For a $2\times 2$ matrix $A$, I always compute the eigenvalues by setting the determinant of $A-\lambda I$ to zero, and then solving the resulting characteristic equation

But for an $n\times n$ matrix, the characteristic equation is an $n$ degree polynomial, and as far as I know, there is no way to find the roots of $n$ degree polynomials for even moderately large $n$ (I think if $n>3$, it is impossible to solve analytically except in special cases, correct?)

So how do we find the eigenvalues of a matrix for large $n$?

  • $\begingroup$ There are formulas to solve third- and fourth-degree equations, but they're very complicated. For higher degrees there are no formulas. In any event, you find the roots by numerical methods. That is, you find approximations, not exact expressions. But there are what you want, usually. Even if you get the exact answer for a quadratic, the first thing you'll do is converrt it to a decimal approximation, right? $\endgroup$ – saulspatz Oct 10 '17 at 18:39
  • $\begingroup$ @saulspatz There are methods for third and fourth degree polynomials which are easier to remember than the formulae (sometimes these can obscure the truth - integers can appear as complicated combinations of cube and square roots and it is not obvious that the answer is an integer). There are short cuts if the roots have nice properties of various kinds. But in general you are right, we are dealing with arbitrary polynomials with coefficients in whatever field/ring/algebra (etc) provides the entries of the matrices. $\endgroup$ – Mark Bennet Oct 10 '17 at 18:48
  • $\begingroup$ @saulspatz There are no solutions to higher equations using radicals, but there are other methods. That said, most of the ways of finding all roots to a polynomial involve numeric methods. $\endgroup$ – Thomas Andrews Oct 10 '17 at 18:48
  • $\begingroup$ @ThomasAndrews Yes, I know that, but it seemed to me that the OP must be thinking about solutions in radicals. $\endgroup$ – saulspatz Oct 10 '17 at 19:00
  • $\begingroup$ @ThomasAndrews, I'm not even sure what a radical is. $\endgroup$ – user56834 Oct 11 '17 at 6:25

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