# Cayley-Hamilton Theorem

**Definition**

Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation.

That is, if *p* ( _{} ) = det ( *A* - _{} *I* ) is the characteristic polynomial of the matrix *A*, then according to Cayley-Hamilton Theorem *p* ( *A* ) = 0

*For example,*

If *p* ( _{} ) = det ( *A* - _{} *I* ) = _{}^{2} - 15 _{} + 56 is the characteristic polynomial, then by Cayley-Hamilton Theorem *A*^{2} - 15 *A* + 56 *I* = 0

The follwing examples will make the matter more clear.

# Practice Problems

**If**

*Example 1.*then show that *A*^{3} = 7 *A* - 6 *I*

*Solution*

the characteristic polynomial of *A* is

Now, by Cayley-Hamilton Theorem

*A*^{2} - 3 *A* + 2 *I* = 0 ------> (1)

From (1): *A*^{2} = 3 *A* - 2 *I*

multiply *A* on both sides of above equation

*A*^{3} = 3 *A*^{2} - 2 *A* -------->(2)

Put *A*^{2} = 3 *A* - 2 *I* in equation (2)

(2) _{} *A*^{3} = 3 ( 3 *A* - 2 *I* ) - 2 *A*

*A*^{3} = 9 *A* - 6 *I* - 2 *A*

*A*^{3} = 7 *A* - 6 *I* *Answer*

** Example 2.**If

then show that *A*^{5} = 981 *A* - 540 *I*

*Solution*

The characteristic polynomial of *A* is

Now, by Cayley-Hamilton Theorem

*A*^{2} - 6 *A* + 3 *I* = 0 ------> (1)

From (1): *A*^{2} = 6 *A* - 3 *I*

multiply *A* on both sides of above equation

*A*^{3} = 6 *A*^{2} - 3 *A* -------->(2)

Put *A*^{2} = 6 *A* - 3 *I* in equation (2)

(2) _{} *A*^{3} = 6 ( 6 *A* - 3 *I* ) - 3 *A*

*A*^{3} = 36 *A* - 18 *I* - 3 *A*

*A*^{3} =33 *A* - 18 *I*

multiply *A*^{2} on both sides of above equation

*A*^{5} = 33 *A*^{3} - 18 *A*^{2} -------->(3)

Put *A*^{2} = 6 *A* - 3 *I* and *A*^{3} =33 *A* - 18 *I* in equation (3)

(3) _{} *A*^{5} = 33 ( 33 *A* - 18 *I* ) - 18 ( 6 *A* - 3 *I* )

*A*^{5} = 1089 *A* - 594 *I* - 108 *A* + 54 *I*

*A*^{5} = 981 *A* - 540 *I* *Answer*