LA-6.5

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Fundamental

Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic polynomial—one of the most elegant and useful results in linear algebra.

The Big Idea: If $\chi_A(\lambda) = \det(\lambda I - A)$ is the characteristic polynomial, then substituting the matrix $A$ for $\lambda$ gives $\chi_A(A) = 0$ (the zero matrix).

Learning Objectives

State and understand the Cayley-Hamilton theorem
Understand the proof using adjugate matrix approach
Use Cayley-Hamilton to compute matrix inverses
Express matrix powers as lower-degree polynomials
Connect Cayley-Hamilton to the minimal polynomial
Apply to solve matrix equations
Understand the theoretical significance of the theorem
Compute matrix functions using Cayley-Hamilton

Prerequisites

Characteristic polynomial and its properties (LA-6.2)
Matrix polynomials and evaluation
Determinants and adjugate matrices (LA-5)
Matrix multiplication and powers
Eigenvalues and diagonalization concepts

1. The Cayley-Hamilton Theorem

Definition 6.10: Matrix Polynomial

A matrix polynomial is an expression $p(A) = c_n A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I$ where $c_i$ are scalars. Note: the constant term is $c_0 I$ , not just $c_0$ .

Theorem 6.10: Cayley-Hamilton Theorem

Let $A \in M_n(F)$ be an $n \times n$ matrix over a field $F$ . If $\chi_A(\lambda) = \det(\lambda I - A)$ is the characteristic polynomial, then:

\chi_A(A) = 0

The matrix $A$ satisfies its own characteristic polynomial.

Remark 6.5: Interpretation

If $\chi_A(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0$ , then:

A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I = 0

Every power of $A$ beyond $n-1$ can be expressed in terms of lower powers!

Example 6.6: 2×2 Verification

For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ :

Step 1: Compute characteristic polynomial:

\chi_A(\lambda) = \det(\lambda I - A) = \det\begin{pmatrix} \lambda - 1 & -2 \\ -3 & \lambda - 4 \end{pmatrix} = \lambda^2 - 5\lambda - 2

Step 2: Verify $A^2 - 5A - 2I = 0$ :

A^2 = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}, \quad 5A = \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix}, \quad 2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}

A^2 - 5A - 2I = \begin{pmatrix} 7-5-2 & 10-10-0 \\ 15-15-0 & 22-20-2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \checkmark

Example 6.7: 3×3 Example

For $A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix}$ (diagonal):

$\chi_A(\lambda) = (\lambda - 2)(\lambda - 3)(\lambda - 5) = \lambda^3 - 10\lambda^2 + 31\lambda - 30$

Check: $A^3 - 10A^2 + 31A - 30I = 0$ ✓

2. Proof of Cayley-Hamilton

Theorem 6.11: Adjugate Identity

For any matrix $A$ , the adjugate matrix satisfies:

(\lambda I - A) \cdot \text{adj}(\lambda I - A) = \det(\lambda I - A) \cdot I = \chi_A(\lambda) \cdot I

Proof:

Proof of Cayley-Hamilton using adjugate:

Step 1: The adjugate $\text{adj}(\lambda I - A)$ is a matrix whose entries are polynomials in $\lambda$ of degree at most $n-1$ . Write:

\text{adj}(\lambda I - A) = B_{n-1}\lambda^{n-1} + B_{n-2}\lambda^{n-2} + \cdots + B_1\lambda + B_0

Step 2: From the adjugate identity:

(\lambda I - A)(B_{n-1}\lambda^{n-1} + \cdots + B_0) = \chi_A(\lambda) \cdot I

Step 3: Expand the left side and equate coefficients of each power of $\lambda$ .

Step 4: Multiply each coefficient equation by the appropriate power of $A$ and sum:

\chi_A(A) = 0

∎

Remark 6.6: Alternative Proof via Jordan Form

If $A = PJP^{-1}$ where $J$ is Jordan form, then $\chi_A(A) = P\chi_A(J)P^{-1}$ . Since each Jordan block $J_k(\lambda_i)$ satisfies $(J_k(\lambda_i) - \lambda_i I)^k = 0$ , the characteristic polynomial annihilates $J$ , hence $A$ .

3. Applications

Corollary 6.3: Finding Matrix Inverse

If $A$ is invertible with $\chi_A(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0$ , then $c_0 = (-1)^n \det(A) \neq 0$ and:

A^{-1} = -\frac{1}{c_0}(A^{n-1} + c_{n-1}A^{n-2} + \cdots + c_1 I)

Proof:

From Cayley-Hamilton: $A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I = 0$

Rearrange: $A(A^{n-1} + c_{n-1}A^{n-2} + \cdots + c_1 I) = -c_0 I$

Divide by $-c_0$ : $A \cdot \left(-\frac{1}{c_0}(A^{n-1} + \cdots + c_1 I)\right) = I$

∎

Example 6.8: Computing Inverse via Cayley-Hamilton

For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ with $\chi_A(\lambda) = \lambda^2 - 5\lambda - 2$ :

Cayley-Hamilton: $A^2 - 5A - 2I = 0$

Rearrange: $A(A - 5I) = 2I$ , so $A^{-1} = \frac{1}{2}(A - 5I)$

A^{-1} = \frac{1}{2}\begin{pmatrix} 1-5 & 2 \\ 3 & 4-5 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}

Corollary 6.4: Matrix Powers

For any $k \geq n$ , $A^k$ can be expressed as a polynomial of degree at most $n-1$ in $A$ :

A^k = a_{n-1}A^{n-1} + a_{n-2}A^{n-2} + \cdots + a_1 A + a_0 I

Example 6.9: Computing High Powers

For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ with $\chi_A(\lambda) = (\lambda - 1)^2 = \lambda^2 - 2\lambda + 1$ :

Cayley-Hamilton: $A^2 = 2A - I$

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

Compute $A^4$ : $A^4 = A \cdot A^3 = A(3A - 2I) = 3A^2 - 2A = 3(2A - I) - 2A = 4A - 3I$

Pattern: $A^n = nA - (n-1)I$ for this specific matrix.

4. Connection to Minimal Polynomial

Definition 6.11: Minimal Polynomial

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

Theorem 6.12: Minimal Polynomial Divides Characteristic

For any matrix $A$ :

$m_A(\lambda)$ divides $\chi_A(\lambda)$
$m_A$ and $\chi_A$ have the same roots (eigenvalues)
$\deg(m_A) \leq \deg(\chi_A) = n$

Proof:

By Cayley-Hamilton, $\chi_A(A) = 0$ . By definition, $m_A$ is the minimal polynomial that annihilates $A$ . By the division algorithm, $\chi_A = m_A \cdot q + r$ where $\deg(r) < \deg(m_A)$ . Since $\chi_A(A) = 0$ and $m_A(A) = 0$ , we have $r(A) = 0$ . By minimality of $m_A$ , $r = 0$ , so $m_A | \chi_A$ .

∎

Example 6.10: Minimal vs Characteristic

Diagonal matrix: $D = \text{diag}(2, 2, 3)$

$\chi_D(\lambda) = (\lambda - 2)^2(\lambda - 3)$

$m_D(\lambda) = (\lambda - 2)(\lambda - 3)$ (no repeated factors needed)

Example 6.11: Jordan Block

For Jordan block $J_3(2)$ :

$\chi_{J_3}(\lambda) = (\lambda - 2)^3$

$m_{J_3}(\lambda) = (\lambda - 2)^3$ (minimal equals characteristic)

Corollary 6.5: Diagonalizability Criterion

A matrix $A$ is diagonalizable if and only if its minimal polynomial has no repeated roots.

5. Common Mistakes

❌ Confusing χ(A) with χ(λ)

χ_A(λ) is a scalar polynomial in λ. χ_A(A) substitutes the matrix A for λ, interpreting constants as multiples of I. The result is the zero MATRIX.

❌ Forgetting I in constant term

The constant term $c_0$ becomes $c_0 I$ , NOT just the scalar $c_0$ .

❌ Thinking det(A - AI) = 0 trivially

This is NOT what Cayley-Hamilton says! We evaluate the polynomial χ_A at the matrix A, NOT compute det(A - A·I) = det(0) = 0.

❌ Wrong sign in characteristic polynomial

Convention matters: $\chi_A(\lambda) = \det(\lambda I - A)$ vs $\det(A - \lambda I)$ . They differ by $(-1)^n$ .

6. Worked Examples

Example 6.12: Rotation Matrix

For the 90° rotation $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ :

$\chi_A(\lambda) = \lambda^2 + 1$ (eigenvalues are $\pm i$ )

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

This confirms: rotating by 90° twice gives rotation by 180°, which is $-I$ !

Also: $A^{-1} = -A$ (rotating by -90° = -1 times rotating by 90°)

Example 6.13: Nilpotent Matrix

For $N = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ :

$\chi_N(\lambda) = \lambda^3$

Cayley-Hamilton: $N^3 = 0$ ✓

For nilpotent matrices, Cayley-Hamilton says some power is zero.

Example 6.14: Companion Matrix

The companion matrix of $p(\lambda) = \lambda^3 - 2\lambda^2 + 3\lambda - 1$ :

C = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & -3 \\ 0 & 1 & 2 \end{pmatrix}

The characteristic polynomial equals $p$ , and Cayley-Hamilton gives $C^3 = 2C^2 - 3C + I$ .

7. Matrix Functions via Cayley-Hamilton

Theorem 6.13: Matrix Function Reduction

For any analytic function $f$ and $n \times n$ matrix $A$ , $f(A)$ can be expressed as a polynomial of degree at most $n-1$ in $A$ :

f(A) = a_{n-1}A^{n-1} + \cdots + a_1 A + a_0 I

Example 6.15: Matrix Exponential

For $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ with $A^2 = -I$ :

e^{tA} = \sum_{k=0}^{\infty} \frac{t^k A^k}{k!}

Using $A^2 = -I$ , $A^3 = -A$ , $A^4 = I$ , ...:

e^{tA} = \cos(t) I + \sin(t) A = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}

This is the rotation matrix by angle $t$ !

Remark 6.7: Hamilton-Cayley for Quaternions

Hamilton originally proved a version for quaternions: every quaternion satisfies a quadratic polynomial. Cayley extended this to matrices.

8. Theoretical Significance

Algebra of Matrices

Cayley-Hamilton shows that $M_n(F)$ is a finite-dimensional algebra over $F$ . Every element satisfies a polynomial of degree $n$ .

Module Theory Connection

Via the correspondence between linear transformations and $F[x]$ -modules, Cayley-Hamilton follows from structure theorems for finitely generated modules over PIDs.

Generalizations

Extensions exist for matrices over commutative rings, bounded operators on Banach spaces, and more general algebraic structures.

9. Key Takeaways

The Theorem

$\chi_A(A) = 0$

Inverse Formula

$A^{-1}$ as polynomial in $A$

Power Reduction

$A^k$ as deg ≤ n-1 polynomial

Minimal Poly

$m_A | \chi_A$

10. Challenge Problems

Challenge 1

Prove Cayley-Hamilton for diagonalizable matrices directly (without using the adjugate proof).

Challenge 2

For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , find $A^{100}$ using Cayley-Hamilton.

Challenge 3

Show that if $A^2 = A$ (idempotent), then the only eigenvalues are 0 and 1.

Challenge 4

Prove: If $AB = BA$ , then $A$ and $B$ satisfy a common polynomial.

11. Conceptual Questions

Q: Why doesn't χ_A(A) = det(A - A·I) = det(0) = 0 work?

This "proof" confuses scalar and matrix arithmetic. det(λI - A) is a polynomial in λ. We substitute the matrix A, not the scalar, getting a matrix polynomial evaluation.

Q: What's the geometric meaning?

The characteristic polynomial encodes how A scales along each eigendirection. Cayley-Hamilton says applying these scaling factors to A itself produces zero—each eigenvector gets multiplied by (λᵢ - λᵢ) = 0.

Q: When is minimal poly = char poly?

When the Jordan form has exactly one block for each eigenvalue (i.e., geometric multiplicity = 1 for all eigenvalues).

Quick Reference

Concept	Formula / Property
Cayley-Hamilton	$\chi_A(A) = 0$
Matrix inverse	$A^{-1} = -\frac{1}{c_0}(A^{n-1} + c_{n-1}A^{n-2} + \cdots)$
Power reduction	$A^n = -(c_{n-1}A^{n-1} + \cdots + c_0 I)$
Minimal polynomial	$m_A \| \chi_A$ , same roots
Diagonalizability	iff $m_A$ has no repeated roots

12. More Practice Problems

Problem 1

Verify Cayley-Hamilton for $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ .

Answer: $\chi_A = (\lambda-2)(\lambda-3) = \lambda^2 - 5\lambda + 6$ . Check $A^2 - 5A + 6I = 0$ .

Problem 2

For $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ , find $A^{-1}$ using Cayley-Hamilton.

Answer: $\chi_A = \lambda^2 + 1$ , so $A^2 = -I$ . Thus $A^{-1} = -A$ .

Problem 3

Express $A^5$ in terms of $A, I$ for a 2×2 matrix with $\chi_A = \lambda^2 - 3\lambda + 2$ .

Hint: First express $A^2 = 3A - 2I$ , then compute $A^3, A^4, A^5$ recursively.

Problem 4

If $\chi_A = \lambda^3 - 6\lambda^2 + 11\lambda - 6$ , express $A^3$ in lower terms.

Answer: $A^3 = 6A^2 - 11A + 6I$

Problem 5

Find the minimal polynomial of $D = \text{diag}(1, 1, 2, 2, 2)$ .

Answer: $m_D = (\lambda - 1)(\lambda - 2)$ . (Compare: $\chi_D = (\lambda-1)^2(\lambda-2)^3$ )

13. Advanced Applications

Example 6.16: Control Theory: Controllability

In control theory, for system $\dot{x} = Ax + Bu$ , the controllability matrix is:

\mathcal{C} = [B, AB, A^2B, \ldots, A^{n-1}B]

By Cayley-Hamilton, higher powers $A^k B$ for $k \geq n$ can be expressed using the first $n$ terms.

Example 6.17: Differential Equations

For the system $\mathbf{x}' = A\mathbf{x}$ , the solution is $e^{At}\mathbf{x}_0$ .

Using Cayley-Hamilton, $e^{At}$ is a polynomial of degree $n-1$ in $A$ :

e^{At} = \alpha_0(t)I + \alpha_1(t)A + \cdots + \alpha_{n-1}(t)A^{n-1}

where $\alpha_i(t)$ are determined by eigenvalues and their multiplicities.

Example 6.18: Markov Chains: Steady State

For a Markov transition matrix $P$ , powers $P^k$ approach the steady-state matrix.

Using Cayley-Hamilton, we can express $P^k$ as a polynomial in $P$ , which helps analyze convergence rates.

Example 6.19: Recurrence Relations

The Fibonacci sequence $F_{n+2} = F_{n+1} + F_n$ can be written using:

\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}

$\chi_A = \lambda^2 - \lambda - 1$ , so $A^n$ can be computed efficiently using Cayley-Hamilton.

14. Alternative Proof Approaches

Proof via Diagonalization

For diagonalizable $A = PDP^{-1}$ with $D = \text{diag}(\lambda_1, \ldots, \lambda_n)$ :

\chi_A(A) = P\chi_A(D)P^{-1} = P \cdot \text{diag}(\chi_A(\lambda_1), \ldots, \chi_A(\lambda_n)) \cdot P^{-1}

Since each $\lambda_i$ is a root of $\chi_A$ , we have $\chi_A(\lambda_i) = 0$ , giving $\chi_A(A) = 0$ .

Proof via Jordan Form

For any matrix, $A = PJP^{-1}$ where $J$ is Jordan form. Each Jordan block $J_k(\lambda_i)$ satisfies $(J_k - \lambda_i I)^k = 0$ . Since $\chi_A$ contains $(\lambda - \lambda_i)^{a_i}$ where $a_i \geq k$ , we get $\chi_A(J) = 0$ .

Proof via Density Argument

Diagonalizable matrices are dense in $M_n(\mathbb{C})$ . Since Cayley-Hamilton holds for all diagonalizable matrices and $\chi_A(A)$ is continuous in $A$ , it holds for all matrices by continuity.

Cayley-Hamilton Examples Gallery

Matrix	Char Poly	Cayley-Hamilton
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$	$\lambda^2 - (a+d)\lambda + (ad-bc)$	$A^2 - \text{tr}(A)A + \det(A)I = 0$
$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$	$\lambda^2$	$A^2 = 0$
$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$	$(\lambda-1)^2$	$(A-I)^2 = 0$
$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$	$\lambda^2 + 1$	$A^2 = -I$

Verification Checklist

When using Cayley-Hamilton:

Correctly compute

\chi_A(\lambda)

Replace

\lambda^k

with

A^k

Constants become scalar × I

Result is zero MATRIX

For inverse:

\det(A) \neq 0

Sign convention consistent

15. Detailed Worked Examples

Example 6.20: Complete 2×2 Verification

For $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ , verify Cayley-Hamilton.

Step 1: Compute $\chi_A(\lambda)$ :

\chi_A(\lambda) = \det\begin{pmatrix} \lambda - 3 & -1 \\ -2 & \lambda - 4 \end{pmatrix} = (\lambda-3)(\lambda-4) - 2 = \lambda^2 - 7\lambda + 10

Step 2: Compute $A^2$ :

A^2 = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 11 & 7 \\ 14 & 18 \end{pmatrix}

Step 3: Verify $A^2 - 7A + 10I = 0$ :

\begin{pmatrix} 11 & 7 \\ 14 & 18 \end{pmatrix} - 7\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} + 10\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \checkmark

Example 6.21: Finding Inverse via Cayley-Hamilton

For $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ , find $A^{-1}$ using Cayley-Hamilton.

From $A^2 - 7A + 10I = 0$ :

A^2 - 7A = -10I

A(A - 7I) = -10I

A \cdot \left(-\frac{1}{10}(A - 7I)\right) = I

Therefore:

A^{-1} = -\frac{1}{10}(A - 7I) = -\frac{1}{10}\begin{pmatrix} 3-7 & 1 \\ 2 & 4-7 \end{pmatrix} = \begin{pmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{pmatrix}

Example 6.22: Computing A^{100}

For $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ , find $A^{100}$ .

$\chi_A(\lambda) = (\lambda - 2)^2 = \lambda^2 - 4\lambda + 4$

Cayley-Hamilton: $A^2 = 4A - 4I$

For this matrix, we can show: $A^n = 2^{n-1}(2I + nN)$ where $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ .

A^{100} = 2^{99}\begin{pmatrix} 2 & 100 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2^{100} & 100 \cdot 2^{99} \\ 0 & 2^{100} \end{pmatrix}

Example 6.23: 3×3 Matrix Power

For $A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ , express $A^4$ in terms of $A^2, A, I$ .

$\chi_A(\lambda) = (\lambda - 1)^3 = \lambda^3 - 3\lambda^2 + 3\lambda - 1$

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

Compute $A^4$ :

A^4 = A \cdot A^3 = A(3A^2 - 3A + I) = 3A^3 - 3A^2 + A

Substitute $A^3$ :

A^4 = 3(3A^2 - 3A + I) - 3A^2 + A = 6A^2 - 8A + 3I

16. Applications Summary

Matrix Inverse

Express $A^{-1}$ as polynomial in $A$ :

$A^{-1} = -\frac{1}{\det(A)}(A^{n-1} + c_{n-1}A^{n-2} + \cdots)$

Matrix Powers

Reduce high powers:

$A^k$ = polynomial of degree ≤ n-1 in $A$

Matrix Exponential

Compute $e^{At}$ :

$e^{At} = \sum \alpha_i(t)A^i$ with $i < n$

Recurrence Relations

Solve linear recurrences:

Transform to matrix form, apply Cayley-Hamilton

More Quick Examples

Projection

If $P^2 = P$

$\chi_P = \lambda(\lambda-1)$

Involution

If $A^2 = I$

$\chi_A = (\lambda-1)(\lambda+1)$

Nilpotent

If $N^k = 0$

$\chi_N = \lambda^n$

Minimal vs Characteristic Polynomial

Property	Minimal $m_A$	Characteristic $\chi_A$
Definition	Smallest degree with $m_A(A)=0$	$\det(\lambda I - A)$
Degree	≤ n	= n
Roots	Eigenvalues (each once or more)	Eigenvalues with mult
Relationship	$m_A \| \chi_A$ , same roots
Diagonalizable iff	No repeated roots	Geometric = Algebraic mult

Study Tips

Start with 2×2: Master the formula $A^2 - \text{tr}(A)A + \det(A)I = 0$ .
Remember the substitution: Replace $\lambda$ with $A$ , constant $c$ with $cI$ .
For inverse: Rearrange to isolate $A \cdot (\text{something}) = \det(A) \cdot I$ .
For powers: Express $A^n$ using lower powers, then substitute recursively.
Minimal polynomial: It divides char poly and has the same roots.
Check your work: Verify by direct multiplication when possible.

Using Cayley-Hamilton: Decision Tree

Goal: Find A^-1?

Compute $\chi_A$ , rearrange $\chi_A(A) = 0$ to solve for $I$ , divide by $A$ .

Goal: Compute A^k for large k?

Express $A^n$ in terms of lower powers, substitute recursively.

Goal: Verify Cayley-Hamilton?

Compute $\chi_A(\lambda)$ , compute all powers of $A$ , substitute and verify = 0.

Goal: Find minimal polynomial?

Start with divisors of $\chi_A$ , test which smallest degree polynomial annihilates $A$ .

Final Summary

In this comprehensive module on the Cayley-Hamilton theorem, you learned:

The theorem: $\chi_A(A) = 0$ — every matrix satisfies its characteristic polynomial
How to use it to compute matrix inverses
How to reduce high powers of matrices to degree ≤ n-1
The connection to minimal polynomial ( $m_A | \chi_A$ )
Applications in differential equations, control theory, and recurrence relations

Historical Notes

Arthur Cayley (1821-1895): British mathematician who first stated the theorem for 2×2 and 3×3 matrices in 1858, verifying it by direct computation. He wrote: "I have not thought it necessary to undertake the labor of a formal proof of the theorem in the general case of a matrix of any degree."

William Rowan Hamilton (1805-1865): Irish mathematician who proved a version for quaternions in 1853. Quaternions can be represented as certain 2×2 complex matrices.

Ferdinand Frobenius (1849-1917): German mathematician who gave the first rigorous general proof in 1878.

Modern significance: The theorem is fundamental to matrix theory, control theory, signal processing, and computational linear algebra.

Exam Preparation

What You Should Know

• Statement of Cayley-Hamilton
• How to verify for small matrices
• Using it to find A^-1
• Reducing high powers of A
• Connection to minimal polynomial

Common Exam Questions

• Verify Cayley-Hamilton for given matrix
• Find A^-1 using Cayley-Hamilton
• Express A^k as lower-degree polynomial
• True/False on theorem properties
• Minimal vs characteristic polynomial

Memorization Aids

The Theorem

"A matrix satisfies its own characteristic equation"

Inverse Formula

"Solve χ(A)=0 for I, divide by A"

Power Reduction

"Aⁿ expressed using Aⁿ⁻¹, ..., I only"

Minimal Polynomial

"Smallest degree, divides char poly"

Learning Path

6.4 Jordan Form

Current

6.5 Cayley-Hamilton

7.1 Inner Products

What's Next?

With Cayley-Hamilton mastered, you've completed the Eigenvalues chapter! Next up:

Inner Product Spaces: Add geometric structure—lengths, angles, orthogonality
Orthonormal Bases: Gram-Schmidt and the power of perpendicularity
Spectral Theorem: Symmetric/Hermitian matrices are orthogonally diagonalizable
SVD: The singular value decomposition—works for ANY matrix

Module Summary

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic polynomial. This elegant result has powerful applications for computing matrix inverses, reducing high powers, and connecting to the minimal polynomial.

13+

Theorems

15+

Examples

Quiz Questions

FAQs

Key Takeaways

• Cayley-Hamilton: $p_A(A) = 0$ — every matrix satisfies its char poly
• Applications: computing $A^{-1}$ , reducing high powers $A^n$
• Minimal poly divides characteristic poly and has same roots
• Matrix functions: $e^A$ , $\sin(A)$ can be reduced to polynomials

Common Mistakes

❌ Thinking $p_A(x) = 0$ gives eigenvalues → this gives characteristic equation

❌ Confusing minimal with characteristic polynomial — minimal has smallest degree

❌ Substituting scalar operations directly — matrix multiplication is not commutative

Frequently Asked Questions

What does Cayley-Hamilton actually say?

Every square matrix A satisfies its characteristic polynomial: if χ_A(λ) = det(λI - A), then χ_A(A) = 0 (the zero matrix, not the number zero).

How do I use it to find A^{-1}?

If χ_A(λ) = λⁿ + c₁λⁿ⁻¹ + ... + cₙ, then Aⁿ + c₁Aⁿ⁻¹ + ... + cₙI = 0. Rearrange and multiply by A^{-1} to get A^{-1} = -(Aⁿ⁻¹ + c₁Aⁿ⁻² + ... + cₙ₋₁I)/cₙ.

What's the minimal polynomial?

The monic polynomial m(λ) of smallest degree such that m(A) = 0. It always divides the characteristic polynomial. A is diagonalizable iff minimal poly has no repeated roots.

Why can't I just plug A into det(A - λI)?

det(A - λI) is a polynomial in the scalar λ. We substitute the matrix A for λ, interpreting constants as multiples of I. The result is a matrix, not a scalar.

How is this related to Jordan form?

Cayley-Hamilton can be proven using Jordan form: each Jordan block satisfies (J - λI)^k = 0 for appropriate k, and the full Jordan matrix satisfies the char poly.

Can I use this for infinite-dimensional operators?

Not directly—Cayley-Hamilton is for finite-dimensional spaces. Extensions exist for certain classes of operators in functional analysis.

Why is it called Cayley-Hamilton?

Named after Arthur Cayley (who stated it for 2×2 and 3×3 matrices in 1858) and William Rowan Hamilton (who proved a version for quaternions).

What if the characteristic polynomial has complex roots?

The theorem holds regardless—χ_A(A) = 0 even if eigenvalues are complex. The computation uses matrix arithmetic, not finding roots.

Cayley-Hamilton Theorem

1. The Cayley-Hamilton Theorem

2. Proof of Cayley-Hamilton

3. Applications

4. Connection to Minimal Polynomial

5. Common Mistakes

❌ Confusing χ(A) with χ(λ)

❌ Forgetting I in constant term

❌ Thinking det(A - AI) = 0 trivially

❌ Wrong sign in characteristic polynomial

6. Worked Examples

7. Matrix Functions via Cayley-Hamilton

8. Theoretical Significance

Algebra of Matrices

Module Theory Connection

Generalizations

9. Key Takeaways

The Theorem

Inverse Formula

Power Reduction

Minimal Poly

10. Challenge Problems

11. Conceptual Questions

Quick Reference

12. More Practice Problems

13. Advanced Applications

14. Alternative Proof Approaches

Proof via Diagonalization

Proof via Jordan Form

Proof via Density Argument

Cayley-Hamilton Examples Gallery

Verification Checklist

15. Detailed Worked Examples

16. Applications Summary

Matrix Inverse

Matrix Powers

Matrix Exponential

Recurrence Relations

More Quick Examples

Projection

Involution

Nilpotent

Minimal vs Characteristic Polynomial

Study Tips

Using Cayley-Hamilton: Decision Tree

Final Summary

More Conceptual Questions

Historical Notes

Exam Preparation

What You Should Know

Common Exam Questions

Memorization Aids

Learning Path

What's Next?

Module Summary

Key Takeaways

Common Mistakes

Related Topics

Frequently Asked Questions

What does Cayley-Hamilton actually say?

How do I use it to find A^{-1}?

What's the minimal polynomial?

Why can't I just plug A into det(A - λI)?

How is this related to Jordan form?

Can I use this for infinite-dimensional operators?

Why is it called Cayley-Hamilton?

What if the characteristic polynomial has complex roots?