108386519 c^2, -4.986933805092816*^6 +
74.24759881256949 c^2, -0.00447586740976628}, \
{160.0595864532318, -0.00447586740976628,
160.0595864532318, -0.00447586740976628, -520.8333333333335}};
MatrixRank[R]
P = FullSimplify[Det[R]]
s1 = NSolve[P == 0 && 0 < c < 10000]
s2 = Flatten[c /. s1]
NN = Flatten[NullSpace[R /. c -> s2[[1]]]];


I have a matrix R which is having a dependency on c. I took the Det[R] and found the Roots of the determinant function. I substituted the back to find the unknown coefficients. Essentially I am talking about Ax=0, But could not able to find the x

• There is noise in that code. Please provide a minimal working example. What is s3? Why do you define fn? Jul 30, 2018 at 10:21
• NullSpace[R] gives an empty list. Why do you say you can't find the NullSpace ? Jul 30, 2018 at 10:25
• Can Nullspace be nonempty, Even we can able to find the Root of a determinant function? Jul 30, 2018 at 10:27
• Computing the determinant to check wether a matrix is invertible is (i) highly inefficient and (ii) extremely inaccurate as a lot of numerical error can be accumulated. Jul 30, 2018 at 11:53

It is a problem of not having enough digits in the default machine precision. After running your original code, try

P /. s1
(*not zeroes*)


The fact that, after replacing the solution into the equation, one does not get something close to zero is the hint.

One way to force Mathematica to use more digits is as follows:

newR = ReplaceAll[R, x_?NumericQ :> Rationalize[x, 1/100000000000000]];
newS = NSolve[Det[newR] == 0 && 0 < c < 10000, WorkingPrecision -> 60];
Flatten[NullSpace[newR /. newS[[1]]]]
(*1D nullspace*)

• Thanks, I Implemented your suggestion and its working now. Jul 30, 2018 at 15:36

The presence of machine numbers and use of Det means there will be some amount of numerical error which in turn means one cannot necessarily get away with the default Tolerance for NullSpace. Dropping it somewhat does give a nontrivial null space generator.

NullSpace[R /. c -> s2[[1]], Tolerance -> 10^(-10)]

(* Out[62]= {{0.638092335503, -0.000643559092341,
0.658783883462, -0.000664427898733, 0.398548755651}} *)


This is not terribly good though (and I am not sure why). Instead I will recommend using the singular values decomposition. The

{uu, ww, vv} = SingularValueDecomposition[R /. c -> s2[[2]]];
nullvec = Last[Transpose[vv]]

(* Out[84]= {0.718279572307, -0.000713124517697, -0.695719341907, \
0.000690726186956, 0.00693308400591} *)

Check:

(R /. c -> s2[[2]]).Last[Transpose[vv]]

(* Out[85]= {-4.82763022225*10^-7, 6.63228964986*10^-10,
4.6759873662*10^-7, -4.65453959914*10^-10, -4.65985650067*10^-9} *)
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