Editing Cayley–Hamilton theorem (section)

===Determinant and inverse matrix===
{{see also|Determinant#Relation to eigenvalues and trace|Characteristic polynomial#Properties}}
For a general {{math|''n''&thinsp;×&thinsp;''n''}} [[invertible matrix]] {{mvar|A}}, i.e., one with nonzero determinant, {{mvar|A}}<sup>−1</sup> can thus be written as an {{nowrap|{{math|(''n'' −&thinsp;1)}}-th}} order [[polynomial expression]] in {{mvar|A}}: As indicated, the Cayley–Hamilton theorem amounts to the identity {{Equation box 1
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|equation = <math>p(A)=A^n+c_{n-1}A^{n-1}+\cdots+c_1A+(-1)^n\det(A)I_n =0.</math>
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The coefficients {{math|''c''<sub>''i''</sub>}} are given by the [[elementary symmetric polynomial]]s of the [[eigenvalue]]s of {{mvar|A}}. Using [[Newton identities]], the elementary symmetric polynomials can in turn be expressed in terms of [[power sum symmetric polynomial]]s of the eigenvalues: 
<math display="block">s_k = \sum_{i=1}^n \lambda_i^k = \operatorname{tr}(A^k),</math> 
where {{math|tr(''A''<sup>''k''</sup>)}} is the [[Trace (linear algebra)|trace]] of the matrix {{mvar|A<sup>''k''</sup>}}. Thus, we can express {{math|''c''<sub>''i''</sub>}} in terms of the trace of powers of {{mvar|A}}.

In general, the formula for the coefficients {{math|''c''<sub>''i''</sub>}} is given in terms of complete exponential [[Bell polynomials]] as<ref group="nb">See Sect. 2 of {{harvtxt|Krivoruchenko|2016}}. An explicit expression for the coefficients
{{math|''c''<sub>''i''</sub>}} is provided by {{harvtxt| Kondratyuk | Krivoruchenko |1992}}:
<math display="block">c_i = \sum_{ k_1,k_2,\ldots ,k_n}\prod_{l=1}^{n} \frac{(-1)^{k_l+1}}{l^{k_l} k_l!}\operatorname{tr}(A^l)^{k_l},</math>
where the sum is taken over the sets of all integer partitions {{math|''k''<sub>''l''</sub> ≥ 0}} satisfying the equation 
<math display="block">\sum_{l=1}^{n}lk_{l} = n-i.</math></ref> 
<math display="block">c_{n-k} = \frac{(-1)^{k}}{k!} B_k(s_1, -1! s_2, 2! s_3, \ldots, (-1)^{k-1}(k-1)! s_k).</math>

In particular, the determinant of {{math|''A''}} equals {{math|(−1)<sup>''n''</sup>''c''<sub>0</sub>}}. Thus, the determinant can be written as the [[trace identity]]:
<math display="block">\det(A) = \frac{1}{n!} B_n(s_1, -1! s_2, 2! s_3, \ldots, (-1)^{n-1}(n-1)! s_n).</math>

Likewise, the characteristic polynomial can be written as
<math display="block">-(-1)^n\det(A)I_n = A(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_{1}I_n),</math>
and, by multiplying both sides by {{math|''A''<sup>−1</sup>}} (note {{math|1=−(−1)<sup>''n''</sup> = (−1)<sup>''n''−1</sup>}}), one is led to an expression for the [[inverse matrix|inverse]] of {{mvar|A}} as a trace identity,
<math display="block"> 
\begin{align}
A^{-1} & = \frac{(-1)^{n-1}}{\det A}(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_{1}I_n), \\[5pt]
 & = \frac{1}{\det A}\sum_{k=0}^{n-1} (-1)^{n+k-1}\frac{A^{n-k-1}}{k!} B_k(s_1, -1! s_2, 2! s_3, \ldots, (-1)^{k-1}(k-1)! s_k). 
\end{align}
</math>

Another method for obtaining these coefficients {{math|''c''<sub>''k''</sub>}} for a general {{math|''n''&thinsp;×&thinsp;''n''}} matrix, provided no root be zero, relies on the following alternative [[Matrix exponential#The determinant of the matrix exponential|expression for the determinant]],
<math display="block">p(\lambda)= \det (\lambda I_n -A) = \lambda^n \exp (\operatorname{tr} (\log (I_n - A/\lambda))).</math>
Hence, by virtue of the [[Mercator series]],
<math display="block">p(\lambda)= \lambda^n \exp \left( -\operatorname{tr} \sum_{m=1}^\infty {({A\over\lambda})^m \over m} \right),</math>
where the exponential ''only'' needs be expanded to order {{math|''λ''<sup>−''n''</sup>}}, since {{math|''p''(''λ'')}} is of order {{math|''n''}}, the net negative powers of {{math|''λ''}} automatically vanishing by the C–H theorem. (Again, this requires a ring containing the [[rational number]]s.) [[Derivative|Differentiation]] of this expression with respect to {{mvar|λ}} allows one to express the coefficients of the characteristic polynomial for general {{mvar|n}} as determinants of {{math|''m''&thinsp;×&thinsp;''m''}} matrices,<ref group="nb">See, e.g., p.&nbsp;54 of {{harvnb|Brown|1994}}, which solves [[Jacobi's formula]], 
<math display="block">\frac{\partial p(\lambda)}{\partial \lambda} = p(\lambda) \sum^\infty _{m=0}\lambda ^{-(m+1)} \operatorname{tr}A^m = p(\lambda) ~ \operatorname{tr} \frac{I}{\lambda I -A}\equiv\operatorname{tr} B~, </math>
where {{mvar|B}} is the adjugate matrix of the next section.

There also exists an equivalent, related recursive algorithm introduced by [[Urbain Le Verrier]] and [[Dmitry Konstantinovich Faddeev]]—the [[Faddeev–LeVerrier algorithm]], which reads

<math display="block">\begin{align}
M_0 &\equiv O      & c_n &= 1                                                            \qquad &(k=0) \\[5pt]
M_k &\equiv AM_{k-1} -\frac{1}{k-1}(\operatorname{tr}(AM_{k-1})) I \qquad \qquad & c_{n-k} &= -\frac 1 k \operatorname{tr}(AM_k) \qquad &k=1,\ldots ,n ~.
\end{align}</math>

(see, e.g., {{harvnb|Gantmacher|1960|p=88}}.) Observe {{math|''A''<sup>−1</sup> {{=}} −''M''<sub>''n''</sub> /''c''<sub>0</sub>}} as the recursion terminates.
See the algebraic proof in the following section, which relies on the modes of the adjugate, {{math|''B''<sub>''k''</sub> ≡ ''M''<sub>''n''−''k''</sub>}}.
Specifically, <math>(\lambda I-A) B = I p(\lambda)</math> and the above derivative of {{mvar|p}} when one traces it yields

<math display="block">\lambda p' -n p =\operatorname{tr} (AB)~,</math> ({{harvnb|Hou|1998}}), and the above recursions, in turn.</ref>
<math display="block">c_{n-m} = \frac{(-1)^m}{m!}
\begin{vmatrix}
\operatorname{tr}A & m-1 & 0 & \cdots \\
\operatorname{tr}A^2 &\operatorname{tr}A & m-2 &\cdots \\
 \vdots & \vdots & & & \vdots \\
\operatorname{tr}A^{m-1} &\operatorname{tr}A^{m-2}& \cdots & \cdots & 1 \\
\operatorname{tr}A^m &\operatorname{tr}A^{m-1}& \cdots & \cdots & \operatorname{tr}A \end{vmatrix} ~.</math>

; Examples
For instance, the first few Bell polynomials are {{math|''B''<sub>0</sub>}} = 1, {{math|1=''B''<sub>1</sub>(''x''<sub>1</sub>) = ''x''<sub>1</sub>}}, {{math|1=''B''<sub>2</sub>(''x''<sub>1</sub>, ''x''<sub>2</sub>) = ''x''{{su|b=1|p=2}} + ''x''<sub>2</sub>}}, and {{math|1=''B''<sub>3</sub>(''x''<sub>1</sub>, ''x''<sub>2</sub>, ''x''<sub>3</sub>) = ''x''{{su|b=1|p=3}} + 3 ''x''<sub>1</sub>''x''<sub>2</sub> + ''x''<sub>3</sub>}}.

Using these to specify the coefficients {{math|''c<sub>i</sub>''}} of the characteristic polynomial of a {{math|2&thinsp;×&thinsp;2}} matrix yields

<math display="block">\begin{align}
c_2 = B_0 = 1, \\[4pt]
c_1 = \frac{-1}{1!} B_1(s_1) = - s_1 = - \operatorname{tr}(A), \\[4pt]
c_0 = \frac{1}{2!} B_2(s_1, -1! s_2) = \frac{1}{2}(s_1^2 - s_2) = \frac{1}{2}((\operatorname{tr}(A))^2 - \operatorname{tr}(A^2)).
\end{align}</math>

The coefficient {{math|''c''<sub>0</sub>}} gives the determinant of the {{math|2&thinsp;×&thinsp;2}} matrix, {{math|''c''<sub>1</sub>}} minus its trace, while its inverse is given by
<math display="block"> A^{-1} = \frac{-1}{\det A}(A + c_1 I_2) = \frac{-2(A - \operatorname{tr}(A) I_2)}{(\operatorname{tr}(A))^2 - \operatorname{tr}(A^2)}.</math>

It is apparent from the general formula for ''c''<sub>''n''−''k''</sub>, expressed in terms of Bell polynomials, that the expressions
<math display="block">-\operatorname{tr}(A)\quad \text{and} \quad \tfrac 1 2 (\operatorname{tr}(A)^2 - \operatorname{tr}(A^2))</math>

always give the coefficients {{math|''c''<sub>''n''−1</sub>}} of {{math|''λ''<sup>''n''−1</sup>}} and {{math|''c''<sub>''n''−2</sub>}} of {{math|''λ''<sup>''n''−2</sup>}} in the characteristic polynomial of any {{math|''n''&thinsp;×&thinsp;''n''}} matrix, respectively. So, for a {{math|3&thinsp;×&thinsp;3}} matrix {{mvar|A}}, the statement of the Cayley–Hamilton theorem can also be written as
<math display="block">A^3- (\operatorname{tr}A)A^2+\frac{1}{2}\left((\operatorname{tr}A)^2-\operatorname{tr}(A^2)\right)A-\det(A)I_3=O,</math>
where the right-hand side designates a {{math|3&thinsp;×&thinsp;3}} matrix with all entries reduced to zero. Likewise, this determinant in the {{math|1=''n'' = 3}} case, is now
<math display="block">\begin{align}
\det(A) &= \frac{1}{3!} B_3(s_1, -1! s_2, 2! s_3) = \frac{1}{6}(s_1^3 + 3 s_1 (-s_2) + 2 s_3) \\[5pt]
 &= \frac{1}{6} \left [ (\operatorname{tr}A)^3-3\operatorname{tr}(A^2)(\operatorname{tr}A)+2\operatorname{tr}(A^3) \right ].
\end{align}</math>
This expression gives the negative of coefficient {{math|''c''<sub>''n''−3</sub>}} of {{math|''λ''<sup>''n''−3</sup>}} in the general case, as seen below.

Similarly, one can write for a {{math|4&thinsp;×&thinsp;4}} matrix {{mvar|A}},
<math display="block"> A^4-(\operatorname{tr}A)A^3 + \tfrac{1}{2}\left[(\operatorname{tr}A)^2-\operatorname{tr}(A^2)\right]A^2 - \tfrac{1}{6}\left[ (\operatorname{tr}A)^3-3\operatorname{tr}(A^2)(\operatorname{tr}A)+2\operatorname{tr}(A^3)\right] A + \det(A)I_4 = O,</math>

where, now, the determinant is {{math|''c''<sub>''n''−4</sub>}},

<math display="block">\tfrac{1}{24}\! \left [ (\operatorname{tr}A)^4-6 \operatorname{tr}(A^2)(\operatorname{tr}A)^2 + 3\left(\operatorname{tr}(A^2)\right)^2 + 8\operatorname{tr}(A^3)\operatorname{tr}(A) -6\operatorname{tr}(A^4) \right ],</math>

and so on for larger matrices. The increasingly complex expressions for the coefficients {{math|''c''<sub>''k''</sub>}} is deducible from [[Newton's identities]] or the [[Faddeev–LeVerrier algorithm]].