adjugate matrix

In linear algebra, the adjugate or classical adjoint of a square matrix is the transpose of its cofactor matrix and is denoted by . It is also occasionally known as adjunct matrix, or "adjoint", though the latter today normally refers to a different concept, the adjoint operator which is the conjugate transpose of the matrix.

The product of a matrix with its adjugate gives a diagonal matrix (entries not on the main diagonal are zero) whose diagonal entries are the determinant of the original matrix:
:$\mathbf \operatorname(\mathbf) = \det(\mathbf) \mathbf,$
where is the identity matrix of the same size as . Consequently, the multiplicative inverse of an invertible matrix can be found by dividing its adjugate by its determinant.

Definition

The adjugate of is the transpose of the cofactor matrix of ,
:$\operatorname(\mathbf) = \mathbf^\mathsf.$

In more detail, suppose is a unital commutative ring and is an matrix with entries from . The -'' minor'' of , denoted , is the determinant of the matrix that results from deleting row and column of . The cofactor matrix of is the matrix whose entry is the '' cofactor'' of , which is the -minor times a sign factor:
:$\mathbf = \left((-1)^ \mathbf_\right)_.$
The adjugate of is the transpose of , that is, the matrix whose entry is the cofactor of ,
:$\operatorname(\mathbf) = \mathbf^\mathsf = \left((-1)^ \mathbf_\right)_.$

Important consequence

The adjugate is defined so that the product of with its adjugate yields a diagonal matrix whose diagonal entries are the determinant . That is,
:$\mathbf \operatorname(\mathbf) = \operatorname(\mathbf) \mathbf = \det(\mathbf) \mathbf,$
where is the identity matrix. This is a consequence of the Laplace expansion of the determinant.

The above formula implies one of the fundamental results in matrix algebra, that is invertible if and only if is an invertible element of . When this holds, the equation above yields
:$\begin \operatorname(\mathbf) &= \det(\mathbf) \mathbf^, \\ \mathbf^ &= \det(\mathbf)^ \operatorname(\mathbf). \end$

Examples

1 × 1 generic matrix

Since the determinant of a 0 x 0 matrix is 1, the adjugate of any 1 × 1 matrix ( complex scalar) is $\mathbf = \begin 1 \end$. Observe that $\mathbf \operatorname(\mathbf) = \mathbf \mathbf = (\det \mathbf) \mathbf .$

2 × 2 generic matrix

The adjugate of the 2 × 2 matrix
:$\mathbf = \begin a & b \\ c & d \end$
is
:$\operatorname(\mathbf) = \begin d & -b \\ -c & a \end.$
By direct computation,
:$\mathbf \operatorname(\mathbf) = \begin ad - bc & 0 \\ 0 & ad - bc \end = (\det \mathbf)\mathbf.$
In this case, it is also true that ((A)) = (A) and hence that ((A)) = A.

3 × 3 generic matrix

Consider a 3 × 3 matrix
:$\mathbf = \begin a_ & a_ & a_ \\ a_ & a_ & a_ \\ a_ & a_ & a_ \end.$
Its cofactor matrix is
:$\mathbf = \begin +\begin a_ & a_ \\ a_ & a_ \end & -\begin a_ & a_ \\ a_ & a_ \end & +\begin a_ & a_ \\ a_ & a_ \end \\ \\ -\begin a_ & a_ \\ a_ & a_ \end & +\begin a_ & a_ \\ a_ & a_ \end & -\begin a_ & a_ \\ a_ & a_ \end \\ \\ +\begin a_ & a_ \\ a_ & a_ \end & -\begin a_ & a_ \\ a_ & a_ \end & +\begin a_ & a_ \\ a_ & a_ \end \end,$
where
:$\begin a_ & a_ \\ a_ & a_ \end = \det\!\begin a_ & a_ \\ a_ & a_ \end .$

Its adjugate is the transpose of its cofactor matrix,
:$\operatorname(\mathbf) = \mathbf^\mathsf = \begin +\begin a_ & a_ \\ a_ & a_ \end & -\begin a_ & a_ \\ a_ & a_ \end & +\begin a_ & a_ \\ a_ & a_ \end \\ & & \\ -\begin a_ & a_ \\ a_ & a_ \end & +\begin a_ & a_ \\ a_ & a_ \end & -\begin a_ & a_ \\ a_ & a_ \end \\ & & \\ +\begin a_ & a_ \\ a_ & a_ \end & -\begin a_ & a_ \\ a_ & a_ \end & +\begin a_ & a_ \\ a_ & a_ \end \end.$

3 × 3 numeric matrix

As a specific example, we have
:$\operatorname\!\begin -3 & 2 & -5 \\ -1 & 0 & -2 \\ 3 & -4 & 1 \end = \begin -8 & 18 & -4 \\ -5 & 12 & -1 \\ 4 & -6 & 2 \end.$
It is easy to check the adjugate is the inverse times the determinant, .

The in the second row, third column of the adjugate was computed as follows. The (2,3) entry of the adjugate is the (3,2) cofactor of A. This cofactor is computed using the submatrix obtained by deleting the third row and second column of the original matrix A,
:$\begin -3 & -5 \\ -1 & -2 \end.$
The (3,2) cofactor is a sign times the determinant of this submatrix:
:$(-1)^\operatorname\!\begin-3&-5\\-1&-2\end = -(-3 \cdot -2 - -5 \cdot -1) = -1,$
and this is the (2,3) entry of the adjugate.

Properties

For any matrix , elementary computations show that adjugates have the following properties:
* $\operatorname(\mathbf) = \mathbf$, where $\mathbf$ is the identity matrix.
* $\operatorname(\mathbf) = \mathbf$, where $\mathbf$ is the zero matrix, except that if $n=1$ then $\operatorname(\mathbf) = \mathbf$.
* $\operatorname(c \mathbf) = c^\operatorname(\mathbf)$ for any scalar .
* $\operatorname(\mathbf^\mathsf) = \operatorname(\mathbf)^\mathsf$.
* $\det(\operatorname(\mathbf)) = (\det \mathbf)^$.
* If is invertible, then $\operatorname(\mathbf) = (\det \mathbf) \mathbf^$. It follows that:
** is invertible with inverse .
** .
* is entrywise polynomial in . In particular, over the real or complex numbers, the adjugate is a smooth function of the entries of .

Over the complex numbers,
* $\operatorname(\overline\mathbf) = \overline$, where the bar denotes complex conjugation.
* $\operatorname(\mathbf^*) = \operatorname(\mathbf)^*$, where the asterisk denotes conjugate transpose.

Suppose that is another matrix. Then
:$\operatorname(\mathbf) = \operatorname(\mathbf)\operatorname(\mathbf).$
This can be 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 and ,
:$\operatorname(\mathbf)\operatorname(\mathbf) = (\det \mathbf)\mathbf^(\det \mathbf)\mathbf^ = (\det \mathbf)(\mathbf)^ = \operatorname(\mathbf).$
Because every non-invertible matrix is the limit of invertible matrices, continuity of the adjugate then implies that the formula remains true when one of or is not invertible.

A corollary of the previous formula is that, for any non-negative integer ,
:$\operatorname(\mathbf^k) = \operatorname(\mathbf)^k.$
If is invertible, then the above formula also holds for negative .

From the identity
:$(\mathbf + \mathbf)\operatorname(\mathbf + \mathbf)\mathbf = \det(\mathbf + \mathbf)\mathbf = \mathbf\operatorname(\mathbf + \mathbf)(\mathbf + \mathbf),$
we deduce
:$\mathbf\operatorname(\mathbf + \mathbf)\mathbf = \mathbf\operatorname(\mathbf + \mathbf)\mathbf.$

Suppose that commutes with . Multiplying the identity on the left and right by proves that
:$\det(\mathbf)\operatorname(\mathbf)\mathbf = \det(\mathbf)\mathbf\operatorname(\mathbf).$
If is invertible, this implies that also commutes with . Over the real or complex numbers, continuity implies that commutes with even when 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 with at least 2''n'' + 1 elements (e.g. a 5 × 5 matrix over the integers modulo 11). is a polynomial in ''t'' with degree at most ''n'', so it has at most ''n'' roots. Note that the ''ij'' th entry of is a polynomial of at most order ''n'', and likewise for . These two polynomials at the ''ij'' th entry agree on at least ''n'' + 1 points, as we have at least ''n'' + 1 elements of the field where is invertible, and we have proven the identity for invertible matrices. Polynomials of degree ''n'' which agree on ''n'' + 1 points must be identical (subtract them from each other and you have ''n'' + 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 has one of the following properties, then does as well:
* Upper triangular,
* Lower triangular,
* Diagonal,
* Orthogonal,
* Unitary,
* Symmetric,
* Hermitian,
* Skew-symmetric,
* Skew-Hermitian,
* Normal.

If is invertible, then, as noted above, there is a formula for in terms of the determinant and inverse of . When is not invertible, the adjugate satisfies different but closely related formulas.
* If , then .
* If , then . (Some minor is non-zero, so is non-zero and hence has rank at least one; the identity implies that the dimension of the nullspace of is at least , so its rank is at most one.) It follows that , where is a scalar and and are vectors such that and .

Column substitution and Cramer's rule

Partition into column vectors:
:$\mathbf = \begin\mathbf_1 & \cdots & \mathbf_n\end.$
Let be a column vector of size . Fix and consider the matrix formed by replacing column of by :
:$(\mathbf \stackrel \mathbf)\ \stackrel\ \begin \mathbf_1 & \cdots & \mathbf_ & \mathbf & \mathbf_ & \cdots & \mathbf_n \end.$
Laplace expand the determinant of this matrix along column . The result is entry of the product . Collecting these determinants for the different possible yields an equality of column vectors
:$\left(\det(\mathbf \stackrel \mathbf)\right)_^n = \operatorname(\mathbf)\mathbf.$

This formula has the following concrete consequence. Consider the linear system of equations
:$\mathbf\mathbf = \mathbf.$
Assume that is non-singular. Multiplying this system on the left by and dividing by the determinant yields
:$\mathbf = \frac.$
Applying the previous formula to this situation yields Cramer's rule,
:$x_i = \frac,$
where is the th entry of .

Characteristic polynomial

Let the characteristic polynomial of be
:p(s) = \det(s\mathbf - \mathbf) = \sum_^n p_i s^i \in R
The first divided difference of is a symmetric polynomial of degree ,
:\Delta p(s, t) = \frac = \sum_ p_ s^j t^k \in R , t
Multiply by its adjugate. Since by the Cayley–Hamilton theorem, some elementary manipulations reveal
:$\operatorname(s\mathbf - \mathbf) = \Delta p(s\mathbf, \mathbf).$

In particular, the resolvent of is defined to be
:$R(z; \mathbf) = (z\mathbf - \mathbf)^,$
and by the above formula, this is equal to
:$R(z; \mathbf) = \frac.$

Jacobi's formula

The adjugate also appears in Jacobi's formula for the derivative of the determinant. If is continuously differentiable, then
:$\frac(t) = \operatorname\left(\operatorname(\mathbf(t)) \mathbf'(t)\right).$
It follows that the total derivative of the determinant is the transpose of the adjugate:
:$d(\det \mathbf)_ = \operatorname(\mathbf_0)^.$

Cayley–Hamilton formula

Let be the characteristic polynomial of . The Cayley–Hamilton theorem states that
:$p_(\mathbf) = \mathbf.$
Separating the constant term and multiplying the equation by gives an expression for the adjugate that depends only on and the coefficients of . These coefficients can be explicitly represented in terms of traces of powers of using complete exponential Bell polynomials. The resulting formula is
:$\operatorname(\mathbf) = \sum_^ \mathbf^ \sum_ \prod_^ \frac\operatorname(\mathbf^\ell)^,$
where is the dimension of , and the sum is taken over and all sequences of satisfying the linear Diophantine equation
:$s+\sum_^\ell k_\ell = n - 1.$

For the 2 × 2 case, this gives
:$\operatorname(\mathbf)=\mathbf_2(\operatorname\mathbf) - \mathbf.$
For the 3 × 3 case, this gives
:$\operatorname(\mathbf)=\frac\mathbf_3\!\left( (\operatorname\mathbf)^2-\operatorname\mathbf^2\right) - \mathbf(\operatorname\mathbf) + \mathbf^2 .$
For the 4 × 4 case, this gives
:$\operatorname(\mathbf)= \frac\mathbf_4\!\left( (\operatorname\mathbf)^3 - 3\operatorname\mathbf\operatorname\mathbf^2 + 2\operatorname\mathbf^ \right) - \frac\mathbf\!\left( (\operatorname\mathbf)^2 - \operatorname\mathbf^2\right) + \mathbf^2(\operatorname\mathbf) - \mathbf^3.$

The same formula follows directly from the terminating step of the Faddeev–LeVerrier algorithm, which efficiently determines the characteristic polynomial of .

Relation to exterior algebras

The adjugate can be viewed in abstract terms using exterior algebras. Let be an -dimensional vector space. The exterior product defines a bilinear pairing
:$V \times \wedge^ V \to \wedge^n V.$
Abstractly, $\wedge^n V$ is isomorphic to , and under any such isomorphism the exterior product is a perfect pairing. Therefore, it yields an isomorphism
:$\phi \colon V\ \xrightarrow\ \operatorname(\wedge^ V, \wedge^n V).$
Explicitly, this pairing sends to $\phi_$, where
:$\phi_\mathbf(\alpha) = \mathbf \wedge \alpha.$
Suppose that is a linear transformation. Pullback by the st exterior power of induces a morphism of spaces. The adjugate of is the composite
:$V\ \xrightarrow\ \operatorname(\wedge^ V, \wedge^n V)\ \xrightarrow\ \operatorname(\wedge^ V, \wedge^n V)\ \xrightarrow\ V.$

If is endowed with its canonical basis , and if the matrix of in this basis is , then the adjugate of is the adjugate of . To see why, give $\wedge^ \mathbf^n$ the basis
:$\_^n.$
Fix a basis vector of . The image of under $\phi$ is determined by where it sends basis vectors:
:$\phi_(\mathbf_1 \wedge \dots \wedge \hat\mathbf_k \wedge \dots \wedge \mathbf_n) = \begin (-1)^ \mathbf_1 \wedge \dots \wedge \mathbf_n, &\text\ k = i, \\ 0 &\text \end$
On basis vectors, the st exterior power of is
:$\mathbf_1 \wedge \dots \wedge \hat\mathbf_j \wedge \dots \wedge \mathbf_n \mapsto \sum_^n (\det A_) \mathbf_1 \wedge \dots \wedge \hat\mathbf_k \wedge \dots \wedge \mathbf_n.$
Each of these terms maps to zero under $\phi_$ except the term. Therefore, the pullback of $\phi_$ is the linear transformation for which
:$\mathbf_1 \wedge \dots \wedge \hat\mathbf_j \wedge \dots \wedge \mathbf_n \mapsto (-1)^ (\det A_) \mathbf_1 \wedge \dots \wedge \mathbf_n,$
that is, it equals
:$\sum_^n (-1)^ (\det A_)\phi_.$
Applying the inverse of $\phi$ shows that the adjugate of is the linear transformation for which
:$\mathbf_i \mapsto \sum_^n (-1)^(\det A_)\mathbf_j.$
Consequently, its matrix representation is the adjugate of .

If is endowed with an inner product and a volume form, then the map can be decomposed further. In this case, can be understood as the composite of the Hodge star operator and dualization. Specifically, if is the volume form, then it, together with the inner product, determines an isomorphism
:$\omega^\vee \colon \wedge^n V \to \mathbf.$
This induces an isomorphism
:$\operatorname(\wedge^ \mathbf^n, \wedge^n \mathbf^n) \cong \wedge^ (\mathbf^n)^\vee.$
A vector in corresponds to the linear functional
:$(\alpha \mapsto \omega^\vee(\mathbf \wedge \alpha)) \in \wedge^ (\mathbf^n)^\vee.$
By the definition of the Hodge star operator, this linear functional is dual to . That is, equals .

Higher adjugates

Let be an matrix, and fix . The th higher adjugate of is an $\binom \!\times\! \binom$ matrix, denoted , whose entries are indexed by size subsets and of . Let and denote the complements of and , respectively. Also let $\mathbf_$ denote the submatrix of containing those rows and columns whose indices are in and , respectively. Then the entry of is
:$(-1)^\det \mathbf_,$
where and are the sum of the elements of and , respectively.

Basic properties of higher adjugates include:
* .
* .
* .
* .
* $\operatorname_r(\mathbf)C_r(\mathbf) = C_r(\mathbf)\operatorname_r(\mathbf) = (\det \mathbf)I_$, where denotes the  th compound matrix.

Higher adjugates may be defined in abstract algebraic terms in a similar fashion to the usual adjugate, substituting $\wedge^r V$ and $\wedge^ V$ for $V$ and $\wedge^ V$, respectively.

Iterated adjugates

Iteratively taking the adjugate of an invertible matrix A times yields

:$\overbrace^k(\mathbf)=\det(\mathbf)^\mathbf^,$
:$\det(\overbrace^k(\mathbf))=\det(\mathbf)^.$

For example,
:$\operatorname(\operatorname(\mathbf)) = \det(\mathbf)^ \mathbf.$
:$\det(\operatorname(\operatorname(\mathbf))) = \det(\mathbf)^.$

See also

* Cayley–Hamilton theorem
* Cramer's rule
* Trace diagram
* Jacobi's formula
* Faddeev–LeVerrier algorithm

References

Bibliography

* Roger A. Horn and Charles R. Johnson (2013), ''Matrix Analysis'', Second Edition. Cambridge University Press,
* Roger A. Horn and Charles R. Johnson (1991), ''Topics in Matrix Analysis''. Cambridge University Press,

External links

Matrix Reference ManualOnline matrix calculator (determinant, track, inverse, adjoint, transpose)
Compute Adjugate matrix up to order 8
*

{{Matrix classes

Matrix theory
Linear algebra