The Cayley-Hamilton Theorem

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. This remarkable result connects matrices to their characteristic polynomials in a fundamental way.

TheoremCayley-Hamilton Theorem

Let $A$ be an $n \times n$ matrix with characteristic polynomial: $p_A(\lambda) = \det(A - \lambda I) = c_0 + c_1\lambda + \cdots + c_{n-1}\lambda^{n-1} + c_n\lambda^n$

Then $A$ satisfies its characteristic equation: $p_A(A) = c_0I + c_1A + \cdots + c_{n-1}A^{n-1} + c_nA^n = 0$

That is, substituting matrix $A$ for variable $\lambda$ in $p_A(\lambda)$ yields the zero matrix.

This theorem implies that any power $A^k$ for $k \geq n$ can be expressed as a linear combination of $I, A, A^2, \ldots, A^{n-1}$ . The space of polynomials in $A$ has dimension at most $n$ .

ExampleVerifying Cayley-Hamilton

Let $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ .

Characteristic polynomial: $p_A(\lambda) = \det(A - \lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3$

Cayley-Hamilton predicts: $A^2 - 4A + 3I = 0$

Check: $A^2 = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$ , $4A = \begin{bmatrix} 8 & 4 \\ 4 & 8 \end{bmatrix}$ , $3I = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

$A^2 - 4A + 3I = \begin{bmatrix} 5-8+3 & 4-4+0 \\ 4-4+0 & 5-8+3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ ✓

TheoremMinimal Polynomial

For matrix $A$ , the minimal polynomial $m_A(\lambda)$ is the monic polynomial of smallest degree such that $m_A(A) = 0$ .

Properties:

$m_A(\lambda)$ divides every polynomial $q(\lambda)$ with $q(A) = 0$
$m_A(\lambda)$ divides the characteristic polynomial $p_A(\lambda)$
$m_A(\lambda)$ and $p_A(\lambda)$ have the same roots (eigenvalues)
$\deg(m_A) \leq n$
$A$ is diagonalizable if and only if $m_A(\lambda)$ factors into distinct linear factors

TheoremApplications of Cayley-Hamilton

Computing Matrix Inverse: If $A$ is invertible and $p_A(\lambda) = \det(A - \lambda I) = c_0 + c_1\lambda + \cdots + \lambda^n$ , then: $c_0I + c_1A + \cdots + c_{n-1}A^{n-1} + A^n = 0$

Since $\det(A) = (-1)^n c_0 \neq 0$ , we have $c_0 \neq 0$ , so: $A^{-1} = -\frac{1}{c_0}(c_1I + c_2A + \cdots + c_{n-1}A^{n-2} + A^{n-1})$

Computing Matrix Powers: To find $A^k$ for large $k$ , use Cayley-Hamilton to reduce to degree $< n$ .

ExampleUsing Cayley-Hamilton for Powers

For $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ , the characteristic polynomial is $p_A(\lambda) = (\lambda-1)^2 = \lambda^2 - 2\lambda + 1$ .

By Cayley-Hamilton: $A^2 - 2A + I = 0$ , so $A^2 = 2A - I$ .

Then: $A^3 = A \cdot A^2 = A(2A - I) = 2A^2 - A = 2(2A - I) - A = 3A - 2I$

Generally: $A^n = nA - (n-1)I$ for $n \geq 1$ .

Remark

The Cayley-Hamilton theorem has profound implications: it shows that the algebraic structure of a matrix is completely captured by a polynomial equation. This connection between linear algebra and polynomial algebra underlies much of matrix theory, from Jordan form to matrix functions.