Editing Determinant (section)

=== Trace ===
The [[Trace (linear algebra)|trace]] tr(''A'') is by definition the sum of the diagonal entries of {{mvar|A}} and also equals the sum of the eigenvalues. Thus, for complex matrices {{mvar|A}},
:<math>\det(\exp(A)) = \exp(\operatorname{tr}(A))</math>

or, for real matrices {{mvar|A}},
:<math>\operatorname{tr}(A) = \log(\det(\exp(A))).</math>

Here exp({{mvar|A}}) denotes the [[matrix exponential]] of {{mvar|A}}, because every eigenvalue {{mvar|λ}} of {{mvar|A}} corresponds to the eigenvalue exp({{mvar|λ}}) of exp({{mvar|A}}). In particular, given any [[matrix logarithm|logarithm]] of {{mvar|A}}, that is, any matrix {{mvar|L}} satisfying
:<math>\exp(L) = A</math>

the determinant of {{mvar|A}} is given by
:<math>\det(A) = \exp(\operatorname{tr}(L)).</math>

For example, for {{math|1=''n'' = 2}}, {{math|1=''n'' = 3}}, and {{math|1=''n'' = 4}}, respectively,
:<math>\begin{align}
  \det(A) &= \frac{1}{2}\left(\left(\operatorname{tr}(A)\right)^2 -  \operatorname{tr}\left(A^2\right)\right), \\
  \det(A) &= \frac{1}{6}\left(\left(\operatorname{tr}(A)\right)^3 - 3\operatorname{tr}(A) ~ \operatorname{tr}\left(A^2\right) + 2 \operatorname{tr}\left(A^3\right)\right), \\
  \det(A) &= \frac{1}{24}\left(\left(\operatorname{tr}(A)\right)^4 - 6\operatorname{tr}\left(A^2\right)\left(\operatorname{tr}(A)\right)^2 + 3\left(\operatorname{tr}\left(A^2\right)\right)^2 + 8\operatorname{tr}\left(A^3\right)~\operatorname{tr}(A) - 6\operatorname{tr}\left(A^4\right)\right).
\end{align}</math>

cf. [[Cayley–Hamilton theorem#Illustration for specific dimensions and practical applications|Cayley-Hamilton theorem]]. Such expressions are deducible from combinatorial arguments, [[Newton's identities#Computing coefficients|Newton's identities]], or the [[Faddeev–LeVerrier algorithm]]. That is, for generic {{mvar|n}}, {{math|det''A'' {{=}} (−1)<sup>''n''</sup>''c''<sub>0</sub>}} the signed constant term of the [[characteristic polynomial]], determined recursively from
:<math>c_n = 1; ~~~c_{n-m} = -\frac{1}{m}\sum_{k=1}^m c_{n-m+k} \operatorname{tr}\left(A^k\right) ~~(1 \le m \le n)~.</math>

In the general case, this may also be obtained from<ref>A proof can be found in the Appendix B of {{cite journal | last1 = Kondratyuk | first1 = L. A. | last2 = Krivoruchenko | first2 = M. I. | year = 1992 | title = Superconducting quark matter in SU(2) color group | journal = Zeitschrift für Physik A | volume = 344 | issue = 1| pages = 99–115 | doi = 10.1007/BF01291027 | bibcode = 1992ZPhyA.344...99K | s2cid = 120467300 }}</ref>
:<math>\det(A) = \sum_{\begin{array}{c}k_1,k_2,\ldots,k_n \geq 0\\k_1+2k_2+\cdots+nk_n=n\end{array}}\prod_{l=1}^n \frac{(-1)^{k_l+1}}{l^{k_l}k_l!} \operatorname{tr}\left(A^l\right)^{k_l},</math>

where the sum is taken over the set of all integers {{math|''k<sub>l</sub>'' ≥ 0}} satisfying the equation
:<math>\sum_{l=1}^n lk_l = n.</math>

The formula can be expressed in terms of the complete exponential [[Bell polynomial]] of ''n'' arguments ''s''<sub>''l''</sub> = −(''l'' – 1)! tr(''A''<sup>''l''</sup>) as
:<math>\det(A) = \frac{(-1)^n}{n!} B_n(s_1, s_2, \ldots, s_n).</math>

This formula can also be used to find the determinant of a matrix {{math|''A<sup>I</sup><sub>J</sub>''}} with multidimensional indices {{math|1=''I'' = (''i''<sub>1</sub>, ''i''<sub>2</sub>, ..., ''i<sub>r</sub>'')}} and {{math|1=''J'' = (''j''<sub>1</sub>, ''j''<sub>2</sub>, ..., ''j<sub>r</sub>'')}}. The product and trace of such matrices are defined in a natural way as
:<math>(AB)^I_J = \sum_K A^I_K B^K_J, \operatorname{tr}(A) = \sum_I A^I_I.</math>

An important arbitrary dimension {{mvar|n}} identity can be obtained from the [[Mercator series]] expansion of the logarithm when the expansion converges.  If every eigenvalue of ''A'' is less than 1 in absolute value,
:<math>\det(I + A) = \sum_{k=0}^\infty \frac{1}{k!} \left(-\sum_{j=1}^\infty \frac{(-1)^j}{j} \operatorname{tr}\left(A^j\right)\right)^k\,,</math>

where {{math|''I''}} is the identity matrix.  More generally, if
:<math>\sum_{k=0}^\infty \frac{1}{k!} \left(-\sum_{j=1}^\infty \frac{(-1)^j s^j}{j}\operatorname{tr}\left(A^j\right)\right)^k\,,</math>

is expanded as a formal [[power series]] in {{mvar|s}} then all coefficients of {{mvar|s}}<sup>{{mvar|m}}</sup> for {{math|''m'' &gt; ''n''}} are zero and the remaining polynomial is {{math|det(''I'' + ''sA'')}}.