$-\lambda^3 + 2\lambda^2 + 3\lambda - 2 = 0$. Any way to solve this for eigenvalues in a reasonable amount of time without resorting to software? $\lambda = \pm 1, \pm 2$ do not work
@ペガサスSeiya Geometry as you know it is not much a part of research. But there definitely are differential geometry and algebraic geometry.
@SillyGoose You can talk about isometries of $\Bbb C^2$ with its standard hermitian metric. This will be $U(2)$. And then the $S$ signifies that they have to have determinant $1$ (which makes sense for linear maps, independent of matrix representation).
@D.C.theIII No. If it is one of my problems, I know that you can find at least one rational root. Are you sure it's really the correct characteristic polynomial?
At a certain point, I don't see any reason not to use Wolfram or Matlab to give you eigenvalues/eigenvectors. You have to know the mechanics, but then why bother laboring over it if you're not in an actual class?
Fred is $\pm$ the sum of the principal $2\times 2$ minors — one of my linear algebra students named it 20+ years ago, and I stuck with it.
The non-leading coefficient is $\pm$ trace because it will be the sum of the eigenvalues.
The linear (in this case) coefficient will be the sum of the principal $2\times 2$ minors because it will be the sum of the products of pairs of eigenvalues. And determinant is obvious.
All these notions are independent of basis, so hold for any matrix representation of the linear map.
