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Cayley Hamilton Theorem
Cayley Hamilton Theorem was given in the year 1858 by two mathematicians, Arthur Cayley and William Rowan Hamilton. The Cayley Hamilton Theorem states that all real and complex square matrices will satisfy their own characteristic polynomial equation. This implies that when a square matrix is transformed into a polynomial then this polynomial will be equal to 0.
Cayley Hamilton Theorem is a very important result that is used in advanced linear algebra to simplify linear transformations. In this article, we will learn more about the Cayley Hamilton theorem, its statement, proof, and associated examples.
1.  What is Cayley Hamilton Theorem? 
2.  Cayley Hamilton Theorem Formula 
3.  Cayley Hamilton Theorem Proof 
4.  Cayley Hamilton Theorem Applications 
5.  FAQs on Cayley Hamilton Theorem 
What is Cayley Hamilton Theorem?
According to the Cayley Hamilton theorem, a square matrix will satisfy its own characteristic polynomial equation. A characteristic polynomial is associated with the determinant of a matrix and the eigenvalues of the matrix will be the roots of this polynomial. Suppose a square matrix A is given with n rows and n columns. The characteristic polynomial of this matrix is given as det\(\left ( \lambda I_{n} A\right )\). Here, \(I_{n}\) is the identity matrix, \(\lambda\) is a scalar quantity and det signifies the determinant operation.
Cayley Hamilton Theorem Statement
The Cayley Hamilton theorem states that the characteristic polynomial expression of a real or complex square matrix will be equal to the zero matrix. The characteristic polynomial p(\(\lambda\)) = det\(\left ( \lambda I_{n} A\right )\) can be decomposed as p(\(\lambda\)) = \(a_{n}\lambda ^{n} + a_{n1}\lambda ^{n1}+...+a_{1}\lambda+a_{0}\lambda ^{0}\). This is a monic polynomial where the leading coefficient, i.e, the coefficient of the highest degree variable, will be equal to 1. Thus, \(a_{n}\) = 1. Here, \(a_{n1}\),...,\(a_{1}\), \(a_{0}\) are coefficients of the variables \(\lambda ^{n1}\),..., \(\lambda ^{1}\), \(\lambda ^{0}\) respectively.
We have p(\(\lambda\)) = \(a_{n}\lambda ^{n} + a_{n1}\lambda ^{n1}+...+a_{1}\lambda+a_{0}\lambda ^{0}\)
p(\(\lambda\)) = \(\lambda ^{n} + a_{n1}\lambda ^{n1}+...+a_{1}\lambda+a_{0}\)
On replacing \(\lambda\) with the matrix, A, the polynomial can be written as follows:
p(A) = \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\)
Now according to the Cayley Hamilton Theorem, this polynomial will be 0.
Thus, p(A) = \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\) = 0 or p(A) = 0
Cayley Hamilton Theorem Example
Suppose a matrix is given as A = \(\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}\). The characteristic polynomial is \(\lambda ^{2}5\lambda 2\). Use the matrix A in place of the variable to get p(A) = A^{2}  5A  2I = \(\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}^{2}  5\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}  2\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}\). Now perform the required computations to get the final value as 0.
Cayley Hamilton Theorem Formula
The Cayley Hamilton theorem formula is extremely useful is performing complicated calculations with speed and accuracy. It can also be used to determine the inverse of a matrix. The formula is given as follows:
Suppose the characteristic polynomial of an n × n square matrix, A, is given as p(\(\lambda\)) = \(\lambda ^{n} + a_{n1}\lambda ^{n1}+...+a_{1}\lambda+a_{0}\)
Then p(A) = \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\) = 0
Thus, p(A) = 0.
To determine the inverse multiply this equation with A^{1}.
\(A ^{n1} + a_{n1}A ^{n2}+...+a_{1}I_{n}+a_{0}A^{1}\) = 0
A^{1} = (\(\frac{A ^{n1} + a_{n1}A ^{n2}+...+a_{1}I_{n}}{a_{0}}\))
Cayley Hamilton Theorem 2 × 2
To apply the Cayley Hamilton Theorem to a 2 × 2 square matrix, the first step is to determine the characteristic polynomial expression.
General form of the characteristic polynomial p(\(\lambda\)) = \(\lambda ^{n} + a_{n1}\lambda ^{n1}+...+a_{1}\lambda+a_{0}\)
As n = 2, thus,
p(\(\lambda\)) = \(\lambda ^{2} + a_{1}\lambda + a_{0} \)
For a 2 × 2 square matrix this polynomial is written as
p(\(\lambda\)) = \(\lambda ^{2}  S_{1}\lambda + S_{0} \)
where, \(S_{1}\) = sum of the diagonal elements and \(S_{0}\) = determinant of the 2 × 2 square matrix.
Now according to the Cayley Hamilton theorem, if \(\lambda\) is substituted with a square matrix then the characteristic polynomial will be 0. The formula can be written as
\(B ^{2}  S_{1}B + S_{0}I \) = 0
here, B is a 2 × 2 square matrix.
Cayley Hamilton Theorem 3 × 3
For a 3 × 3 square matrix the characteristic polynomial is given as p(\(\lambda\)) = \(\lambda ^{3}  T_{2} \lambda ^{2} + T_{1}\lambda  T_{0} \)
where, \(T_{2}\) = sum of the main diagonal elements, \(T_{1}\) = sum of the minors of the main diagonal elements, \(T_{0}\) = determinant of the 3 × 3 square matrix.
On applying the Cayley Hamilton theorem, the formula is given as
\(C ^{3}  T_{2}C^{2} + T_{1}C  T_{0}I \) = 0
here, C is a 3 × 3 square matrix.
Cayley Hamilton Theorem Example 3 × 3 Matrix
Verify the Cayley Hamilton Theorem for the matrix C = \(\begin{bmatrix} 2 & 1 & 2\\ 1& 2& 1\\ 1& 1 & 2 \end{bmatrix}\).
Solution: The characteristic polynomial is given as \(C ^{3}  T_{2}C^{2} + T_{1}C  T_{0} \).
\(T_{2}\) = 2 + 2 + 2 = 6
\(T_{1}\) = 3 + 2 + 3 = 8
\(T_{0}\) = C = 3
Now p(C) = C^{3 } 6C^{2} + 8C  3I
C^{3} = \(\begin{bmatrix} 29 & 28 & 38\\ 22& 23& 28\\ 22& 22 & 29 \end{bmatrix}\)
6C^{2} = \(\begin{bmatrix} 42 & 36 & 54\\ 30& 36& 36\\ 30& 30 & 42 \end{bmatrix}\)
8C = \(\begin{bmatrix} 16 & 8 & 16\\ 8& 16& 8\\ 8& 8 & 16 \end{bmatrix}\)
3I = \(\begin{bmatrix} 3 & 0 & 0\\ 0& 3& 0\\ 0& 0 & 3 \end{bmatrix}\)
Now substituting these values in p(C) we get \(\begin{bmatrix} 0 & 0 & 0\\ 0& 0& 0\\ 0& 0 & 0 \end{bmatrix}\). Thus, the Cayley Hamilton Theorem has been verified.
Cayley Hamilton Theorem Proof
In pure mathematics, several methods are available that can be used to prove the Cayley Hamilton Theorem. However, the easiest method is by using substitution. Let matrix A = \(\begin{bmatrix} a & b\\ c& d \end{bmatrix}\). According to the Cayley Hamilton theorem, p(A) = A^{2} − (a + d)A + (ad − bc)I = 0. The proof of this theorem is given as follows:
A^{2} = \(\begin{bmatrix} a^{2}+bc & ab+bd\\ ac+cd& bc +d^{2} \end{bmatrix}\)
(a + d)A = \(\begin{bmatrix} a(a + d) & b(a + d)\\ c(a + d)& d(a + d) \end{bmatrix}\) = \(\begin{bmatrix} a^{2} + ad & ba + bd)\\ ca + cd& da + d^{2} \end{bmatrix}\)
(ad − bc)I = \(\begin{bmatrix} adbc& 0\\ 0& adbc \end{bmatrix}\)
A^{2} − (a + d)A + (ad − bc)I = \(\begin{bmatrix} a^{2}+bc & ab+bd\\ ac+cd& bc +d^{2} \end{bmatrix}\)  \(\begin{bmatrix} a^{2} + ad & ba + bd)\\ ca + cd& da + d^{2} \end{bmatrix}\) + \(\begin{bmatrix} adbc& 0\\ 0& adbc \end{bmatrix}\)
A^{2} − (a + d)A + (ad − bc)I = \(\begin{bmatrix} 0& 0\\ 0& 0 \end{bmatrix}\)
Similarly, the Cayley Hamilton theorem proof can be given for higher order square matrices.
Cayley Hamilton Theorem Applications
The Cayley Hamilton Theorem forms an important concept that is widely used in the proofs of many theorems in pure mathematics. Some of the important applications of this theorem are listed below:
 The Cayley Hamilton Theorem is used to define vital concepts in control theory such as the controllability of linear systems.
 In commutative algebra, Nakayama's lemma can be proved using a generalization of the Cayley Hamilton Theorem.
 The Theorem of Jacobson can be proved using the Cayley Hamilton Theorem.
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Important Notes on Cayley Hamilton Theorem
 The Cayley Hamilton Theorem states that square matrices (both real and complex) will satisfy their own characteristic polynomial equation.
 Mathematically, the Cayley Hamilton theorem can be expressed as p(A) = \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\) = 0 where A is an n × n square matrix.
 The characteristic equation for a 2 × 2 matrix is given as \(B ^{2}  S_{1}B + S_{0}I \) = 0 and for a 3 × 3 matrix is written as \(C ^{3}  T_{2}C^{2} + T_{1}C  T_{0}I \) = 0.
Examples on Cayley Hamilton Theorem

Example 1: Using the Cayley Hamilton Theorem find A^{1} if A = \(\begin{bmatrix} 1 & 4\\ 2& 3 \end{bmatrix}\)
Solution: A = \(\begin{bmatrix} 1 & 4\\ 2& 3 \end{bmatrix}\)
A = 3  8 =  5. As A \(\neq\) 0 thus, A^{1} exists.
According to the Cayley Hamilton theorem \(B ^{2}  S_{1}B + S_{0}I \) = 0
\(S_{1}\) = 1 + 3 = 4, \(S_{0}\) = A = 5
Thus, the characteristic equation is p(A) = A^{2}  4A  5I = 0
Multiplying both sides by A^{1},
A^{1}(A^{2}  4A  5I) = 0; (as A^{1}A = I)
A  4I  5A^{1} = 0
A^{1} = (A  4I) / 5
A^{1} = \(\frac{1}{5}\)\(\left ( \begin{bmatrix} 1 & 4\\ 2 & 3 \end{bmatrix}  4\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\right )\) = \(\frac{1}{5}\)\(\left ( \begin{bmatrix} 3 & 4\\ 2 & 1 \end{bmatrix} \right )\)
Answer: A^{1} = \(\frac{1}{5}\)\(\left ( \begin{bmatrix} 3 & 4\\ 2 & 1 \end{bmatrix} \right )\) 
Example 2: Using the Cayley Hamilton Theorem find A^{4} if A = \(\begin{bmatrix} 1 & 2\\ 2& 1 \end{bmatrix}\)
Solution: According to the Cayley Hamilton Theorem \(B ^{2}  S_{1}B + S_{0}I \) = 0
\(S_{1}\) = 1  1 = 0, \(S_{0}\) = A = 1  4 = 5
Thus, the characteristic equation is p(A) = A^{2}  5I = 0
A^{2} = 5I
Multiply A^{2} on both sides
A^{4} = A^{2}(5I)
A^{4}= (5I)(5I)
A^{4} = 25I = \(\begin{bmatrix} 25 & 0\\ 0& 25 \end{bmatrix}\)
Answer: A^{4} = \(\begin{bmatrix} 25 & 0\\ 0& 25 \end{bmatrix}\) 
Example 3: Verify the Cayley Hamilton Theorem for A = \(\begin{bmatrix} 1 & 3 & 7\\ 4 & 2 &3 \\ 1& 2 & 1 \end{bmatrix}\)
Solution: The characteristic equation is given as \(C ^{3}  T_{2}C^{2} + T_{1}C  T_{0}I \)
\(T_{2}\) = 1 + 2 + 1 = 4, \(T_{1}\) =  4  6 10 =  20, \(T_{0}\) = A = 35
Thus, the characteristic expression is p(A) = A^{3}  4A^{2} 20A  35I
A^{3} = \(\begin{bmatrix} 135 & 152 & 232\\ 140 & 163 &208 \\ 60& 76 & 111 \end{bmatrix}\)
4A^{2} = 4 \(\begin{bmatrix} 20 & 23 & 23\\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{bmatrix}\) = \(\begin{bmatrix} 80 & 92 & 92\\ 60 & 88 & 148 \\ 40 & 36 & 56 \end{bmatrix}\)
20A = 20 \(\begin{bmatrix} 1 & 3 & 7\\ 4 & 2 &3 \\ 1& 2 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 20 &60 &140 \\ 80 & 40 & 60\\ 20 & 40 & 20 \end{bmatrix}\)
35I = \(\begin{bmatrix} 35 & 0 &0 \\ 0 & 35 & 0\\ 0 & 0 & 35 \end{bmatrix}\)
Substituting these values in the characteristic expression,
\(\begin{bmatrix} 135 & 152 & 232\\ 140 & 163 &208 \\ 60& 76 & 111 \end{bmatrix}\)  \(\begin{bmatrix} 80 & 92 & 92\\ 60 & 88 & 148 \\ 40 & 36 & 56 \end{bmatrix}\)  \(\begin{bmatrix} 20 &60 &140 \\ 80 & 40 & 60\\ 20 & 40 & 20 \end{bmatrix}\)  \(\begin{bmatrix} 35 & 0 &0 \\ 0 & 35 & 0\\ 0 & 0 & 35 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 0 &0 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\)
Thus, p(A) = A^{3}  4A^{2} 20A  35I = 0
Hence, the Cayley Hamilton theorem is proved.
FAQs on Cayley Hamilton Theorem
What is the Cayley Hamilton Theorem?
The Cayley Hamilton Theorem states that a square matrix will satisfy its own characteristic equation. It is represented as p(A) = \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\) = 0 where A is an n × n square matrix.
What is Cayley Hamilton Theorem Example?
Suppose there is a matrix given by A = \(\begin{bmatrix} 1 & 1\\ 1& 3 \end{bmatrix}\). Then according to the Cayley Hamilton theorem the characteristic equation of this matrix, p(A) = A^{2}  4A + 2, will be equal to 0.
Who Gave the Cayley Hamilton Theorem?
The Cayley Hamilton Theorem was given by two mathematicians Arthur Cayley and William Rowan Hamilton in 1858.
What is the Cayley Hamilton Theorem Equation?
The characteristic polynomial expression of a square matrix, A, given by \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\), will be equal to the zero matrix. Thus, the Cayley Hamilton theorem equation is p(A) = \(A ^{n} + a_{n1}A ^{n1}+...+a_{1}A+a_{0}I_{n}\) = 0.
How to Use Cayley Hamilton Theorem?
The Cayley Hamilton Theorem can be used to find the inverse of square matrices as it reduces calculations. Furthermore, it can be used to calculate the values of matrices that are raised to a large exponent.
How to Prove the Cayley Hamilton Theorem?
There are many different techniques that can be used to prove the Cayley Hamilton theorem. However, the simplest way is by using the method of substitution.
What are the Applications of the Cayley Hamilton Theorem?
Cayley Hamilton theorem is widely used in pure mathematics to give proofs of theorems such as the Theorem of Jacobson. It is also used in the field of engineering to define certain control systems.
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