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.
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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.
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.
Cayley Hamilton Theorem
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 and 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) = A n + a n-1 A n-1 + ….. + a 1 A +a 0 I n = 0
OR
p(A) = 0, where A is an n×n square matrix
And the characteristic polynomial for (B) is [Tex]\lambda^2 – 4\lambda + 1 [/Tex]
Now, we can use the Cayley-Hamilton Theorem by substituting (B) into the polynomial:
On calculating, we get
[Tex]B^2 = \begin 2 & 1 \\ 3 & 2 \end \times \begin 2 & 1 \\ 3 & 2 \end = \begin 7 & 4 \\ 12 & 7 \end [/Tex]
[Tex]4B = 4 \times \begin 2 & 1 \\ 3 & 2 \end = \begin 8 & 4 \\ 12 & 8 \end [/Tex]
Now, put these into the expression:
Combining the matrices:
So, in this example, by substituting the matrix (B) into the characteristic polynomial [Tex]\lambda^2 – 4\lambda + 1 [/Tex] , we get p(B) = 0, confirming the Cayley-Hamilton Theorem for this specific matrix.
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:
[Tex]p(\lambda) = \lambda^n + a_\lambda^ + \ldots + a_1\lambda + a_0 [/Tex]
Then, substituting the matrix \(A\) into this polynomial, denoted as p(A), results in:
[Tex]p(A) = A^n + a_A^ + \ldots + a_1A + a_0I_n = 0 [/Tex]
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:
[Tex]A^ + a_A^ + \ldots + a_1I_n + a_0A^ = 0 [/Tex]
Solving for A -1 , it is expressed as:
[Tex] A^ = -[A^ + a_A^ + \ldots + a_1I_n]a_0 [/Tex]
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.
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 λ 2 + a 1 λ + a 0 , where a 1 and a 0 are coefficients. Since we’re dealing with a 2 × 2 matrix (meaning n = 2, the polynomial simplifies to λ 2 + a 1 λ + a 0 .
In the specific case of a 2 × 2 matrix, we can represent the characteristic polynomial as λ 2 – S 1 λ + S 0 , where S is the sum of the diagonal elements and S 0 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 B 2 – S 1 B + S 0 I = 0. Here, (B) is the 2 × 2 square matrix, (B 2 ) is the matrix multiplied by itself, S 1 is the sum of the diagonal elements of (B), S 0 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 [Tex]C = \begin 3 & 2 \\ 1 & 4 \end [/Tex]
Solution:
To find characteristic polynomial, the general form is λ 2 – S 1 λ + S 0 , where S 1 is the sum of the diagonal elements, and S 0 is the determinant.
For matrix (C):
S 1 = 3 + 4 = 7
S 0 = (3 × 4) – (2 × 1) = 10
The characteristic polynomial is λ 2 – 7λ + 10
Now, applying the Cayley Hamilton Theorem:
[Tex]C^2 – 7C + 10I = \begin 3 & 2 \\ 1 & 4 \end^2 – 7 \times \begin 3 & 2 \\ 1 & 4 \end + 10 \times \begin 1 & 0 \\ 0 & 1 \end [/Tex]
[Tex]C^2 = \begin 3 & 2 \\ 1 & 4 \end \times \begin 3 & 2 \\ 1 & 4 \end = \begin 11 & 14 \\ 4 & 18 \end [/Tex]
[Tex]7C = 7 \times \begin 3 & 2 \\ 1 & 4 \end = \begin 21 & 14 \\ 7 & 28 \end [/Tex]
[Tex]10I = 10 \times \begin 1 & 0 \\ 0 & 1 \end = \begin 10 & 0 \\ 0 & 10 \end [/Tex]
Now, substitute these into the Cayley Hamilton Theorem equation:
[Tex]C^2 – 7C + 10I = \begin 11 & 14 \\ 4 & 18 \end – \begin 21 & 14 \\ 7 & 28 \end + \begin 10 & 0 \\ 0 & 10 \end [/Tex]
After performing the subtraction, you should find that the result is the zero matrix:
[Tex] \begin 0 & 0 \\ 0 & 0 \end [/Tex]
This confirms the Cayley Hamilton Theorem for the given 2 × 2 matrix example.
For a 3 × 3 square matrix, the Cayley-Hamilton Theorem relates the matrix to its characteristic polynomial, which is expressed as p(λ) = λ 3 – T 2 λ 2 + T 1 λ – T 0 . In this formula, T 2 is the sum of the main diagonal elements, T 1 is the sum of the minors of the main diagonal elements, and T 0 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:
C 3 – T 2 C 2 + T 1 C – T 0 I = 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)
Solution:
To prove the Cayley Hamilton Theorem, the easiest method is by substitution. Consider a matrix (A) given as:
According to the Cayley Hamilton theorem, the expression [Tex]p(A) = A^2 – (a + d)A + (ad – bc)I [/Tex] should equal the zero matrix. The proof unfolds as follows:
[Tex]A^2 = \begin a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end [/Tex]
[Tex](a + d)A = \begin a(a + d) & b(a + d) \\ c(a + d) & d(a + d) \end = \begin a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \end [/Tex]
⇒ [Tex](ad – bc)I = \begin ad – bc & 0 \\ 0 & ad – bc \end [/Tex]
Now, combine these matrices in the expression [Tex]A^2 – (a + d)A + (ad – bc)I [/Tex]
[Tex]\begin a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end – \begin a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \end + \begin ad – bc & 0 \\ 0 & ad – bc \end [/Tex]
After performing the subtraction, the result is the zero matrix:
This confirms the Cayley Hamilton Theorem for a 2 × 2 matrix. The proof can be extended similarly for higher-order square matrices.
Various uses of Cayley Hamilton Theorem are,
Example 1: Let (F) be a 2 × 2 matrix given by F = \begin 1 & 2 \\ 3 & 4 \end . Find (F 3 ) using the Cayley Hamilton Theorem.
Solution:
Given Matrix:
F = [Tex]\begin 1 & 2 \\ 3 & 4 \end [/Tex]
Characteristic polynomial of (F) is found by solving det(?I – F) = 0
For matrix (F),
det(λI – F) = [Tex]\left(\begin \lambda – 1 & -2 \\ -3 & \lambda – 4 \end\right) [/Tex]
(λ- 1)(λ – 4) – (-2)(-3) = λ 2 – 5λ + 10
According to the Cayley Hamilton Theorem, F 2 – 5F + 10I = 0
Multiply both sides of equation by (F)
F.(F 2 – 5F + 10I) = F 3 – 5F 2 + 10F = 0
Now, solving for (F 3 )
F 3 = 5F 2 – 10F
⇒ F 3 = 5(F 2 ) – 10F = 5(5F – 10I) – 10F
⇒ F 3 = 15F – 50I
⇒ F 3 = [Tex]\begin 15 & -30 \\ 45 & 60 \end [/Tex]
Example 2: Consider a 3 × 3 matrix (D): D = \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end
Solution:
Characteristic polynomial of matrix (D)
Characteristic polynomial is obtained by finding the determinant of (λI – D), where (I) is the identity matrix.
[Tex]\text(\lambda I – D) = \text\left(\begin \lambda – 2 & -1 & 0 \\ 0 & \lambda – 2 & -1 \\ 0 & 0 & \lambda – 2 \end\right) [/Tex]
= (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
D 3 – 6D 2 + 12D – 8I = [Tex]\begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end^3 – 6 \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end^2 + 12 \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end – 8 \begin 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end [/Tex]
First, find (D 2 ) and (D 3 )
[Tex]D^2 = \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end \times \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end = \begin 4 & 4 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end \\D^3 = D \times D^2 = \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end \times \begin 4 & 4 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end = \begin 8 & 12 & 8 \\ 0 & 8 & 12 \\ 0 & 0 & 8 \end [/Tex]
Now, substitute these into D 3 – 6D 2 + 12D – 8I
[Tex]\begin 8 & 12 & 8 \\ 0 & 8 & 12 \\ 0 & 0 & 8 \end – 6 \begin 4 & 4 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end + 12 \begin 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end – 8 \begin 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end \\\begin 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end [/Tex]
This confirms the Cayley Hamilton Theorem for the given matrix (D).
Problem 1: Let matrix [Tex]\bold < J = \begin2 & 1 \\ 0 & 2 \end >[/Tex]
Problem 2: Consider a 3 × 3 matrix (K): [Tex]\bold 5 & 1 & 0 \\ 0 & 5 & 1 \\ 0 & 0 & 5 \end>[/Tex]
Cayley Hamilton Theorem is a math rule that says every square matrix satisfies its own characterstics equation.
Cayley Hamilton Theorem was introduced by mathematician Arthur Cayley and William Rowan Hamilton.
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.
Cayley Hamilton Theorem is used in math to understand and work with matrices. It helps solve problems involving these mathematical structures.
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.