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I was trying to find square root of this $3×3$ matrix $$\begin{pmatrix} 12&8&31\\ 7&14&28\\ 9&35&6\\ \end{pmatrix}$$ I searched Wikipedia and some site for helpful information and also try to $${ \begin{pmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{pmatrix} }^3 = \begin{pmatrix} 12&8&31\\ 7&14&28\\ 9&35&6\\ \end{pmatrix}$$ Though I know that a similar question has been posted for a $2×2$ matrix, the workings are different so I'll need some help on this Thanks

2 Answers2

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And here is how to get that result with SageMath

sage: M = matrix(QQbar,3,3,[12,8,31,7,14,28,9,35,6])                            
sage: a,b=M.diagonalization()                                                   
sage: c=diagonal_matrix([x.sqrt() for x in a.diagonal()])                       
sage: b*c*~b                                                                    
[ 2.994275474024927? + 0.151836493602799?*I  1.197978250858210? + 2.057588565728163?*I  3.000734901853716? - 2.183155335351262?*I]
[0.777514011271777? + 0.1253357713833934?*I  3.460596511446746? + 1.698468161085322?*I  2.737777768777000? - 1.802119184349875?*I]
[0.9410435413715063? - 0.201547233197786?*I  3.344874508620062? - 2.731235901474129?*I  2.789009650240784? + 2.897912794483287?*I]
F. C.
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There are eight (complex) square roots of this matrix, since there are three distinct, nonzero eigenvalues. The closed-form expressions for these will be quite unpleasant, as the characteristic polynomial is an irreducible cubic. A numerical approximation for one of them is

$$ \left[\begin{array}{ccc} 2.994275474+ 0.1518364930 \;{i} & 1.197978251+ 2.057588565 \;{i} & 3.000734902- 2.183155336 \;{i} \\ 0.7775140113+ 0.1253357713 \;{i} & 3.460596511+ 1.698468161 \;{i} & 2.737777769- 1.802119185 \;{i} \\ 0.9410435414- 0.2015472331 \;{i} & 3.344874508- 2.731235902 \;{i} & 2.789009650+ 2.897912794 \;{i} \end{array}\right] $$

Robert Israel
  • 448,999