Characteristic Polynomial

The characteristic polynomial of a square matrix A is a polynomial function f(λ) defined as:

f(λ) = det(A – λI_n)

Here, 

• A is an n × n matrix
• I_n is the n × n identity matrix
• λ is a scalar variable (known as eigenvalues)
• det denotes the determinant

The characteristic polynomial is primarily used to find the eigenvalues of a matrix as its roots correspond directly to the eigenvalues of the given matrix.

For a 2×2 Matrix

Let us find the characteristic polynomial of the matrix ${A=\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}}$

Using the formula, we have

f(λ) = det(A – λI₂) 

Here, (A – λI₂)

= ${\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}-\lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}}$

= ${\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}-\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}}$

= ${\begin{pmatrix} 2-\lambda & 1 \\ 1 & 3-\lambda \end{pmatrix}}$

Now, det(A – λI₂)

= ${\det \begin{pmatrix} 2-\lambda & 1 \\ 1 & 3-\lambda \end{pmatrix}}$

= ${\begin{vmatrix} 2-\lambda & 1 \\ 1 & 3-\lambda \end{vmatrix}}$

= (2 – λ)(3 – λ) – (1)(1)

= 6 – 5λ + λ² – 1

= λ² – 5λ +5

Thus, the characteristic polynomial is f(λ) = λ² – 5λ +5

For a 3 × 3 Matrix

Let us find the characteristic polynomial of the matrix ${A=\begin{pmatrix} 4 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 4 \end{pmatrix}}$ 

As we know, the formula of the characteristic polynomial is f(λ) = det(A – λI₂)

Given, 

${A=\begin{pmatrix} 4 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 4 \end{pmatrix}}$

${I=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}}$

Here, (A – λI₂)

= ${\begin{pmatrix} 4 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 4 \end{pmatrix}-\lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}}$

= ${\begin{pmatrix} 4 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 4 \end{pmatrix}-\begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix}}$

= ${\begin{pmatrix} 4-\lambda & 2 & 0 \\ 2 & 4-\lambda & 2 \\ 0 & 2 & 4-\lambda \end{pmatrix}}$

Now, det(A – λI₂)

= ${\det \begin{pmatrix} 4-\lambda & 2 & 0 \\ 2 & 4-\lambda & 2 \\ 0 & 2 & 4-\lambda \end{pmatrix}}$

= ${\left( 4-\lambda \right) \begin{vmatrix} 4-\lambda & 2 \\ 2 & 4-\lambda \end{vmatrix}-2\begin{vmatrix} 2 & 2 \\ 0 & 4-\lambda \end{vmatrix}}$

= (4 – λ)[(4 – λ)(4 – λ) – (2)(2)] – 2[(2)(4 – λ) – (0)(2)]

= (4 – λ)(16 – 4λ – 4λ + λ² – 4) – 2(8 – 2λ)

= (4 – λ)(λ² – 8λ + 12) – 16 + 4λ

= 4λ² – 32λ + 48 – λ³ + 8λ² – 12λ – 16 + 4λ

= -λ³ + 12λ² – 40λ + 32

Thus, the characteristic polynomial is f(λ) = -λ³ + 12λ² – 40λ + 32

Finding Roots 

The roots of the characteristic polynomial are the eigenvalues of the matrix. To find the eigenvalues of a square matrix, we follow the theorem given below:

Statement

If A is an n × n matrix, and f(λ) = det(A – λI_n) is its characteristic polynomial, then a number λ₀ is an eigenvalue of A if and only if f(λ₀) = 0

Proof 

As we know, the matrix equation (A – λ₀I_n)x = 0 has a nontrivial solution if and only if det (A – λ₀I_n) = 0 (by the matrix theorem)

Thus, 

λ₀ is an eigenvalue of the matrix A ⇔ (A – λ₀I_n)x = 0 has a nontrivial solution

This means, 

A – λ₀I_n is not invertible

⇒ det(A – λ₀I_n) = 0

⇒ f(λ₀) = 0

Also, if f(λ₀) = 0

⇒ det(A – λ₀I_n) = 0, indicating A – λ₀I_n is not invertible

⇒ (A – λ₀I_n)x = 0 has a nontrivial solution

⇒ λ₀ is an eigenvalue of A

Here, we observe that eigenvalues are the scaling factor of the matrix, and the linear transformation of this factor is known as the eigenvector.  

Eigenvectors are the vector quantities that take the form Ax = λx, where λ is the eigenvalue. 

For a 2 × 2 Matrix

Let us find the eigenvalues and eigenvectors of the matrix ${A=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}}$

As we know, the characteristic polynomial is given by 

f(λ) = det(A – λI₂)

A – λI₂ = ${\begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix}}$

The determinant is f(λ) = ${\det \begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix}}$

= (4 – λ)(3 – λ) – (2)(1) 

= 12 – 4λ – 3λ + λ² – 2

= λ² – 7λ + 10

Solve the quadratic equation λ² – 7λ + 10 = 0 by the quadratic formula, we get

${\lambda =\dfrac{-\left( -7\right) \pm \sqrt{\left( -7\right) ^{2}-4\times 1\times 10}}{2\times 1}}$

= ${\dfrac{7\pm \sqrt{49-40}}{2}}$

= ${\dfrac{7\pm 3}{2}}$

Thus, the eigenvalues are λ₁= 5 and λ₂= 2

Now, for λ₁= 5, (A – 5I₂)x = 0

⇒ ${\begin{pmatrix} 4-5 & 1 \\ 2 & 3-5 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}=0}$

⇒ ${\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}=0}$

⇒ The system of equations are: -x₁ + x₂= 0 and 2x₁ – 2x₂ = 0

Thus, the eigenvector is: x₁ = ${\begin{pmatrix} 1 \\ 1 \end{pmatrix}}$

 Similarly, for λ₂= 2, (A – 2I₂)x = 0

⇒ ${\begin{pmatrix} 4-2 & 1 \\ 2 & 3-2 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}=0}$

⇒ ${\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}=0}$

⇒ The system of equations are: 2x₁ + x₂= 0 and 2x₁ + x₂ = 0

Thus, the eigenvector is: x₂ = ${\begin{pmatrix} 1 \\ -2 \end{pmatrix}}$

Hence, the eigenvalues are λ₁= 5 and λ₂= 2 and their corresponding eigenvectors are x₁ = ${\begin{pmatrix} 1 \\ 1 \end{pmatrix}}$ and x₂ = ${\begin{pmatrix} 1 \\ -2 \end{pmatrix}}$ respectively.

For a 3 × 3 Matrix

To solve higher-degree characteristic polynomials, such as 3 × 3 or 4 × 4 Matrices, we use the rational root theorem.

This theorem provides a way to test for potential eigenvalues manually. Once a root is found, the polynomial can be reduced using polynomial long division to find other roots.

Let us find the eigenvalues of the matrix: 

${B=\begin{pmatrix} 6 & 2 & 1 \\ -2 & 3 & 0 \\ 1 & 0 & -1 \end{pmatrix}}$

By expanding cofactors along the first row, the characteristic polynomial is

f(λ) = det(B – λI) = ${\det \begin{pmatrix} 6 & 2 & 1 \\ -2 & 3 & 0 \\ 1 & 0 & -1 \end{pmatrix}}$

Expanding along the first row, 

f(λ) = ${\left( 6-\lambda \right) \det \begin{pmatrix} 3-\lambda & 0 \\ 0 & -1-\lambda \end{pmatrix}-2\det \begin{pmatrix} -2 & 0 \\ 1 & -1-\lambda \end{pmatrix}+1\det \begin{pmatrix} -2 & 3-\lambda \\ 1 & 0 \end{pmatrix}}$

= (6 – λ)[(3 – λ)(3 – λ) – (0)(0)] – 2[(-2)(-1 – λ) – (1)(0)] + 1[(-2)(0) – (1)(3 – λ)]

= (6 – λ)(λ² – 2λ – 3) – 2(2 + 2λ) + (λ – 3)

= 6λ² – 12λ – 18 – λ³ + 2λ² + 3λ – 4 – 4λ + λ – 3

= -λ³ + 6λ² + 2λ² – 12λ + 3λ – 4λ + λ – 18 – 3

= -λ³ + 8λ² – 14λ – 25

Now, the constant term of f(λ) is -25, and the leading coefficient is -1

By the rational root theorem, the possible integer roots are ±1, ±5, and ±25

At λ = -1, f(-1) = -(-1)³ + 8(-1)² – 14(-1) – 25 = 1 + 8 + 14 – 25 = -2, not a root

At λ = 5, f(5) = (5)³ + 8(5)² – 14(5) – 25 = -125 + 200 – 70 – 25 = 0

Thus, λ = 5 is a root.

By using polynomial long division to divide f(λ) = -λ³ + 8λ² – 14λ – 25 by (λ – 5), we get

f(λ) = (λ – 5)(-λ² + 3λ + 5)

Solving the quadratic -λ² + 3λ + 5 = 0 using the quadratic formula, we get

​${\lambda =\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Here, a = -1, b = 3, and c = 5

Thus, ${\lambda =\dfrac{-3\pm \sqrt{3^{2}-4\left( -1\right) \left( 5\right) }}{2\left( -1\right) }}$

= ${\lambda =\dfrac{-3\pm \sqrt{9+20}}{-2}}$

= ${\lambda =\dfrac{-3\pm \sqrt{29}}{-2}}$

⇒ the eigenvalues are λ₁ = ${\lambda =\dfrac{3-\sqrt{29}}{2}}$ and λ₂ = ${\lambda =\dfrac{3+\sqrt{29}}{2}}$

Finding Degree and Coefficients

To find the degree and coefficient of characteristic polynomials of a square matrix, we follow the theorem given below:

Statement

If A is an n × n matrix, and f(λ) = det(A – λI_n) is its characteristic polynomial, then the degree of f(λ) is n and f(λ) takes the form of 

f(λ) = (-1)ⁿλⁿ + (-1)^{n – 1}Tr(A)λ^{n – 1} + … + det(A)

This means,

• The coefficient of λ^{n – 1} is ± Tr(A)
• The constant term is det(A)

The other coefficients are just numbers without names.

Proof

Since f(0) = det(A – 0I_n) = det(A)

Thus, the constant term is always det(A)

Let us consider a 2 × 2 matrix as A = ${\begin{pmatrix} a & b \\ c & d \end{pmatrix}}$ 

Then, f(λ) = det(A – λI₂) 

⇒ f(λ) = ${\begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix}}$

⇒ f(λ) = (a – λ)(d – λ) – bc

⇒ f(λ) = λ² – (a+d)λ + (ad – bc) 

⇒ f(λ) = λ² – Tr(A)λ + det(A)

Thus, for n = 2, the characteristic polynomial can be obtained by

f(λ) = λ² – Tr(A)λ + det(A)

⇒ f(λ) = (-1)ⁿλⁿ + (-1)^{n – 1}Tr(A)λ^{n – 1} + det(A), proved.

Relation to Cayley-Hamilton Theorem

Statement

For a n × n matrix A, it satisfies its characteristic equation. 

Mathematically, if the characteristic polynomial of A is:

f(λ) = (-1)ⁿ[λⁿ + c₁λ^{n – 1} + c₂λ^{n – 2} + … + c_{n – 1}λ + c_n], then 

f(A) = (-1)ⁿ[Aⁿ + c₁A^{n – 1} + c₂A^{n – 2} + … + c_{n – 1}A + c_nI_n] = 0

Here,

• I_n is the identity matrix
• 0 is the n × n zero matrix

Proof

To prove this, we use the adjoint and determinant properties.

The cofactor expansion of a determinant is M ⋅ adj(M) = det(M) ⋅ I, where adj(M) is the transpose of the cofactor matrix.

By substituting M = A – λI,

(A – λI) ⋅ adj(A – λI) = det(A – λI) ⋅ I …..(i)

As we know, the characteristic polynomial is f(λ) = det(A – λI) …..(ii)

Then, (A – λI) ⋅ adj(A – λI) = p(λ) ⋅ I

Now, let us expand the adjoint matrix.

Since adj(A – λI) is a matrix polynomial of degree n – 1, it can be written as:

adj(A – λI) = B₀ + B₁λ + B₂λ² + ⋯ + B_{n – 1}λ^{n – 1}, where B₀, B₁, … , B_{n – 1} are n × n matrices

Substituting adj(A – λI) into the equation, we get

(A – λI)(B₀ + B₁λ + B₂λ² + ⋯ + B_{n – 1}λ^{n – 1}) = f(λ) ⋅ I

Comparing the coefficients of powers of λ, we derive n + 1 equations for B_k, we get

• For λ_n, the coefficient on both sides is equal
• For lower powers of λ, terms like AB_k – B_{k – 1} match with the constant terms c_kI

From the above equations, substituting λ = A into f(λ), we get

f(A) = Aⁿ + c₁A^{n – 1} + c₂A^{n – 2} + … + c_{n – 1}A + c_nI_n = 0

Note: If we substitute λ = A directly into p(λ) = det(A – λI), this approach does not work, as P(A) is a matrix, not a scalar.