nLab relation between determinant and trace

Statement

If $A$ is an $n \times n$ square matrix, then the determinant of its exponential equals the exponential of its trace

$$\det\big(\exp(A)\big) \;=\; \exp\big(tr(A)\big) \,.$$

More generally, the determinant of $A$ is a polynomial in the traces of the powers of $A$:

For $2 \times 2$-matrices:

$$det(A) \;=\; \tfrac{1}{2}\left( tr(A)^2 - tr(A^2) \right)$$

For $3 \times 3$-matrices:

$$det(A) \;=\; \tfrac{1}{6} \left( (tr(A))^3 - 3 tr(A^2) tr(A) + 2 tr(A^3) \right)$$

For $4 \times 4$-matrices:

$$det(A) \;=\; \tfrac{1}{24} \left( (tr(A))^4 - 6 tr(A^2)(tr(A))^2 + 3 (tr(A^2))^2 + 8 tr(A^3) tr(A) - 6 tr(A^4) \right)$$

Generally for $n \times n$-matrices (Kondratyuk-Krivoruchenko 92, appendix B):

(1)$$det(A) \;=\; \underset{ { k_1,\cdots, k_n \in \mathbb{N} } \atop { \underoverset{\ell = 1}{n}{\sum} \ell k_\ell = n } }{\sum} \underoverset{ l = 1 }{ n }{\prod} \frac{ (-1)^{k_l + 1} }{ l^{k_l} k_l ! } \left(tr(A^l)\right)^{k_l}$$

Proof of (1)

It is enough to prove this for semisimple matrices $A$ (matrices that are diagonalizable upon passing to the algebraic closure of the ground field) because this subset of matrices is Zariski dense (using for example the nonvanishing of the discriminant of the characteristic polynomial) and the set of $A$ for which the equation holds is Zariski closed.

Thus, without loss of generality we may suppose that $A$ is diagonal with $n$ eigenvalues $\lambda_1, \ldots, \lambda_n$ along the diagonal, where the statement can be rewritten as follows. Letting $p_k = tr(A^k) = \lambda_1^k + \ldots + \lambda_n^k$, the following identity holds:

$$\prod_{i=1}^n \lambda_i = \underset{ { k_1,\cdots, k_n \in \mathbb{N} } \atop { \underoverset{\ell = 1}{n}{\sum} \ell k_\ell = n } }{\sum} \underoverset{ l = 1 }{ n }{\prod} \frac{ (-1)^{k_l + 1} }{ l^{k_l} k_l ! } p_l^{k_l}$$

This of course is just a polynomial identity, one closely related to various of the Newton identities that concern symmetric polynomials in indeterminates $x_1, \ldots, x_n$. Thus we again let $p_k = x_1^k + \ldots + x_n^k$, and define the elementary symmetric polynomials $\sigma_k = \sigma_k(x_1, \ldots, x_n)$ via the generating function identity

$$\sum_{k \geq 0} \sigma_k t^k = \prod_{i=1}^n (1 + x_i t).$$

Then we compute

$$\array{ \sum_{k \geq 0} \sigma_k t^k & = & \prod_{i=1}^n (1 + x_i t) \\ & = & \exp\left(\sum_{i=1}^n \log(1 + x_i t)\right) \\ & = & \exp\left(\sum_{i=1}^n \sum_{k \geq 1} (-1)^{k+1} \frac{x_i^k}{k} t^k \right)\\ & = & \exp\left( \sum_{k \geq 1} (-1)^{k+1} \frac{p_k}{k} t^k\right) }$$

and simply match coefficients of $t^n$ in the initial and final series expansions, where we easily compute

$$x_1 x_2 \ldots x_n = \sum_{n = k_1 + 2k_2 + \ldots + n k_n} \prod_{l=1}^n \frac1{(k_l)!} \left(\frac{p_l}{l}\right)^{k_l} (-1)^{k_l+1} \,.$$

This completes the proof.

Related concepts

• determinant, trace

• characteristic form

References

The proof of (1) is spelled out in

• L. A. Kondratyuk, I. Krivoruchenko, Appendix B of: Superconducting quark matter in $SU(2)$ colour group, Zeitschrift für Physik A Hadrons and Nuclei March 1992, Volume 344, Issue 1, pp 99–115 (doi:10.1007/BF01291027)