Characteristic Polynomial

The characteristic polynomial of a matrix is a polynomial derived from a square matrix that helps determine the matrix's eigenvalues.

For a square matrix A of size n×n, the characteristic polynomial is defined as:

f(𝛌) = |A-𝛌I|
or
f(𝛌) = det(A-𝛌I)

Where:

• A is a matrix whose characteristic equation needs to be found.
• 𝛌 represents roots of the characteristic equation.
• I represents the identity matrix having the same order as A.
• Dot represents the determinant of a matrix.

Key Points

• The roots of the equation det( A - 𝛌I) = 0 give the eigenvalues of the matrix.
• The degree of the equation is n, signifying that it has n roots, which correspond to eigenvalues of the matrix.
• This concept plays a significant role in determining the eigenvalues of a matrix, which are essential in various applications across mathematics and engineering.

Let us look at examples of characteristic polynomials of 2×2 and 3×3 matrices for better understanding.

Characteristic Polynomial for 2×2 Matrix

As we know, the characteristic polynomial is given by the function f(𝛌) = |A-𝛌I|. Let us take matrix A as follows, for instance, as follows,

A = \begin{bmatrix}2 & 3\\1 & 4\end{bmatrix}

Identity matrix of order 2×2 is given by,

I = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Substituting these values in the expression for the characteristic polynomial, we get,

f(\lambda) = \begin{vmatrix}2-\lambda & 3\\1 & 4-\lambda\end{vmatrix}

⇒ f(𝛌) = (2-𝛌)(4-𝛌) - 3

⇒ f(𝛌) = 8 - 2𝛌 - 4𝛌 + 𝛌² - 3

⇒ f(𝛌) = 𝛌² - 6𝛌 + 5

Thus, the above equation represents the characteristic polynomial of the matrix A taken as an example.

Characteristic Polynomial for 3×3 Matrix

Let us take an example of a 3×3 matrix as follows,

A = \begin{bmatrix}1 & 2 & 3\\5 & 2 & 4\\2 & 3 & 3\\\end{bmatrix}

Identity matrix of order 3×3 is given by,

I = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\\end{bmatrix}

Substituting these values in the expression for the characteristic polynomial, we get,

f(\lambda) = \begin{vmatrix}1-\lambda & 2 & 3\\5 & 2-\lambda & 4\\2 & 3 & 3 - \lambda\end{vmatrix}

⇒ f(\lambda) = (1-\lambda)\begin{vmatrix}2-\lambda & 4\\3 & 4-\lambda\end{vmatrix} - 2\begin{vmatrix}5 & 4\\2 & 3-\lambda\end{vmatrix}+3\begin{vmatrix}5 & 2-\lambda\\2 & 3\end{vmatrix}

⇒ f(𝛌) = (1 - 𝛌){(2 - 𝛌)(3 - 𝛌) - 12} - 2{5(3 - 𝛌) - 8} + 3{15 - 2(2 - 𝛌)}

⇒ f(𝛌) = (1 - 𝛌)(𝛌² - 5𝛌 + 6 - 12) - 2(15 - 5𝛌 - 8) + 3(15 - 4 + 2𝛌)

⇒ f(𝛌) = (1 - 𝛌)(𝛌² - 5𝛌 - 6) - 2(7 - 5𝛌) + 3(11 + 2𝛌)

⇒ f(𝛌) = 𝛌² - 5𝛌 - 6 - 𝛌³ + 5𝛌² + 6𝛌 - 14 + 10𝛌 + 33 + 6𝛌

⇒ f(𝛌) = -𝛌³ + 6𝛌² +17𝛌 + 13

Thus, the above equation represents the characteristic polynomial of the matrix A taken as an example. It is observed from the above examples that the characteristic polynomial has the same degree as the order of the given matrix, i.e., a 2×2 matrix would have a characteristic polynomial of degree 2, and that of a 3×3 matrix would have a degree 3.

Roots of Characteristic Equation

Roots of the characteristic equation give eigenvalues of the matrix. Considering the characteristic polynomial obtained for the 2×2 example matrix above, we get the following characteristic equation,

⇒ 𝛌² - 6𝛌 + 5 = 0

Finding the roots of the above equation by the quadratic formula, we get,

⇒ 𝛌 = 1, 5

Related Articles

• Application of Eigenvalues
• Eigenvalues and Eigenvectors

Examples of Characteristic Polynomial

Example 1: Find the characteristic polynomial for the matrix given as A = \begin{bmatrix}1 & 4\\5 & 2\end{bmatrix}.

Solution:

We know that,

Characteristic polynomial for a matrix is given as, p(𝛌) = |A - 𝛌I|

p(\lambda) = \begin{vmatrix}1-\lambda & 4\\5 & 2 -\lambda\end{vmatrix}

⇒ p(𝛌) = (1 - 𝛌)(2 - 𝛌) - 20
⇒ p(𝛌) = 𝛌² - 3𝛌 + 2 - 20
⇒ p(𝛌) = 𝛌² - 3𝛌 - 18

Thus, characteristic polynomial for the given matrix has been obtained as above.

Example 2: Find the characteristic equation for the given matrix as A = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}.

Solution:

We know that, characteristic equation is given as, |A - 𝛌I| = 0. Thus, we get,

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

⇒ (1-𝛌)(2-𝛌)(3-𝛌) = 0
⇒ -𝛌³ + 6𝛌² - 11𝛌 + 6 = 0

Hence, we have obtained the characteristic equation for the given matrix.

Practice Problems on Characteristic Polynomial

Question 1: Find the characteristic polynomial for the given matrixA = \begin{bmatrix}5 & 7\\3 & 2\end{bmatrix}.

Question 2: Write the characteristic equation for the matrixA = \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\\\end{bmatrix}.

Question 3: Find the eigenvalues of the matrix whose characteristic polynomial is given by f(𝛌) = 𝛌³ + 6𝛌² + 18𝛌 + 12.

Question 4: What is the determinant of the matrix whose characteristic polynomial is given by f(𝛌) = 𝛌² - 5𝛌 + 6?

Question 5: Find the characteristic polynomial for the matrix given byA = \begin{bmatrix}1 & 2\\3 & 4\\\end{bmatrix}.