The eigenvalues of a $2\times2$ matrix can be expressed in terms of the trace and determinant.
$\lambda_\pm = \frac{1}{2}\left(\textrm{tr} \pm \sqrt{\textrm{tr}^2-4\det}\right)$
Is there a similar formula for higher dimensional matrices?
Approach
The trace and determinant of a matrix are equal to the trace and determinant of the matrix in Jordan normal form. For a matrix in Jordan canonical form, $\textrm{tr } =\sum \lambda$ and $\det =\prod \lambda $.
Substituting these latter two identities into the first results in an identity, which is encouraging. I'm not sure how to check this assumption for larger matrices. I'm not sure how generate more than two eigenvalues from the first formula. For the $3\times3$ case, the first formula seems to break down.
Best Answer
For a $2\times2$ matrix, $\operatorname{tr}$ and $\det$ are the matrix invariants that are the coefficients of the characteristic polynomial.
For a $3\times3$ matrix there are the same invariants and another one, given by $$ \frac{1}{2}\left[(\operatorname{tr}A)^{2}-\operatorname{tr}(A^{2})\right] $$ but expressing the eigenvalues in terms of invariant means to solve a cubic equation.
For higher dimensions there are other invariants, but solving a polynomial equation cannot be done by a general formula for $n\geq5$.