I have a square matrix whose dimensions is 9 cross 9, when I extract the rank of the matrix R, I am getting rank as 6. I have constructed R matrix by minimizing the Lagrangian Lg with respect to a[1]..a[7], d[1] and λ. My objective is to get the roots of β when the rank of the matrix is full. In a way, I am solving Ax=0.

Lg=-24.3551 a[1]^2-0.00707144 a[1] a[2]-389.64 a[2]^2+0.00500023 a[1] a[3]-0.00707116 a[2] a[3]-1972.53 a[3]^2-1.32815*10^-10 a[2] a[4]-1.33452*10^-9 a[3] a[4]-6234.18 a[4]^2+37.99 a[1] a[5]+394.314 a[2] a[5]-192.796 a[3] a[5]+0.000393434 a[4] a[5]-120.672 a[5]^2+12.9494 a[1] a[6]-387.102 a[2] a[6]-3532.97 a[3] a[6]+0.000319477 a[4] a[6]+0.00011462 a[5] a[6]-1710.82 a[6]^2+9.77634 a[1] a[7]-228.736 a[2] a[7]+960.941 a[3] a[7]-0.000176814 a[4] a[7]-0.00061614 a[5] a[7]-0.000460789 a[6] a[7]-5559.35 a[7]^2+β^4 (0.250029 a[1]^2+0.250002 a[2]^2+0.25 a[3]^2+0.25 a[4]^2-0.389965 a[1] a[5]-0.252998 a[2] a[5]+0.024435 a[3] a[5]-3.09273*10^-9 a[4] a[5]+0.216672 a[5]^2-0.132924 a[1] a[6]+0.248371 a[2] a[6]+0.44777 a[3] a[6]-1.93707*10^-8 a[4] a[6]-3.23022*10^-8 a[5] a[6]+0.280169 a[6]^2-0.100353 a[1] a[7]+0.146761 a[2] a[7]-0.12179 a[3] a[7]+7.35786*10^-10 a[4] a[7]+1.69118*10^-8 a[5] a[7]+9.20443*10^-9 a[6] a[7]+0.125315 a[7]^2)+λ (-0.500022 a[1]+0.707114 a[2]-0.500002 a[3]+7.32034*10^-8 a[5]+4.35461*10^-8 a[6]+8.5048*10^-8 a[7]-1. d[1])-5921.76 d[1]^2+4.9348 β^4 d[1]^2;
  • $\begingroup$ rationalize your coefficients? $\endgroup$ – AccidentalFourierTransform Aug 28 '19 at 12:26
  • $\begingroup$ How to rationalize the coefficients? $\endgroup$ – acoustics Aug 28 '19 at 12:49
  • 1
    $\begingroup$ 243551/10000 instead of 24.3551, etc. More easily, use Rationalize[Lg,0] instead of Lg. $\endgroup$ – AccidentalFourierTransform Aug 28 '19 at 12:53
  • $\begingroup$ It works, but too many digits are coming as a output $\endgroup$ – acoustics Aug 28 '19 at 13:06
  • $\begingroup$ What do you mean by "get the roots of $\beta$ when the rank of the matrix is full"? Your $9 \times 9$ matrix $R$ seems to have $rank(R) = 6 < 9$, so it does NOT have full rank. If a matrix has full rank, then it is invertible and $Rx = 0$ has only the trivial solution $x=0$. What do you mean? May be, see also math.stackexchange.com/questions/628925/… . $\endgroup$ – Mauricio Fernández Aug 29 '19 at 9:29

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