From the Adjugate Identity to Cayley–Hamilton
The characteristic polynomial of a square matrix is built from a scalar variable, but the Cayley–Hamilton theorem makes a claim about evaluating that polynomial at the matrix itself. The key is to connect these two settings without treating a matrix-valued polynomial as though its variable could simply be replaced by a matrix. Instead, we expand the adjugate identity in powers of a scalar variable, compare coefficients, and combine the resulting equations so that the adjugate coefficients cancel.
Let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(n\geq 1\) and \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). The characteristic polynomial, with the convention used here, is \(p_{\mathbf{A}}(t)=\det(\mathbf{A}-t\mathbf{I}_n)\). If \(p_{\mathbf{A}}(t)=c_0+c_1t+\cdots+c_nt^n\), then its evaluation at \(\mathbf{A}\) means the matrix \(p_{\mathbf{A}}(\mathbf{A})=c_0\mathbf{I}_n+c_1\mathbf{A}+\cdots+c_n\mathbf{A}^n\). This is a polynomial with scalar coefficients; it is not a substitution into the adjugate identity.
We prove the theorem from the adjugate identity. For a square matrix \(\mathbf{M}\), its adjugate \(\operatorname{adj}\mathbf{M}\) is the transpose of its cofactor matrix. The adjugate identity states that \(\mathbf{M}\operatorname{adj}\mathbf{M}=(\det\mathbf{M})\mathbf{I}_n\). Applied with the scalar indeterminate \(t\) to \(\mathbf{M}=\mathbf{A}-t\mathbf{I}_n\), it gives the polynomial-matrix identity \((\mathbf{A}-t\mathbf{I}_n)\operatorname{adj}(\mathbf{A}-t\mathbf{I}_n)=p_{\mathbf{A}}(t)\mathbf{I}_n\).
Each entry of the adjugate is a cofactor, hence a determinant of an \((n-1)\times(n-1)\) submatrix. Its entries are polynomials in \(t\), of degree at most \(n-1\). This also holds when \(n=1\): the adjugate of a one-by-one matrix is the one-by-one matrix with entry \(1\), so it has degree zero. Consequently we can write, for coefficient matrices \(\mathbf{B}_0,\dots,\mathbf{B}_{n-1}\),
We also write \(p_{\mathbf{A}}(t)=\sum_{k=0}^{n}c_kt^k\). The determinant formula shows that \(c_n=(-1)^n\), since the degree-\(n\) term comes from choosing \(-t\) from each diagonal position. The proof will compare the coefficients on the two sides of the adjugate identity; each \(c_k\) is a scalar and each \(\mathbf{B}_k\) is a matrix.
The Coefficient Equations
Multiply the adjugate expansion on the left by \(\mathbf{A}-t\mathbf{I}_n\). The coefficient of \(t^0\) is \(\mathbf{A}\mathbf{B}_0\). For \(1\leq k\leq n-1\), the coefficient of \(t^k\) is \(\mathbf{A}\mathbf{B}_k-\mathbf{B}_{k-1}\). The coefficient of \(t^n\) is \(-\mathbf{B}_{n-1}\). Equality of polynomials means equality of the coefficient matrices, so
When \(n=1\), the middle range is empty; the first and last equations still apply. To obtain powers of \(\mathbf{A}\), multiply the equation for index \(k\) on the left by \(\mathbf{A}^k\). The first equation uses \(\mathbf{A}^0=\mathbf{I}_n\), and the last equation is multiplied on the left by \(\mathbf{A}^n\). This gives the scalar-coefficient terms \(c_k\mathbf{A}^k\) on the right-hand sides.
Proof. Assume \(\mathbf{A}\in\mathbb{F}^{n\times n}\) with \(n\geq1\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). We must show \(p_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\). The adjugate identity and coefficient comparison give the equations above. Multiply the first by \(\mathbf{A}^0\), each middle equation by \(\mathbf{A}^k\) on the left, and the final equation by \(\mathbf{A}^n\) on the left. Adding the resulting equations yields $$ \begin{aligned} p_{\mathbf{A}}(\mathbf{A}) &=c_0\mathbf{I}_n+\sum_{k=1}^{n-1}c_k\mathbf{A}^k+c_n\mathbf{A}^n\\ &=\mathbf{A}\mathbf{B}_0+ \sum_{k=1}^{n-1}\bigl(\mathbf{A}^{k+1}\mathbf{B}_k-\mathbf{A}^k\mathbf{B}_{k-1}\bigr) -\mathbf{A}^n\mathbf{B}_{n-1}\\ &=\mathbf{O}. \end{aligned} $$ In the final line, the term \(+\mathbf{A}\mathbf{B}_0\) cancels with \(-\mathbf{A}\mathbf{B}_0\) from the \(k=1\) summand; for every \(1\leq k<n-1\), \(+\mathbf{A}^{k+1}\mathbf{B}_k\) cancels with the negative term in the next summand; and \(+\mathbf{A}^n\mathbf{B}_{n-1}\) cancels with the final term. If \(n=1\), there are no middle summands: \(\mathbf{A}\mathbf{B}_0-\mathbf{A}\mathbf{B}_0=\mathbf{O}\) gives the same conclusion. Thus \(p_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\). \(\square\)
The cancellation is the central mechanism: each coefficient equation contributes one positive term and one negative term involving the adjugate coefficients, and neighboring equations cancel in sequence. The proof does not assume that the matrices \(\mathbf{A}\) and \(\mathbf{B}_k\) commute. It uses left multiplication consistently, so the order of each product is specified.
Worked Examples of the Coefficient Cancellation
Worked Example: A One-by-One Matrix
Let \(\mathbf{A}=\begin{bmatrix}5\end{bmatrix}\). The characteristic polynomial is \(p_{\mathbf{A}}(t)=\det\begin{bmatrix}5-t\end{bmatrix}=5-t\), so \(c_0=5\) and \(c_1=-1\). The adjugate is \(\begin{bmatrix}1\end{bmatrix}\), giving \(\mathbf{B}_0=\begin{bmatrix}1\end{bmatrix}\).
The constant and linear coefficient equations are \(\mathbf{A}\mathbf{B}_0=5\mathbf{I}_1\) and \(-\mathbf{B}_0=-\mathbf{I}_1\). Multiplying the first by \(\mathbf{A}^0\) and the second on the left by \(\mathbf{A}\), then adding, gives \(5\mathbf{I}_1-\mathbf{A}=\mathbf{O}\).
Check: \(p_{\mathbf{A}}(\mathbf{A})=\begin{bmatrix}5\end{bmatrix}-\begin{bmatrix}5\end{bmatrix}=\begin{bmatrix}0\end{bmatrix}\). The scalar polynomial does vanish when evaluated at this matrix.
Worked Example: A Two-by-Two Upper-Triangular Matrix
Take \(\mathbf{A}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\). Then $$ p_{\mathbf{A}}(t)=\det\begin{bmatrix}1-t&2\\0&3-t\end{bmatrix} =(1-t)(3-t)=3-4t+t^2. $$ The adjugate is $$ \operatorname{adj}(\mathbf{A}-t\mathbf{I}_2) =\begin{bmatrix}3-t&-2\\0&1-t\end{bmatrix} =\mathbf{B}_0+\mathbf{B}_1t, $$ where \(\mathbf{B}_0=\begin{bmatrix}3&-2\\0&1\end{bmatrix}\) and \(\mathbf{B}_1=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\).
Check the coefficient equations by multiplying: $$ \mathbf{A}\mathbf{B}_0 =\begin{bmatrix}1&2\\0&3\end{bmatrix} \begin{bmatrix}3&-2\\0&1\end{bmatrix} =\begin{bmatrix}3&0\\0&3\end{bmatrix} =3\mathbf{I}_2. $$ For the coefficient of \(t\), $$ \mathbf{A}\mathbf{B}_1-\mathbf{B}_0 =\begin{bmatrix}-1&-2\\0&-3\end{bmatrix} -\begin{bmatrix}3&-2\\0&1\end{bmatrix} =\begin{bmatrix}-4&0\\0&-4\end{bmatrix} =-4\mathbf{I}_2. $$ Finally, \(-\mathbf{B}_1=\mathbf{I}_2\), as required for \(c_2=1\).
The telescoping combination is \(3\mathbf{I}_2-4\mathbf{A}+\mathbf{A}^2\). Since $$ \mathbf{A}^2=\begin{bmatrix}1&8\\0&9\end{bmatrix}, $$ we obtain $$ 3\mathbf{I}_2-4\mathbf{A}+\mathbf{A}^2 =\begin{bmatrix}3&0\\0&3\end{bmatrix} +\begin{bmatrix}-4&-8\\0&-12\end{bmatrix} +\begin{bmatrix}1&8\\0&9\end{bmatrix} =\mathbf{O}. $$ Check: the result is exactly \(p_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\), with all entries equal to zero.
Worked Example: A Three-by-Three Diagonal Matrix
Let \(\mathbf{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}\). The characteristic polynomial and adjugate expansion are $$ p_{\mathbf{A}}(t)=(1-t)(2-t)(3-t)=6-11t+6t^2-t^3, $$ $$ \operatorname{adj}(\mathbf{A}-t\mathbf{I}_3) =\begin{bmatrix}6-5t+t^2&0&0\\0&3-4t+t^2&0\\0&0&2-3t+t^2\end{bmatrix}. $$ Thus \(\mathbf{B}_0=\operatorname{diag}(6,3,2)\), \(\mathbf{B}_1=\operatorname{diag}(-5,-4,-3)\), and \(\mathbf{B}_2=\mathbf{I}_3\), where \(\operatorname{diag}\) denotes the diagonal matrix with the listed entries.
The coefficient equations can be checked entry by entry on the diagonal: $$ \mathbf{A}\mathbf{B}_0=\operatorname{diag}(6,6,6)=6\mathbf{I}_3, $$ $$ \mathbf{A}\mathbf{B}_1-\mathbf{B}_0 =\operatorname{diag}(-5,-8,-9)-\operatorname{diag}(6,3,2) =\operatorname{diag}(-11,-11,-11), $$ $$ \mathbf{A}\mathbf{B}_2-\mathbf{B}_1 =\operatorname{diag}(1,2,3)-\operatorname{diag}(-5,-4,-3) =\operatorname{diag}(6,6,6), \qquad -\mathbf{B}_2=-\mathbf{I}_3. $$ These are the coefficient matrices \(6\mathbf{I}_3,-11\mathbf{I}_3,6\mathbf{I}_3,-\mathbf{I}_3\).
For a direct check, \(\mathbf{A}^2=\operatorname{diag}(1,4,9)\) and \(\mathbf{A}^3=\operatorname{diag}(1,8,27)\). Therefore $$ 6\mathbf{I}_3-11\mathbf{A}+6\mathbf{A}^2-\mathbf{A}^3 =\operatorname{diag}(6-11+6-1,\;6-22+24-8,\;6-33+54-27) =\mathbf{O}. $$ Check: each diagonal entry is \(0\), and all off-diagonal entries are \(0\), so \(p_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\).
A Common Pitfall: Substituting into the Adjugate Identity
The adjugate identity is an identity in a scalar indeterminate \(t\), with matrix coefficients. It is not a license to replace \(t\) by \(\mathbf{A}\) inside every matrix expression. A concrete calculation shows the problem. For the two-by-two matrix above, the coefficient expression for the adjugate is \(\mathbf{B}_0+\mathbf{B}_1t\). A tempting substitution would produce $$ \mathbf{B}_0+\mathbf{B}_1\mathbf{A} =\begin{bmatrix}3&-2\\0&1\end{bmatrix} -\begin{bmatrix}1&2\\0&3\end{bmatrix} =\begin{bmatrix}2&-4\\0&-2\end{bmatrix}, $$ which is not the zero matrix. But the expression \(\mathbf{A}-\mathbf{A}\mathbf{I}_2\) is \(\mathbf{O}\), whose adjugate is \(\operatorname{adj}(\mathbf{O})=\mathbf{O}\) for a two-by-two matrix. These are not equal: replacing the scalar variable inside a matrix expression and evaluating its coefficient expression are not interchangeable operations.
The correction is to keep \(t\) scalar, use the adjugate identity only as a polynomial identity, and compare its coefficients. Then multiply the resulting coefficient equations on the left by the specified powers of \(\mathbf{A}\); the telescoping cancellation proves the desired result without making the invalid substitution.
A related order error is to multiply a coefficient equation on the right. In the two-by-two example, \(\mathbf{A}\mathbf{B}_1-\mathbf{B}_0=-4\mathbf{I}_2\). Right multiplication by \(\mathbf{A}\) gives \((\mathbf{A}\mathbf{B}_1-\mathbf{B}_0)\mathbf{A}=-4\mathbf{A}\), but its terms do not have the form \(\mathbf{A}^2\mathbf{B}_1-\mathbf{A}\mathbf{B}_0\) needed to cancel neighboring equations. Left multiplication gives precisely \(\mathbf{A}(\mathbf{A}\mathbf{B}_1-\mathbf{B}_0)=\mathbf{A}^2\mathbf{B}_1-\mathbf{A}\mathbf{B}_0\). Keeping the order fixed is essential to the proof.
Why the Coefficient Proof Is Useful
Cayley–Hamilton is a general statement: it applies to every square matrix over either \(\mathbb{R}\) or \(\mathbb{C}\), with no assumption that the matrix is diagonalizable or that its characteristic polynomial has distinct roots. The proof also explains why the scalar coefficients of the characteristic polynomial can govern a matrix expression: the adjugate identity supplies coefficient equations that link those scalars to products involving the matrix.
The proof method is worth retaining independently of the theorem. Whenever a polynomial identity has matrix coefficients, coefficient comparison can turn it into a finite list of matrix equations. Here those equations are arranged so that multiplying by successive powers of \(\mathbf{A}\) causes every adjugate coefficient to cancel. The final statement concerns only the scalar coefficients \(c_k\) and powers of \(\mathbf{A}\), exactly the terms in the definition of \(p_{\mathbf{A}}(\mathbf{A})\).
Check Your Understanding
Use coefficient comparison and the order of multiplication carefully in each question.
- For \(\mathbf{A}=\begin{bmatrix}2&0\\0&4\end{bmatrix}\), find \(p_{\mathbf{A}}(t)\) and verify \(p_{\mathbf{A}}(\mathbf{A})=\mathbf{O}\).
- For an \(n\times n\) matrix, state the coefficient of \(t^k\) in \((\mathbf{A}-t\mathbf{I}_n)\operatorname{adj}(\mathbf{A}-t\mathbf{I}_n)\) when \(1\leq k\leq n-1\).
- Explain which two terms cancel when the coefficient equation for index \(k\) is multiplied on the left by \(\mathbf{A}^k\).
- Why does the coefficient proof still apply when \(n=1\), even though there are no middle coefficient equations?
- Identify the invalid step in replacing \(t\) by \(\mathbf{A}\) directly in the adjugate identity.
