Adjugate matrix

In linear algebra, the adjugate or classical adjoint of a square matrix is the transpose of its cofactor matrix.[1] It is also occasionally known as adjunct matrix,[2][3] though this nomenclature appears to have decreased in usage.

The adjugate[4] has sometimes been called the "adjoint",[5] but today the "adjoint" of a matrix normally refers to its corresponding adjoint operator, which is its conjugate transpose.

In more detail, suppose R is a commutative ring and A is an n × n matrix with entries from R. The (i,j)-minor of A, denoted Mij, is the determinant of the (n − 1) × (n − 1) matrix that results from deleting row i and column j of A. The cofactor matrix of A is the n × n matrix C whose (i, j) entry is the (i, j) cofactor of A, which is the (i, j)-minor times a sign factor:

The adjugate of A is the transpose of C, that is, the n×n matrix whose (i,j) entry is the (j,i) cofactor of A,

The adjugate is defined as it is so that the product of A with its adjugate yields a diagonal matrix whose diagonal entries are the determinant det(A). That is,

where I is the n×n 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 A is invertible if and only if det(A) is an invertible element of R. When this holds, the equation above yields

In this case, it is also true that det(adj(A)) = det(A) and hence that adj(adj(A)) = A.

It is easy to check the adjugate is the inverse times the determinant, −6.

The −1 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,

For any n × n matrix A, elementary computations show that adjugates enjoy the following properties.

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 A and B,

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 A or B is not invertible.

A corollary of the previous formula is that, for any non-negative integer k,

Suppose that A commutes with B. Multiplying the identity AB = BA on the left and right by adj(A) proves that

If A is invertible, this implies that adj(A) also commutes with B. Over the real or complex numbers, continuity implies that adj(A) commutes with B even when 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 with at least 2n+1 elements (e.g. a 5×5 matrix over the integers mod 11). det(A+tI) is a polynomial in t with degree at most n, so it has at most n roots. Note that the ijth entry of adj((A+tI)(B)) is a polynomial of at most order n, and likewise for adj(A+tI)adj(B). These two polynomials at the ijth entry agree on at least n+1 points, as we have at least n+1 elements of the field where A+tI 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 na 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 A has one of the following properties, then adj A does as well:

If A is invertible, then, as noted above, there is a formula for adj(A) in terms of the determinant and inverse of A. When A is not invertible, the adjugate satisfies different but closely related formulas.

Let b be a column vector of size n. Fix 1 ≤ in and consider the matrix formed by replacing column i of A by b:

Laplace expand the determinant of this matrix along column i. The result is entry i of the product adj(A)b. Collecting these determinants for the different possible i yields an equality of column vectors

This formula has the following concrete consequence. Consider the linear system of equations

Assume that A is non-singular. Multiplying this system on the left by adj(A) and dividing by the determinant yields

The first divided difference of p is a symmetric polynomial of degree n − 1,

Multiply sIA by its adjugate. Since p(A) = 0 by the Cayley–Hamilton theorem, some elementary manipulations reveal

The adjugate also appears in Jacobi's formula for the derivative of the determinant. If A(t) is continuously differentiable, then

It follows that the total derivative of the determinant is the transpose of the adjugate:

Let pA(t) be the characteristic polynomial of A. The Cayley–Hamilton theorem states that

Separating the constant term and multiplying the equation by adj(A) gives an expression for the adjugate that depends only on A and the coefficients of pA(t). These coefficients can be explicitly represented in terms of traces of powers of A using complete exponential Bell polynomials. The resulting formula is

where n is the dimension of A, and the sum is taken over s and all sequences of kl ≥ 0 satisfying the linear Diophantine equation

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

The adjugate can be viewed in abstract terms using exterior algebras. Let V be an n-dimensional vector space. The exterior product defines a bilinear pairing

Suppose that T : VV is a linear transformation. Pullback by the (n − 1)st exterior power of T induces a morphism of Hom spaces. The adjugate of T is the composite

If V is endowed with an inner product and a volume form, then the map φ can be decomposed further. In this case, φ can be understood as the composite of the Hodge star operator and dualization. Specifically, if ω is the volume form, then it, together with the inner product, determines an isomorphism

By the definition of the Hodge star operator, this linear functional is dual to *v. That is, ω ∘ φ equals v ↦ *v.

where σ(I) and σ(J) are the sum of the elements of I and J, respectively.

Iteratively taking the adjugate of an invertible matrix A k times yields