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I have a matrix M = {{0,1,b},{a,0,1},{1,c,0}}.

For my work, I need to assume that $M$ is singular, i.e. $abc+1=0$. The issue is that I am trying to work out NullSpace[M], it assumes that the determinant is nonzero. I have tried using Refine, but it seems to make no difference, even if I add Assumptions -> {a==1,b==-1,c==1}. Any ideas on how I can make Mathematica interpret the matrix as singular?

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  • $\begingroup$ Can you just replace c with -1/(a b)? $\endgroup$
    – evanb
    Feb 9 at 12:57
  • $\begingroup$ That makes sense, thanks. I guess I kept thinking that I'd see $a,b,c$ in the final expression that I didn't even think of replacing them. Thanks! $\endgroup$
    – Jacques
    Feb 9 at 13:11

1 Answer 1

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Create an explicit matrix-times-vector equation, with extra conditions that the vector not vanish and the determinant vanish.

mat = {{0, 1, b}, {a, 0, 1}, {1, c, 0}};
vec = Array[x, Length[mat]];
Solve[Flatten[{Det[mat], mat . vec, vec[[1]] - 1}] == 0, vec, 
 MaxExtraConditions -> Infinity]

(* Out[183]= {{x[1] -> ConditionalExpression[1, 1 + a b c == 0], 
  x[2] -> ConditionalExpression[a b, 1 + a b c == 0], 
  x[3] -> ConditionalExpression[-a, 1 + a b c == 0]}} *)

Here I forced the first vector component to be 1. In cases where it has to vanish this won't work, but one could do similar but setting a random linear combination of the vector components to 1.

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