For a 2x2 matrix for example, we can see that if $(Trace)^2 - 4*Determinant < 0$ , the matrix has complex eigenvalues. Can this be generalized? Also, if it is not possible to check directly, can we check using some decomposition (to Hessenberg/Tridiagonal form etc.)?
Asked
Active
Viewed 2,224 times
4
-
Well, a Jordan normal form will reveal the existence of complex eigenvalues, but I don't think that's what you're looking for. – AlexR Mar 18 '15 at 12:12
-
Are you familiar with the equation $|A -\lambda I| =0$? – GFauxPas Mar 18 '15 at 12:17
-
@AlexR In order to find the Jordan normal form, don't we first derive all the eigenvalues(on the diagonal)? It seems we cannot trivially check if a matrix has complex Eigen values. – Anurag Reddy Mar 18 '15 at 12:54
-
@GFauxPas Yes. That is how I arrived at $(Trace)^2 − 4∗Determinant < 0$ for 2x2 matrices. – Anurag Reddy Mar 18 '15 at 12:56
-
To add a little context, I am trying to implement Jordan and Schur decomposition. In order to check if Schur converges (to a real matrix), I am interested in finding out if there is any "Technique" which helps me determine if any of the eigenvalues are complex. – Anurag Reddy Mar 18 '15 at 13:01
1 Answers
2
Your $(Trace)^2-4*Det$ is simply the discriminant of the equation $|A-\lambda I|=0$ in the case that $A$ is a $2 \times 2$ matrix. The eigenvalues are, by definitions, the roots of this caracteristic equation that is, in general an equation of degree $n$ and, as far as I know, the only way to know if some eigenvalue is a complex number is to study the roots of this equation.
Generally, given an equation of degree $n$ the only way to know the number of its real solutions (and then by difference to the degree the number of complex solutions) is by means of the Sturm's theorem. See some reference on my answer to How to determine if a quintic polynomial is solvable by radicals.
Emilio Novati
- 62,675