Editing Determinant (section)

== Properties of the determinant in relation to other notions ==

=== Eigenvalues and characteristic polynomial ===

The determinant is closely related to two other central concepts in linear algebra, the [[eigenvalue]]s and the [[characteristic polynomial]] of a matrix. Let <math>A</math> be an <math>n \times n</math> matrix with [[complex number|complex]] entries. Then, by the Fundamental Theorem of Algebra, <math>A</math> must have exactly ''n'' [[eigenvectors|eigenvalues]] <math>\lambda_1, \lambda_2, \ldots, \lambda_n</math>. (Here it is understood that an eigenvalue with [[algebraic multiplicity]] {{mvar|μ}} occurs {{mvar|μ}} times in this list.) Then, it turns out the determinant of {{mvar|A}} is equal to the ''product'' of these eigenvalues,
:<math>\det(A) = \prod_{i=1}^n \lambda_i=\lambda_1\lambda_2\cdots\lambda_n.</math>
The product of all non-zero eigenvalues is referred to as [[pseudo-determinant]].

From this, one immediately sees that the determinant of a matrix <math>A</math> is zero if and only if <math>0</math> is an eigenvalue of <math>A</math>. In other words, <math>A</math> is invertible if and only if <math>0</math> is not an eigenvalue of <math>A</math>.

The characteristic polynomial is defined as<ref>{{harvnb|Lang|1985|loc=§VIII.2}}, {{harvnb|Horn|Johnson|2018|loc=Def. 1.2.3}}</ref>
:<math>\chi_A(t) = \det(t \cdot I - A).</math>
Here, <math>t</math> is the [[indeterminate (variable)|indeterminate]] of the polynomial and <math>I</math> is the identity matrix of the same size as <math>A</math>. By means of this polynomial, determinants can be used to find the [[eigenvalue]]s of the matrix <math>A</math>: they are precisely the [[Root of a polynomial|roots]] of this polynomial, i.e., those complex numbers <math>\lambda</math> such that
:<math>\chi_A(\lambda) = 0.</math>

A [[Hermitian matrix]] is [[positive definite matrix|positive definite]] if all its eigenvalues are positive. [[Sylvester's criterion]] asserts that this is equivalent to the determinants of the submatrices
:<math>A_k := \begin{bmatrix}
  a_{1,1} & a_{1,2} & \cdots & a_{1,k} \\
  a_{2,1} & a_{2,2} & \cdots & a_{2,k} \\
   \vdots &  \vdots & \ddots &  \vdots \\
  a_{k,1} & a_{k,2} & \cdots & a_{k,k}
\end{bmatrix}</math>

being positive, for all <math>k</math> between <math>1</math> and <math>n</math>.<ref>{{harvnb|Horn|Johnson|2018|loc=Observation 7.1.2, Theorem 7.2.5}}</ref>

=== 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'')}}.

=== Upper and lower bounds ===
For a positive definite matrix {{math|''A''}}, the trace operator gives the following tight lower and upper bounds on the log determinant
:<math>\operatorname{tr}\left(I - A^{-1}\right) \le \log\det(A) \le \operatorname{tr}(A - I)</math>

with equality if and only if {{math|1=''A'' = ''I''}}. This relationship can be derived via the formula for the [[Kullback-Leibler divergence]] between two [[multivariate normal]] distributions.

Also,
:<math>\frac{n}{\operatorname{tr}\left(A^{-1}\right)} \leq \det(A)^\frac{1}{n} \leq \frac{1}{n}\operatorname{tr}(A) \leq \sqrt{\frac{1}{n}\operatorname{tr}\left(A^2\right)}.</math>

These inequalities can be proved by expressing the traces and the determinant in terms of the eigenvalues. As such, they represent the well-known fact that the [[harmonic mean]] is less than the [[geometric mean]], which is less than the [[arithmetic mean]], which is, in turn, less than the [[root mean square]].

=== Derivative ===
The Leibniz formula shows that the determinant of real (or analogously for complex) square matrices is a [[polynomial]] function from <math>\mathbf R^{n \times n}</math> to <math>\mathbf R</math>. In particular, it is everywhere [[differentiable]]. Its derivative can be expressed using [[Jacobi's formula]]:<ref>{{harvnb|Horn|Johnson|2018|loc=§&nbsp;0.8.10}}</ref>

:<math>\frac{d \det(A)}{d \alpha} = \operatorname{tr}\left(\operatorname{adj}(A) \frac{d A}{d \alpha}\right).</math>

where <math>\operatorname{adj}(A)</math> denotes the [[adjugate]] of <math>A</math>. In particular, if <math>A</math> is invertible, we have

:<math>\frac{d \det(A)}{d \alpha} = \det(A) \operatorname{tr}\left(A^{-1} \frac{d A}{d \alpha}\right).</math>

Expressed in terms of the entries of <math>A</math>, these are

: <math> \frac{\partial \det(A)}{\partial A_{ij}}= \operatorname{adj}(A)_{ji} = \det(A)\left(A^{-1}\right)_{ji}.</math>

Yet another equivalent formulation is

:<math>\det(A + \epsilon X) - \det(A) = \operatorname{tr}(\operatorname{adj}(A) X) \epsilon + O\left(\epsilon^2\right) = \det(A) \operatorname{tr}\left(A^{-1} X\right) \epsilon + O\left(\epsilon^2\right)</math>,

using [[big O notation]]. The special case where <math>A = I</math>, the identity matrix, yields

:<math>\det(I + \epsilon X) = 1 + \operatorname{tr}(X) \epsilon + O\left(\epsilon^2\right).</math>

This identity is used in describing [[Lie algebra]]s associated to certain matrix [[Lie group]]s. For example, the special linear group <math>\operatorname{SL}_n</math> is defined by the equation <math>\det A = 1</math>. The above formula shows that its Lie algebra is the [[special linear Lie algebra]] <math>\mathfrak{sl}_n</math> consisting of those matrices having trace zero.

Writing a <math>3 \times 3</math> matrix as <math>A = \begin{bmatrix}a & b & c\end{bmatrix}</math> where <math>a, b,c</math> are column vectors of length 3, then the gradient over one of the three vectors may be written as the [[cross product]] of the other two:

: <math>\begin{align}
  \nabla_\mathbf{a}\det(A) &= \mathbf{b} \times \mathbf{c} \\
  \nabla_\mathbf{b}\det(A) &= \mathbf{c} \times \mathbf{a} \\
  \nabla_\mathbf{c}\det(A) &= \mathbf{a} \times \mathbf{b}.
\end{align}</math>