Editing Adjugate matrix (section)

== Properties ==
For any {{math|''n'' × ''n''}} matrix {{math|'''A'''}}, elementary computations show that adjugates have the following properties:
* <math>\operatorname{adj}(\mathbf{I}) = \mathbf{I}</math>, where <math>\mathbf{I}</math> is the [[identity matrix]].
* <math>\operatorname{adj}(\mathbf{0}) = \mathbf{0}</math>, where <math>\mathbf{0}</math> is the [[zero matrix]], except that if <math>n=1</math> then <math>\operatorname{adj}(\mathbf{0}) = \mathbf{I}</math>.
* <math>\operatorname{adj}(c \mathbf{A}) = c^{n - 1}\operatorname{adj}(\mathbf{A})</math> for any scalar {{mvar|c}}.
* <math>\operatorname{adj}(\mathbf{A}^\mathsf{T}) = \operatorname{adj}(\mathbf{A})^\mathsf{T}</math>.
* <math>\det(\operatorname{adj}(\mathbf{A})) = (\det \mathbf{A})^{n-1}</math>.
* If {{math|'''A'''}} is invertible, then <math>\operatorname{adj}(\mathbf{A}) = (\det \mathbf{A}) \mathbf{A}^{-1}</math>.  It follows that:
** {{math|adj('''A''')}} is invertible with inverse {{math|(det '''A''')<sup>−1</sup>'''A'''}}.
** {{math|1=adj('''A'''<sup>−1</sup>) = adj('''A''')<sup>−1</sup>}}.
* {{math|adj('''A''')}} is entrywise [[polynomial]] in {{math|'''A'''}}.  In particular, over the [[real number|real]] or complex numbers, the adjugate is a [[smooth function]] of the entries of {{math|'''A'''}}.

Over the complex numbers,
* <math>\operatorname{adj}(\overline\mathbf{A}) = \overline{\operatorname{adj}(\mathbf{A})}</math>, where the bar denotes [[complex conjugation]].
* <math>\operatorname{adj}(\mathbf{A}^*) = \operatorname{adj}(\mathbf{A})^*</math>, where the asterisk denotes [[conjugate transpose]].

Suppose that {{math|'''B'''}} is another {{math|''n'' × ''n''}} matrix.  Then
:<math>\operatorname{adj}(\mathbf{AB}) = \operatorname{adj}(\mathbf{B})\operatorname{adj}(\mathbf{A}).</math>
This can be [[mathematical proof|proved]] in three ways.  One way, valid for any commutative ring, is a direct computation using the [[Cauchy–Binet formula]].  The second way, valid for the real or complex numbers, is to first observe that for invertible matrices {{math|'''A'''}} and {{math|'''B'''}},
:<math>\operatorname{adj}(\mathbf{B})\operatorname{adj}(\mathbf{A}) = (\det \mathbf{B})\mathbf{B}^{-1}(\det \mathbf{A})\mathbf{A}^{-1} = (\det \mathbf{AB})(\mathbf{AB})^{-1} = \operatorname{adj}(\mathbf{AB}).</math>
Because every non-invertible matrix is the limit of invertible matrices, [[continuous function|continuity]] of the adjugate then implies that the formula remains true when one of {{math|'''A'''}} or {{math|'''B'''}} is not invertible.

A [[corollary]] of the previous formula is that, for any non-negative [[integer]] {{mvar|k}},
:<math>\operatorname{adj}(\mathbf{A}^k) = \operatorname{adj}(\mathbf{A})^k.</math>
If {{math|'''A'''}} is invertible, then the above formula also holds for negative {{mvar|k}}.

From the identity
:<math>(\mathbf{A} + \mathbf{B})\operatorname{adj}(\mathbf{A} + \mathbf{B})\mathbf{B} = \det(\mathbf{A} + \mathbf{B})\mathbf{B} = \mathbf{B}\operatorname{adj}(\mathbf{A} + \mathbf{B})(\mathbf{A} + \mathbf{B}),</math>
we deduce
:<math>\mathbf{A}\operatorname{adj}(\mathbf{A} + \mathbf{B})\mathbf{B} = \mathbf{B}\operatorname{adj}(\mathbf{A} + \mathbf{B})\mathbf{A}.</math>

Suppose that {{math|'''A'''}} [[commuting matrices|commutes]] with {{math|'''B'''}}.  Multiplying the identity {{math|1='''AB''' = '''BA'''}} on the left and right by {{math|adj('''A''')}} proves that
:<math>\det(\mathbf{A})\operatorname{adj}(\mathbf{A})\mathbf{B} = \det(\mathbf{A})\mathbf{B}\operatorname{adj}(\mathbf{A}).</math>
If {{math|'''A'''}} is invertible, this implies that {{math|adj('''A''')}} also commutes with {{math|'''B'''}}.  Over the real or complex numbers, continuity implies that {{math|adj('''A''')}} commutes with {{math|'''B'''}} even when {{math|'''A'''}} is not invertible.

Finally, there is a more general proof than the second proof, which only requires that an ''n'' × ''n'' matrix has entries over a [[field (mathematics)|field]] with at least 2''n''&nbsp;+ 1 elements (e.g. a 5 × 5 matrix over the integers [[modular arithmetic|modulo]] 11). {{math|det('''A'''+''t'''''I''')}} is a polynomial in ''t'' with [[degree of a polynomial|degree]] at most ''n'', so it has at most ''n'' [[root of a polynomial|roots]]. Note that the ''ij''th entry of {{math|adj(('''A'''+''t'''''I''')('''B'''))}} is a polynomial of at most order ''n'', and likewise for {{math|adj('''A'''+''t'''''I''')adj('''B''')}}. These two polynomials at the ''ij''th entry agree on at least ''n''&nbsp;+ 1 points, as we have at least ''n''&nbsp;+ 1 elements of the field where {{math|'''A'''+''t'''''I'''}} is invertible, and we have proven the identity for invertible matrices. Polynomials of degree ''n'' which agree on ''n''&nbsp;+ 1 points must be identical (subtract them from each other and you have ''n''&nbsp;+ 1 roots for a polynomial of degree at most ''n'' – a contradiction unless their difference is identically zero). As the two polynomials are identical, they take the same value for every value of ''t''. Thus, they take the same value when ''t'' = 0. 

Using the above properties and other elementary computations, it is straightforward to show that if {{math|'''A'''}} has one of the following properties, then {{math|adj'''A'''}} does as well:
* [[Upper triangular matrix|upper triangular]],
* [[Lower triangular matrix|lower triangular]],
* [[Diagonal matrix|diagonal]],
* [[Orthogonal matrix|orthogonal]],
* [[Unitary matrix|unitary]],
* [[Symmetric matrix|symmetric]],
* [[Hermitian matrix|Hermitian]],
* [[Normal matrix|normal]].

If {{math|'''A'''}} is [[Skew-symmetric matrix|skew-symmetric]], then {{math|adj('''A''')}} is skew-symmetric for even ''n'' and symmetric for odd ''n''. Similarly, if {{math|'''A'''}} is [[Skew-Hermitian matrix|skew-Hermitian]], then {{math|adj('''A''')}} is skew-Hermitian for even ''n'' and Hermitian for odd ''n''.

If {{math|'''A'''}} is invertible, then, as noted above, there is a formula for {{math|adj('''A''')}} in terms of the determinant and inverse of {{math|'''A'''}}.  When {{math|'''A'''}} is not invertible, the adjugate satisfies different but closely related formulas.
* If {{math|1=rk('''A''') ≤ ''n'' − 2}}, then {{math|1=adj('''A''') = '''0'''}}.
* If {{math|1=rk('''A''') = ''n'' − 1}}, then {{math|1=rk(adj('''A''')) = 1}}.  (Some minor is non-zero, so {{math|adj('''A''')}} is non-zero and hence has [[rank (linear algebra)|rank]] at least one; the identity {{math|1=adj('''A''')'''A''' = '''0'''}} implies that the [[dimension (vector space)|dimension]] of the [[nullspace]] of {{math|adj('''A''')}} is at least {{math|''n''&nbsp;− 1}}, so its rank is at most one.)  It follows that {{math|1=adj('''A''') = ''α'''''xy'''<sup>T</sup>}}, where {{math|''α''}} is a scalar and {{math|'''x'''}} and {{math|'''y'''}} are vectors such that {{math|1='''Ax''' = '''0'''}} and {{math|1='''A'''<sup>T</sup> '''y''' = '''0'''}}.

=== Column substitution and Cramer's rule ===
{{see also|Cramer's rule}}

Partition {{math|'''A'''}} into [[column vector]]s:
:<math>\mathbf{A} = \begin{bmatrix}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{bmatrix}.</math>
Let {{math|'''b'''}} be a column vector of size {{math|''n''}}.  Fix {{math|1 ≤ ''i'' ≤ ''n''}} and consider the matrix formed by replacing column {{math|''i''}} of {{math|'''A'''}} by {{math|'''b'''}}:
:<math>(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})\ \stackrel{\text{def}}{=}\ \begin{bmatrix} \mathbf{a}_1 & \cdots & \mathbf{a}_{i-1} & \mathbf{b} & \mathbf{a}_{i+1} & \cdots & \mathbf{a}_n \end{bmatrix}.</math>
Laplace expand the determinant of this matrix along column {{mvar|i}}.  The result is entry {{mvar|i}} of the product {{math|adj('''A''')'''b'''}}.  Collecting these determinants for the different possible {{mvar|i}} yields an equality of column vectors
:<math>\left(\det(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})\right)_{i=1}^n = \operatorname{adj}(\mathbf{A})\mathbf{b}.</math>

This formula has the following concrete consequence.  Consider the [[linear system of equations]]
:<math>\mathbf{A}\mathbf{x} = \mathbf{b}.</math>
Assume that {{math|'''A'''}} is [[singular matrix|non-singular]].  Multiplying this system on the left by {{math|adj('''A''')}} and dividing by the determinant yields
:<math>\mathbf{x} = \frac{\operatorname{adj}(\mathbf{A})\mathbf{b}}{\det \mathbf{A}}.</math>
Applying the previous formula to this situation yields '''Cramer's rule''',
:<math>x_i = \frac{\det(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})}{\det \mathbf{A}},</math>
where {{math|''x''<sub>''i''</sub>}} is the {{mvar|i}}th entry of {{math|'''x'''}}.

=== Characteristic polynomial ===
Let the [[characteristic polynomial]] of {{math|'''A'''}} be
:<math>p(s) = \det(s\mathbf{I} - \mathbf{A}) = \sum_{i=0}^n p_i s^i \in R[s].</math>
The first [[divided difference]] of {{math|''p''}} is a [[symmetric polynomial]] of degree {{math|''n''&nbsp;− 1}},
:<math>\Delta p(s, t) = \frac{p(s) - p(t)}{s - t} = \sum_{0 \le j + k < n} p_{j+k+1} s^j t^k \in R[s, t].</math>
Multiply {{math|''s'''''I''' − '''A'''}} by its adjugate.  Since {{math|1=''p''('''A''') = '''0'''}} by the [[Cayley–Hamilton theorem]], some elementary manipulations reveal
:<math>\operatorname{adj}(s\mathbf{I} - \mathbf{A}) = \Delta p(s\mathbf{I}, \mathbf{A}).</math>

In particular, the [[resolvent formalism|resolvent]] of {{math|'''A'''}} is defined to be
:<math>R(z; \mathbf{A}) = (z\mathbf{I} - \mathbf{A})^{-1},</math>
and by the above formula, this is equal to
:<math>R(z; \mathbf{A}) = \frac{\Delta p(z\mathbf{I}, \mathbf{A})}{p(z)}.</math>

=== Jacobi's formula ===
{{main|Jacobi's formula}}
The adjugate also appears in [[Jacobi's formula]] for the [[derivative]] of the determinant.  If {{math|'''A'''(''t'')}} is [[continuously differentiable]], then
:<math>\frac{d(\det \mathbf{A})}{dt}(t) = \operatorname{tr}\left(\operatorname{adj}(\mathbf{A}(t)) \mathbf{A}'(t)\right).</math>
It follows that the [[total derivative]] of the determinant is the transpose of the adjugate:
:<math>d(\det \mathbf{A})_{\mathbf{A}_0} = \operatorname{adj}(\mathbf{A}_0)^{\mathsf{T}}.</math>

=== Cayley–Hamilton formula ===
{{main|Cayley–Hamilton theorem}}
Let {{math|''p''<sub>'''A'''</sub>(''t'')}} be the characteristic polynomial of {{math|'''A'''}}.  The [[Cayley–Hamilton theorem]] states that
:<math>p_{\mathbf{A}}(\mathbf{A}) = \mathbf{0}.</math>
Separating the constant term and multiplying the equation by {{math|adj('''A''')}} gives an expression for the adjugate that depends only on {{math|'''A'''}} and the coefficients of {{math|''p''<sub>'''A'''</sub>(''t'')}}.  These coefficients can be explicitly represented in terms of [[trace (linear algebra)|traces]] of powers of {{math|'''A'''}} using complete exponential [[Bell polynomials]].  The resulting formula is
:<math>\operatorname{adj}(\mathbf{A}) = \sum_{s=0}^{n-1} \mathbf{A}^{s} \sum_{k_1, k_2, \ldots, k_{n-1}} \prod_{\ell=1}^{n-1} \frac{(-1)^{k_\ell+1}}{\ell^{k_\ell}k_{\ell}!}\operatorname{tr}(\mathbf{A}^\ell)^{k_\ell},</math>
where {{mvar|n}} is the dimension of {{math|'''A'''}}, and the sum is taken over {{mvar|s}} and all sequences of {{math|''k<sub>l</sub>'' ≥ 0}} satisfying the linear [[Diophantine equation]]
:<math>s+\sum_{\ell=1}^{n-1}\ell k_\ell = n - 1.</math>

For the 2 × 2 case, this gives
:<math>\operatorname{adj}(\mathbf{A})=\mathbf{I}_2(\operatorname{tr}\mathbf{A}) - \mathbf{A}.</math>
For the 3 × 3 case, this gives
:<math>\operatorname{adj}(\mathbf{A})=\frac{1}{2}\mathbf{I}_3\!\left( (\operatorname{tr}\mathbf{A})^2-\operatorname{tr}\mathbf{A}^2\right) - \mathbf{A}(\operatorname{tr}\mathbf{A}) + \mathbf{A}^2 .</math>
For the 4 × 4 case, this gives
:<math>\operatorname{adj}(\mathbf{A})=
\frac{1}{6}\mathbf{I}_4\!\left(
  (\operatorname{tr}\mathbf{A})^3
  - 3\operatorname{tr}\mathbf{A}\operatorname{tr}\mathbf{A}^2
  + 2\operatorname{tr}\mathbf{A}^{3}
\right)
- \frac{1}{2}\mathbf{A}\!\left( (\operatorname{tr}\mathbf{A})^2 - \operatorname{tr}\mathbf{A}^2\right)
+ \mathbf{A}^2(\operatorname{tr}\mathbf{A})
- \mathbf{A}^3.</math>

The same formula follows directly from the terminating step of the [[Faddeev–LeVerrier algorithm]], which efficiently determines the [[characteristic polynomial]] of {{math|'''A'''}}.

In general, adjugate matrix of arbitrary dimension N matrix can be computed by Einstein's convention.
:<math>(\operatorname{adj}(\mathbf{A}))_{i_N}^{j_N} = \frac{1}{(N-1)!} \epsilon_{i_1 i_2 \ldots i_N}  \epsilon^{j_1 j_2 \ldots j_N} A_{j_1}^{i_1} A_{j_2}^{i_2} \ldots A_{j_{N-1}}^{i_{N-1}}
</math>