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?


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.

  • $\begingroup$ Trace and deteminant are still sum an dproduct of eigenvalues. However, the other coefficients of the characteristic polynomials are the mixed symmetric polynomials, such as $\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1$ in the $3\times 3$ case. Those are hardly better described tha the coefficients of $\det(A-\lambda I)$. $\endgroup$ – Hagen von Eitzen May 23 '14 at 22:00
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    $\begingroup$ As mentioned in the answers, a formula for the eigenvalues is hopeless for $n \geq 5$, but one can get formulas for the coefficients of the characteristic polynomial, for instance some of the things mentioned here. $\endgroup$ – George Shakan May 23 '14 at 22:48

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 $n\geq5$.

  • $\begingroup$ Could you please write out the characteristic polynomial explicitly? $\endgroup$ – DrM Jan 20 at 13:18

There is if you generalize in the correct manner. The characteristic equation $\lambda^n+\sum\limits_{i=0}^{n-1}c_i\lambda^i=0$ can be expressed with coefficients in terms of the trace and the determinant of the matrix, but as $n$ grows, this gets extremely laborious. Please see this Wikipedia article.

Of particular interest are $c_{n-1}=-\DeclareMathOperator{\tr}{tr} \tr(M)$ and $c_0=(-1)^n\det(M)$.


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