Editing Quadratic form (section)

== Geometric meaning ==
Using [[Cartesian coordinates]] in three dimensions, let {{math|1='''x''' = (''x'', ''y'', ''z'')<sup>T</sup>}}, and let {{math|''A''}} be a [[symmetric matrix|symmetric]] 3-by-3 matrix. Then the geometric nature of the [[solution set]] of the equation {{math|1='''x'''<sup>T</sup>''A'''''x''' + '''b'''<sup>T</sup>'''x''' = 1}} depends on the eigenvalues of the matrix {{math|''A''}}.

If all [[eigenvalue]]s of {{math|''A''}} are non-zero, then the solution set is an [[ellipsoid]] or a [[hyperboloid]].{{Citation needed|date=February 2017}} If all the eigenvalues are positive, then it is an ellipsoid; if all the eigenvalues are negative, then it is an ''imaginary ellipsoid'' (we get the equation of an ellipsoid but with imaginary radii); if some eigenvalues are positive and some are negative, then it is a hyperboloid.

If there exist one or more eigenvalues {{math|1=''λ''<sub>''i''</sub> = 0}}, then the shape depends on the corresponding {{math|''b''<sub>''i''</sub>}}. If the corresponding {{math|''b''<sub>''i''</sub> ≠ 0}}, then the solution set is a [[paraboloid]] (either elliptic or hyperbolic); if the corresponding {{math|1=''b''<sub>''i''</sub> = 0}}, then the dimension {{math|''i''}} degenerates and does not come into play, and the geometric meaning will be determined by other eigenvalues and other components of {{math|'''b'''}}. When the solution set is a paraboloid, whether it is elliptic or hyperbolic is determined by whether all other non-zero eigenvalues are of the same sign: if they are, then it is elliptic; otherwise, it is hyperbolic.