Editing Cayley–Hamilton theorem (section)

=== A synthesis of the first two proofs ===
In the first proof, one was able to determine the coefficients {{math|''B''<sub>''i''</sub>}} of {{math|''B''}} based on the right-hand fundamental relation for the adjugate only. In fact the first {{math|''n''}} equations derived can be interpreted as determining the quotient {{math|''B''}} of the [[Euclidean division]] of the polynomial {{math|''p''(''t'')''I<sub>n</sub>''}} on the left by the [[monic polynomial]] {{math|''I<sub>n</sub>t'' − ''A''}}, while the final equation expresses the fact that the remainder is zero. This division is performed in the ring of polynomials with matrix coefficients. Indeed, even over a non-commutative ring, Euclidean division by a monic polynomial {{math|''P''}} is defined, and always produces a unique quotient and remainder with the same [[degree of a polynomial|degree]] condition as in the commutative case, provided it is specified at which side one wishes {{math|''P''}} to be a factor (here that is to the left).

To see that quotient and remainder are unique (which is the important part of the statement here), it suffices to write <math>PQ+r = PQ'+r'</math> as <math>P(Q-Q') = r'-r</math> and observe that since {{math|''P''}} is monic, {{math|''P''(''Q''−''Q''′)}} cannot have a degree less than that of {{math|''P''}}, unless {{math|''Q'' {{=}} ''Q''′}}.

But the dividend {{math|''p''(''t'')''I<sub>n</sub>''}} and divisor {{math|''I<sub>n</sub>t'' − ''A''}} used here both lie in the subring {{math|(''R''[''A''])[''t'']}}, where {{math|''R''[''A'']}} is the subring of the matrix ring {{math|''M''(''n'', ''R'')}} generated by {{math|''A''}}: the {{math|''R''}}-linear [[linear span|span]] of all powers of {{math|''A''}}. Therefore, the Euclidean division can in fact be performed within that ''commutative'' polynomial ring, and of course it then gives the same quotient {{math|''B''}} and remainder 0 as in the larger ring; in particular this shows that {{math|''B''}} in fact lies in {{math|(''R''[''A''])[''t'']}}.

But, in this commutative setting, it is valid to set {{math|''t''}} to {{math|''A''}} in the equation

<math display="block">p(t)I_n=(tI_n-A)B;</math>

in other words, to apply the evaluation map

<math display="block">\operatorname{ev}_A:(R[A])[t]\to R[A]</math>

which is a ring homomorphism, giving

<math display="block">p(A)=0\cdot\operatorname{ev}_A(B)=0</math>

just like in the second proof, as desired.

In addition to proving the theorem, the above argument tells us that the coefficients {{math| ''B<sub>i</sub>''}} of {{math|''B''}} are polynomials in {{math|''A''}}, while from the second proof we only knew that they lie in the centralizer {{math|''Z''}} of {{math|''A''}}; in general {{math|''Z''}} is a larger subring than {{math|''R''[''A'']}}, and not necessarily commutative. In particular the constant term {{math|1=''B''<sub>0</sub> = adj(−''A'')}} lies in {{math|''R''[''A'']}}. Since {{math|''A''}} is an arbitrary square matrix, this proves that {{math|adj(''A'')}} can always be expressed as a polynomial in {{math|''A''}} (with coefficients that depend on {{math|''A'')}}.

In fact, the equations found in the first proof allow successively expressing <math>B_{n-1}, \ldots, B_1, B_0</math> as polynomials in {{math|''A''}}, which leads to the identity
{{Equation box 1
|indent =::
|equation =<math>\operatorname{adj}(-A)=\sum_{i=1}^nc_iA^{i-1},</math>
|cellpadding= 6
|border
|border colour = #0070BF
|bgcolor=#FAFFFB}}
valid for all {{math|''n''&thinsp;×&thinsp;''n''}} matrices, where 
<math display="block">p(t)=t^n+c_{n-1}t^{n-1}+\cdots+c_1t+c_0</math> 
is the characteristic polynomial of {{mvar|A}}.

Note that this identity also implies the statement of the Cayley–Hamilton theorem: one may move {{math|adj(−''A'')}} to the right hand side, multiply the resulting equation (on the left or on the right) by {{math|''A''}}, and use the fact that
<math display="block">-A\cdot \operatorname{adj}(-A) = \operatorname{adj}(-A)\cdot (-A) = \det(-A) I_n = c_0I_n.</math>

{{see also|Faddeev–LeVerrier algorithm}}