The Adjugate Identity as a Polynomial Matrix

From the Adjugate Identity to Coefficient Equations

The adjugate identity turns a determinant into a matrix equation. When the matrix contains an indeterminate \(t\), that equation is an identity of polynomials, so we can compare the coefficients of each power of \(t\). The resulting equations relate the coefficients of the determinant polynomial to fixed matrices built from \(\mathbf A\). They are the algebraic ingredients used in a coefficient proof of Cayley–Hamilton.

Let \(\mathbb F\) be \(\mathbb R\) or \(\mathbb C\), let \(n\geq 1\), and let \(\mathbf A\in\mathbb F^{n\times n}\). We treat \(t\) as an indeterminate: it commutes with scalars and matrix entries, but it is not a particular number or matrix. Write \(\mathbf I_n\) for the \(n\times n\) identity matrix, and set \(\mathbf M(t)=t\mathbf I_n-\mathbf A\).

For \(\mathbf M(t)\), each entry is a polynomial of degree at most one. Each cofactor is a determinant of an \((n-1)\times(n-1)\) matrix, so its degree is at most \(n-1\). Thus \(\operatorname{adj}(t\mathbf I_n-\mathbf A)\) is a matrix whose entries are polynomials of degree at most \(n-1\). We can write it using fixed coefficient matrices \(\mathbf B_0,\dots,\mathbf B_{n-1}\):

The determinant \(\det(t\mathbf I_n-\mathbf A)\) is monic of degree \(n\): in the determinant expansion, the term of degree \(n\) comes from choosing \(t\) from each diagonal entry. Write

Here \(\chi_{\mathbf A}(t)\) is just a notation for this determinant polynomial. The course notation \(p_{\mathbf A}(\lambda)=\det(\mathbf A-\lambda\mathbf I_n)\) uses the opposite matrix order, so \(\chi_{\mathbf A}(t)=(-1)^n p_{\mathbf A}(t)\). Keeping the two sign conventions distinct prevents sign errors in the coefficient equations.

Why the Adjugate Identity Holds

Proof. Consider the entry in row \(i\), column \(j\) of \(\mathbf M\operatorname{adj}\mathbf M\). It is \(\sum_k m_{ik}C_{jk}\). If \(i=j\), this is the cofactor expansion of \(\det\mathbf M\) along row \(i\). If \(i\ne j\), it is the determinant expansion along row \(j\) of the matrix formed by replacing row \(j\) of \(\mathbf M\) with row \(i\). That matrix has two equal rows, so its determinant is zero. Therefore the product has \(\det\mathbf M\) on its diagonal and zero off its diagonal, proving \(\mathbf M\operatorname{adj}\mathbf M=(\det\mathbf M)\mathbf I\). Applying the same argument to columns proves \(\operatorname{adj}\mathbf M\,\mathbf M=(\det\mathbf M)\mathbf I\). The determinant expansion, the fact that a determinant with repeated rows or columns is zero, and the cofactor definition apply to polynomial entries as well, so both identities hold over the polynomial ring. \(\square\)

Apply the identity to \(\mathbf M(t)=t\mathbf I_n-\mathbf A\). The left-hand side is a product of matrix polynomials. Because \(t\) is a scalar indeterminate, it commutes with every coefficient matrix, and we may collect terms by powers of \(t\). This is coefficient comparison in a polynomial identity; it does not yet replace \(t\) by a matrix.

The Coefficient Equations

Proof. Multiply the two factors using the adjugate expansion. The terms from \(t\mathbf I_n\) are \(\mathbf B_0t^n+\mathbf B_1t^{n-1}+\cdots+\mathbf B_{n-1}t\), and those from \(-\mathbf A\) are \(-\mathbf A\mathbf B_0t^{n-1}-\cdots-\mathbf A\mathbf B_{n-1}\). Hence

By the adjugate identity, this equals \(\chi_{\mathbf A}(t)\mathbf I_n\). Two matrix polynomials are equal exactly when the coefficients of each power of \(t\) agree entry by entry. The coefficient of \(t^n\) gives \(\mathbf B_0=\mathbf I_n\). For each \(1\leq k\leq n-1\), the coefficient of \(t^{n-k}\) gives \(\mathbf B_k-\mathbf A\mathbf B_{k-1}=c_k\mathbf I_n\). Comparing constant terms gives \(-\mathbf A\mathbf B_{n-1}=c_n\mathbf I_n\). When \(n=1\), the expansion has only \(\mathbf B_0\), and comparing the \(t\) and constant terms gives precisely the stated leading- and constant-term equations. \(\square\)

The identity on the other side, \(\operatorname{adj}(t\mathbf I_n-\mathbf A)(t\mathbf I_n-\mathbf A)=\chi_{\mathbf A}(t)\mathbf I_n\), gives corresponding equations with the order of multiplication reversed: \(\mathbf B_k-\mathbf B_{k-1}\mathbf A=c_k\mathbf I_n\), and \(-\mathbf B_{n-1}\mathbf A=c_n\mathbf I_n\). Stating the order matters: matrix multiplication is not generally commutative.

Worked Coefficient Comparisons

A Common Cofactor Pitfall

The cofactor matrix is not the adjugate. The adjugate is the transpose of the cofactor matrix. In the two-by-two example, using the cofactor matrix itself would give the incorrect expression \(\begin{bmatrix}t-4&3\\2&t-1\end{bmatrix}\). Multiplying it on the left by \(t\mathbf I_2-\mathbf A\) produces

The off-diagonal entries \(t-1\) and \(-(t-4)\) are not zero polynomials, so this product is not a scalar multiple of \(\mathbf I_2\). Transposing the cofactor matrix gives the correct adjugate \(\begin{bmatrix}t-4&2\\3&t-1\end{bmatrix}\), for which the product is \((t^2-5t-2)\mathbf I_2\). Checking the defining identity is an effective way to catch this transposition error.

Why These Equations Matter

The adjugate identity supplies more than the determinant polynomial alone: it gives a sequence of matrix equations linking the adjugate coefficients \(\mathbf B_k\) to \(\mathbf A\) and the scalar coefficients \(c_k\). The order of factors in each equation comes directly from expanding \((t\mathbf I_n-\mathbf A)\operatorname{adj}(t\mathbf I_n-\mathbf A)\). These equations are useful because they can be combined with powers of \(\mathbf A\); the next step is to use that structure in a coefficient proof of Cayley–Hamilton.

For now, the essential discipline is to keep the indeterminate \(t\), the scalar coefficients \(c_k\), and the matrix coefficients \(\mathbf B_k\) distinct. Expand first, compare one power at a time, and verify the constant term separately. This makes the signs and multiplication order visible rather than hiding them in a formal substitution.