Cayley-Hamilton Theorem: Statement, Solved Problems

The Cayley-Hamilton theorem is about the characteristic equation of a square matrix. Using this theorem one can find the inverse of a matrix, the integral power of a matrix, and many more. In this post, we will study this theorem along with some applications.

Cayley-Hamilton Theorem Statement

Statement: Every square matrix satisfies its own characteristic equation.

More specifically, if A is an n×n matrix with characteristic polynomial c₀xⁿ + c₁x^n-1 + …+c_n-1x +c_n, then Cayley Hamilton theorem says that

c₀Aⁿ + c₁A^n-1 + …+c_n-1A +c_n I_n = O

where O denotes the n×n zero matrix and I_n is the n×n identity matrix.

Related Topic: Characteristic Equation of a Matrix

Solved Problems

Q1: Using the Cayley-Hamilton theorem, find the inverse of the following matrix: $A= \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix}$

Answer:

The characteristic equation of the matrix is given by det(A- xI₂) = 0. That is,

$\begin{vmatrix} 2-x & 4 \\ 1 & 3-x \end{vmatrix}=0$

⇒ (2-x)(3-x) – 4⋅1 = 0

⇒ x² -5x+2 = 0.

So by Cayley-Hamilton theorem, A satisfies its own characteristic equation. In other words, we have that

A² -5A+2I₂ = 0

⇒ A(A- 5I₂) = -2I₂

⇒ A ⋅ $\dfrac{-1}{2}$(A- 5I₂) = I₂

Hence by the definition of the inverse of a matrix, the inverse of A is given by

A^-1 = $-\dfrac{1}{2}$(A- 5I₂)

⇒ A^-1 = $-\dfrac{1}{2}$ $\begin{pmatrix} 2-5 & 4 \\ 1 & 3-5 \end{pmatrix}$

⇒ A^-1 = $-\dfrac{1}{2}$ $\begin{pmatrix} -3 & 4 \\ 1 & -2 \end{pmatrix}$.

Q2: Using Cayley-Hamilton theorem, find A50 where $A= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$

Answer:

The characteristic equation of A is given by

$\begin{vmatrix} 1-x & 1 \\ 0 & 1-x \end{vmatrix}=0$

⇒ (1-x)² = 0

⇒ x² -2x+1 = 0.

By Cayley-Hamilton theorem, A² -2A+I₂ = 0.

⇒ A² -A = A -I₂

Thus, A³ -A² = A(A² -A) = A(A -I₂), by previous equation.

⇒ A³ -A² = A² -A

⇒ A³ -A² = A -I₂.

In a similar way, one gets that A⁵⁰ -A⁴⁹ = A -I₂. Therefore, we have the following equations

A2 -A = A -I2 A3 -A2 = A -I2 … A50 -A49 = A -I2.

Adding these equations, we obtain that A⁵⁰ – A = 49A – 49I₂.

⇒ A⁵⁰ = 50A – 49I₂

⇒ A⁵⁰ = $\begin{pmatrix} 50 & 50 \\ 0 & 50 \end{pmatrix}$ – $\begin{pmatrix} 49 & 0 \\ 0 & 49 \end{pmatrix}$

⇒ A⁵⁰ = $\begin{pmatrix} 1 & 50 \\ 0 & 1 \end{pmatrix}$.

Q3: Verify Cayley-Hamilton theorem for $A= \begin{pmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{pmatrix}$

Answer:

The characteristic equation of A is given by

$\begin{vmatrix} 0-x & 0 & 1 \\ 3 & 1-x & 0 \\ -2 & 1 & 4-x \end{vmatrix} = 0$

⇒ $-x \begin{vmatrix} 1-x & 0 \\ 1 & 4-x \end{vmatrix}$ + $ \begin{vmatrix} 3 & 1-x \\ -2 & 1 \end{vmatrix}$ = 0

⇒ x³ – 5x² +6x -5 =0.

To verify Cayley-Hamilton theorem, we need to check A³ – 5A² +6A -5I =0.

Now, A² = AA = $\begin{pmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{pmatrix}$ = $\begin{pmatrix} -2 & 1 & 4 \\ 3 & 1 & 3 \\ -5 & 5 & 14 \end{pmatrix}$

A³ = A²A = $\begin{pmatrix} -2 & 1 & 4 \\ 3 & 1 & 3 \\ -5 & 5 & 14 \end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{pmatrix}$ = $\begin{pmatrix} -5 & 5 & 14 \\ -3 & 4 & 15 \\ -13 & 19 & 51 \end{pmatrix}$

Therefore,

A³ – 5A² +6A -5I

= $\begin{pmatrix} -5 & 5 & 14 \\ -3 & 4 & 15 \\ -13 & 19 & 51 \end{pmatrix}$ – 5 $\begin{pmatrix} -2 & 1 & 4 \\ 3 & 1 & 3 \\ -5 & 5 & 14 \end{pmatrix}$ + 6 $\begin{pmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{pmatrix}$ – 5 $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

= $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

= O, the zero matrix.

So the Cayley-Hamilton theorem is verified for A.