Cayley Hamilton Theorem

Last Updated : 14 Dec, 2023

Cayley Hamilton Theorem was introduced by mathematician Arthur Cayley. It is a crucial idea in matrix algebra. It states that each square matrix adheres to a distinct equation known as the characteristic polynomial. This polynomial, derived from adjustments to the matrix, is essential for comprehending the matrix’s characteristics. The theorem is applied in various mathematical domains, assisting in matrix-related operations like inversion, exponentiation, and control theory. This article covers the meaning of Cayley Hamilton’s Theorem, the Statement of Cayley Hamilton’s Theorem Formula, and the Proof of Cayley Hamilton’s Theorem in 2 × 2 matrix and 3 × 3 matrix.

What is Cayley Hamilton’s Theorem?

The Cayley Hamilton Theorem says that for any square matrix (which has the same number of rows and columns), the matrix will always follow a certain equation. This equation is called the characteristic polynomial. To find this polynomial, you take the determinant of a modified matrix (λIn−A), where λ is a number and In is the identity matrix. The roots of this polynomial are the eigenvalues of the matrix, which are special numbers associated with the matrix. So, in simple terms, the Cayley-Hamilton Theorem tells us that every square matrix behaves according to its own special equation based on its characteristic polynomial.

Cayley Hamilton Theorem Statement

The Cayley Hamilton Theorem says that if you have a square matrix with real or complex numbers, the special polynomial you get from that matrix, called the characteristic polynomial, will turn out to be the zero matrix. This characteristic polynomial is usually written as det(λIn−A), where det represents the determinant, λ is a variable, In is the identity matrix, and A is the given matrix.

This polynomial can be broken down into a simpler form, written as p(λ)=anλn+an−1λn−1+…..+a1λ+a0λ0. It’s a kind of math expression where an is the leading coefficient, which means it’s the number in front of the highest degree variable (λn), and in this case, an is always 1. The other coefficients, an−1,…,a1,a0, are the numbers in front of the other variables in decreasing order of degree, like λn−1,…,λ1,λ0.

p(A) = Aⁿ + a_n-1A^n-1+ ….. + a₁A +a₀I_n = 0

OR

p(A) = 0, where A is an n×n square matrix

Cayley Hamilton Theorem Example

In matrix (B)

$B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$

And the characteristic polynomial for (B) is $\lambda^2 - 4\lambda + 1$

Now, we can use the Cayley-Hamilton Theorem by substituting (B) into the polynomial:

p(B) = B² – 4B + I

On calculating, we get

$B^2 = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 4 \\ 12 & 7 \end{bmatrix}$

$4B = 4 \times \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 4 \\ 12 & 8 \end{bmatrix}$

$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Now, put these into the expression:

$p(B) = \begin{bmatrix} 7 & 4 \\ 12 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 4 \\ 12 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Further subtracting:

$p(B) = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Combining the matrices:

$p(B) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

So, in this example, by substituting the matrix (B) into the characteristic polynomial $\lambda^2 - 4\lambda + 1$ , we get p(B) = 0, confirming the Cayley-Hamilton Theorem for this specific matrix.

Cayley Hamilton Theorem Formula

The Cayley-Hamilton Theorem formula is a useful tool for solving complex calculations quickly and accurately, especially in matrix algebra. It is also employed to find the inverse of a matrix. The formula is as follows:

Characteristic polynomial for an n×n square matrix (A), written as:

$p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \ldots + a_1\lambda + a_0$

Then, substituting the matrix $A$ into this polynomial, denoted as p(A), results in:

$p(A) = A^n + a_{n-1}A^{n-1} + \ldots + a_1A + a_0I_n = 0$

This implies that p(A) = 0

To find the inverse of the matrix, you can multiply this equation by the inverse of (A), denoted as A^-1, giving:

$A^{n-1} + a_{n-1}A^{n-2} + \ldots + a_1I_n + a_0A^{-1} = 0$

Solving for A^-1, it is expressed as:

$A^{-1} = -[A^{n-1} + a_{n-1}A^{n-2} + \ldots + a_1I_n]a_0$

In other words, the Cayley-Hamilton Theorem formula helps in understanding that when a matrix is appplied to its own characteristic polynomial, the result is always zero. This property is then leveraged to find the inverse of the matrix.

Cayley Hamilton Theorem for 2×2 Matrix

When dealing with a 2 × 2 square matrix and applying the Cayley Hamilton Theorem, the first thing is to figure out the characteristic polynomial expression. The general form of this polynomial is written as λ² + a₁λ + a₀, where a₁ and a₀ are coefficients. Since we’re dealing with a 2 × 2 matrix (meaning n = 2, the polynomial simplifies to λ² + a₁λ + a₀.

In the specific case of a 2 × 2 matrix, we can represent the characteristic polynomial as λ² – S₁λ + S₀, where S is the sum of the diagonal elements and S₀ is the determinant of the matrix.

Now, according to the Cayley Hamilton Theorem, if we replace λ with our 2 × 2 square matrix, say (B), the characteristic polynomial becomes B² – S₁B + S₀I = 0. Here, (B) is the 2 × 2 square matrix, (B²) is the matrix multiplied by itself, S₁ is the sum of the diagonal elements of (B), S₀ is the determinant of (B), and (I) is the identity matrix.

In simpler terms, plugging our matrix (B) into this equation should give us the zero matrix, as per the Cayley Hamilton Theorem for 2 × 2 matrices.

Example: Consider a 2 × 2 matrix $C = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$

To find the characteristic polynomial, the general form is λ² – S₁λ + S₀, where S₁ is the sum of the diagonal elements, and S₀ is the determinant.

For matrix (C):

S₁ = 3 + 4 = 7

S₀ = (3 × 4) – (2 × 1) = 10

The characteristic polynomial is λ² – 7λ + 10

Now, applying the Cayley Hamilton Theorem:

$C^2 - 7C + 10I = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}^2 - 7 \times \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} + 10 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$C^2 = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \times \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 11 & 14 \\ 4 & 18 \end{bmatrix}$

$7C = 7 \times \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 21 & 14 \\ 7 & 28 \end{bmatrix}$

$10I = 10 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$

Now, substitute these into the Cayley Hamilton Theorem equation:

$C^2 - 7C + 10I = \begin{bmatrix} 11 & 14 \\ 4 & 18 \end{bmatrix} - \begin{bmatrix} 21 & 14 \\ 7 & 28 \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$

After performing the subtraction, you should find that the result is the zero matrix:

$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

This confirms the Cayley Hamilton Theorem for the given 2 × 2 matrix example.

Cayley Hamilton Theorem for 3×3 Matrix

For a 3 × 3 square matrix, the Cayley-Hamilton Theorem relates the matrix to its characteristic polynomial, which is expressed as p(λ) = λ³ – T₂λ² + T₁λ – T₀. In this formula, T₂ is the sum of the main diagonal elements, T₁ is the sum of the minors of the main diagonal elements, and T₀ is the determinant of the 3 × 3 square matrix.

When we apply the Cayley Hamilton Theorem to a 3 × 3 matrix (C), the resulting formula is:

C³ – T₂C² + T₁C – T₀I = 0

Here, (C) represents the 3 × 3 square matrix, and (I) is the identity matrix. The theorem tells us that if we plug the matrix (C) into this equation, the result will be the zero matrix.

Example: Consider a 3 × 3 matrix (C)

$C = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$

Now, we need to find the characteristic polynomial using the given formula:

p(λ) = λ³ – T₂λ² + T₁λ- T₀

where:

T₂ is the sum of the main diagonal elements,
T₁ is the sum of the minors of the main diagonal elements,
T₀ is the determinant of the 3 × 3 square matrix.

T₂ = 1 + 5 + 9 = 15

T₁ = (1 × 5 × 9) + (2 × 6 × 7) + (3 × 4 × 8) – (3 × 5 \× 7) – (2 ×s 4 × 9) – (1 × 6 × 8) = -24

T₀ = det(C) = 1 × (5 × 9 – 6 × 8) – 2 × (4 × 9 – 6 × 7) + 3 × (4 × 8 – 5 × 7) = 0

Now, the characteristic polynomial is:

p(λ) = λ³ – 15λ² – 24λ

Applying the Cayley-Hamilton Theorem to the matrix (C):

C³ – 15C² – 24C = 0

$C^3 - 15C^2 - 24C = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}^3 - 15 \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}^2 - 24 \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$

Now, perform the matrix operations:

$C^3 = \begin{bmatrix} 468 & 576 & 684 \\ 1062 & 1296 & 1530 \\ 1656 & 2016 & 2376 \end{bmatrix}$

$15C^2 = 15 \times \begin{bmatrix} 30 & 36 & 42 \\ 66 & 81 & 96 \\ 102 & 126 & 150 \end{bmatrix}$

$24C = 24 \times \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$

Now, subtract these matrices:

$\begin{bmatrix} 468 & 576 & 684 \\ 1062 & 1296 & 1530 \\ 1656 & 2016 & 2376 \end{bmatrix} - \begin{bmatrix} 450 & 540 & 630 \\ 990 & 1215 & 1440 \\ 1230 & 1512 & 1794 \end{bmatrix} - \begin{bmatrix} 24 & 48 & 72 \\ 96 & 120 & 144 \\ 168 & 192 & 216 \end{bmatrix}$

Performing the subtraction, you should find that the result is the zero matrix:

$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Cayley-Hamilton Theorem holds for the given 3 × 3 matrix (C).

Proof of Cayley Hamilton Theorem

To prove the Cayley Hamilton Theorem, the easiest method is by substitution. Consider a matrix (A) given as:

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

According to the Cayley Hamilton theorem, the expression $p(A) = A^2 - (a + d)A + (ad - bc)I$ should equal the zero matrix. The proof unfolds as follows:

$A^2 = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix}$

$(a + d)A = \begin{bmatrix} a(a + d) & b(a + d) \\ c(a + d) & d(a + d) \end{bmatrix} = \begin{bmatrix} a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \end{bmatrix}$

$(ad - bc)I = \begin{bmatrix} ad - bc & 0 \\ 0 & ad - bc \end{bmatrix}$

Now, combine these matrices in the expression $A^2 - (a + d)A + (ad - bc)I$

$\begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix} - \begin{bmatrix} a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \end{bmatrix} + \begin{bmatrix} ad - bc & 0 \\ 0 & ad - bc \end{bmatrix}$

After performing the subtraction, the result is the zero matrix:

$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

This confirms the Cayley Hamilton Theorem for a 2 × 2 matrix. The proof can be extended similarly for higher-order square matrices.

Applications of Cayley Hamilton Theorem

Various uses of Cayley Hamilton Theorem are,

Finding Matrix Inverses: The Cayley Hamilton Theorem is employed to simplify the process of calculating the inverse of square matrices.
Matrix Exponentiation: It can be used to efficiently compute the values of matrices raised to large exponents.
Control Theory: In control theory, the Cayley Hamilton Theorem plays a role in defining important concepts, such as checking the controllability of linear systems.
Proof of Jacobson’s Theorem: The Cayley Hamilton Theorem is instrumental in proving Jacobson’s Theorem.
Nakayama’s Lemma in Commutative Algebra: In commutative algebra, a generalization of the Cayley Hamilton Theorem is used to prove Nakayama’s Lemma, a significant result in the field.

Examples on Cayley Hamilton Theorem

Example 1: Let (F) be a 2 × 2 matrix given by F = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} . Find (F³) using the Cayley Hamilton Theorem.

Solution:

Given Matrix:

F = $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Characteristic polynomial of (F) is found by solving det(?I – F) = 0

For matrix (F),

det(λI – F) = ${det}\left(\begin{bmatrix} \lambda - 1 & -2 \\ -3 & \lambda - 4 \end{bmatrix}\right)$

(λ- 1)(λ – 4) – (-2)(-3) = λ² – 5λ + 10

According to the Cayley Hamilton Theorem, F² – 5F + 10I = 0

Multiply both sides of equation by (F)

F.(F² – 5F + 10I) = F³ – 5F² + 10F = 0

Now, solving for (F³)

F³ = 5F² – 10F

F³ = 5(F²) – 10F = 5(5F – 10I) – 10F

F³ = 15F – 50I

F³ = $\begin{bmatrix} 15 & -30 \\ 45 & 60 \end{bmatrix}$

Example 2: Consider a 3 × 3 matrix (D): D = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}

Find the characteristic polynomial of matrix (D).
Applying the Cayley Hamilton Theorem to show that D³ – 6D² + 12D – 8I = 0

Solution:

Characteristic polynomial of matrix (D)

Characteristic polynomial is obtained by finding the determinant of (λI – D), where (I) is the identity matrix.

$\text{det}(\lambda I - D) = \text{det}\left(\begin{bmatrix} \lambda - 2 & -1 & 0 \\ 0 & \lambda - 2 & -1 \\ 0 & 0 & \lambda - 2 \end{bmatrix}\right)$

= (λ – 2)³

So, the characteristic polynomial is p(λ) = (λ – 2)³

Applying the Cayley Hamilton Theorem

Cayley Hamilton Theorem states that for a 3 × 3 matrix (D) with characteristic polynomial (λ – 2)³, substituting (D) into (λ – 2)³ – 3(λ – 2)² + 3(λ – 2) – I = 0 should result in the zero matrix.

Now, substitute (D) into the Cayley Hamilton Equation

D³ – 6D² + 12D – 8I = $\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}^3 - 6 \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}^2 + 12 \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

First, find (D²) and (D³)

$D^2 = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{bmatrix} \\D^3 = D \times D^2 = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} \times \begin{bmatrix} 4 & 4 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 12 & 8 \\ 0 & 8 & 12 \\ 0 & 0 & 8 \end{bmatrix}$

Now, substitute these into D³ – 6D² + 12D – 8I

$\begin{bmatrix} 8 & 12 & 8 \\ 0 & 8 & 12 \\ 0 & 0 & 8 \end{bmatrix} - 6 \begin{bmatrix} 4 & 4 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{bmatrix} + 12 \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

This confirms the Cayley Hamilton Theorem for the given matrix (D).

Practice Questions on Cayley Hamilton Theorem

Problem 1: Let matrix J = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}

Determine the characteristic polynomial of matrix (J).
Use the Cayley Hamilton Theorem to verify that J² – 4J + 4I = 0

Problem 2: Consider a 3 × 3 matrix (K): K = \begin{bmatrix} 5 & 1 & 0 \\ 0 & 5 & 1 \\ 0 & 0 & 5 \end{bmatrix}

Find the characteristic polynomial of matrix (K).
Apply the Cayley Hamilton Theorem to show that K³ – 15K² + 75K – 125I = 0

Cayley Hamilton Theorem – FAQs

1. What is Cayley Hamilton Theorem?

Cayley Hamilton Theorem is a math rule that says every square matrix satisfies its own characterstics equation.

2. Who Introduced Cayley Hamilton Theorem?

Cayley Hamilton Theorem was introduced by mathematician Arthur Cayley and William Rowan Hamilton.

3. What is Cayley Hamilton Theorem Equation?

Cayley Hamilton Theorem Equation is |A|×I – A = 0, where |A| is the determinant of matrix square matrix A, I is the Identity Matrix, and A is given Square Matrix.

4. What is Cayley Hamilton Theorem Used For?

Cayley Hamilton Theorem is used in math to understand and work with matrices. It helps solve problems involving these mathematical structures.

5. Is Cayley Hamilteon Theorem Applicable for All Types of Matrices?

Yes, the Cayley Hamilton Theorem is applicable for all types of matrices as long as they are square matrices, that means they have the same number of rows and columns.

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