# Cayley-Hamilton Theorem

In linear algebra, the Cayleyâ€“Hamilton theorem (termed after the mathematicians Arthur Cayley and William Rowan Hamilton) says that every square matrix over a commutative ring (for instance the real or complex field) satisfies its own characteristic equation. If A is a provided as nÃ—n matrix and In is the nÃ—n identity matrix, then the distinctive polynomial of A is articulated as:

p(x) = det(xInÂ – A)

Where the determinant operation is ‘det’ and for the scalar element of the base ring, the variable is taken as x. As the entries of the matrix are (linear or constant) polynomials in x, the determinant is also an n-th order monic polynomial in x.

## What isÂ Cayleyâ€“Hamilton theorem?

The Cayleyâ€“Hamilton theorem states that substituting the matrix A for x in polynomial,Â p(x) = det(xInÂ – A), results in the zero matrices, such as:

p(A) = 0

It states that a ‘n x n’ matrix A is demolished by its characteristic polynomial det(tI – A), which is monic polynomial of degree n. The powers of A, found by substitution from powers of x, are defined by recurrent matrix multiplication; the constant term of p(x) provides a multiple of the power A0, where power is described as the identity matrix.

The theorem allows An to be articulated as a linear combination of the lower matrix powers of A. If the ring is a field, the Cayleyâ€“Hamilton theorem is equal to the declaration that the smallest polynomial of a square matrix divided by its characteristic polynomial.

## Example of Cayley-Hamilton Theorem

#### 1.) 1 x 1 Matrices

For 1 x 1 matrix A(a1,1) the characteristic polynomial is given byÂ

$$\begin{array}{l}p(\lambda )=\lambda – a\end{array}$$

So, p(A) = (a) – (a1,1) Â = 0 is obvious.

2.) 2 x 2 Matrices

Let us look this through an example

$$\begin{array}{l}A = \begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}\end{array}$$

$$\begin{array}{l}p(\lambda )=det(\lambda I_{2}-A)= det\begin{pmatrix} \lambda -1 & -2\\ -3 & \lambda -4 \end{pmatrix} = (\lambda -1)(\lambda -4)-(-2)(-3)=\lambda ^{2}-5\lambda -2\end{array}$$

The Cayley-Hamilton claims that if, we define

$$\begin{array}{l}p(X) =Â X^{2}-5X-2I_{2}\end{array}$$

then,

$$\begin{array}{l}p(A) =Â A^{2}-5A-2I_{2} =\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}\end{array}$$

We can verify this result by computation

$$\begin{array}{l}A^{2}-5A-2I_{2} =Â \begin{pmatrix} 7 & 10\\ 15 & 22 \end{pmatrix}-\begin{pmatrix} 5 & 10\\ 15 & 20 \end{pmatrix}-\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}\end{array}$$

For a generic 2 x 2 matrix,

$$\begin{array}{l}A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\end{array}$$

The resultant polynomial is given by:

$$\begin{array}{l}P(\lambda )=\lambda^{2}-(a+d)\lambda +(ad-bc)\end{array}$$

So the Cayley-Hamilton theorem states that

$$\begin{array}{l}p(A)=A^{2}-(a+d)A+(ad-bc)I_{2}=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}\end{array}$$

it is always the case, which is evident by working out onÂ  A2.

## Frequently Asked Questions â€“ FAQs

Q1

### What is Cayley Hamilton theorem?

As per Cayley-Hamilton theorem, all the square matrix satisfies its own characteristic equation, over a commutative ring of real or complex field.
Q2

### What is the formula for Cayley-Hamilton Theorem?

If A is a provided as nÃ—n matrix and I is the nÃ—n identity matrix and x is any variable, then the distinctive polynomial of A is articulated as: p(x) = |xI â€“ A|
Q3

### Is this theorem applicable for all types of matrices?

The theorem is only applicable for Square matrix.