How to Find the Eigenvalues of a Matrix

Eigenvalue is a scalar quantity which is associated with a linear transformation belonging to a vector space. In this article students will learn how to determine the eigenvalues of a matrix.

The roots of the linear equation matrix system are known as eigenvalues. It is also considered equivalent to the process of matrix diagonalization.

What are Eigenvalues?

Eigenvalue is explained to be a scalar associated with a linear set of equations which when multiplied by a nonzero vector equals to the vector obtained by transformation operating on the vector.

Let us consider k x k square matrix A and v be a vector, then λ\lambda is a scalar quantity represented in the following way:

AV = λ\lambdaV

Here, λ\lambda is considered to be eigenvalue of matrix A.

Above equation can also be written as:

(A – λ\lambda I) = 0

Where, “I” is the identity matrix of the same order as A.

This equation can be represented in determinant of matrix form.

AλI \left | A – \lambda I \right | = 0

Above relation enables us to calculate eigenvalues λ\lambda easily.

Steps to Find Eigenvalues of a Matrix

In order to find eigenvalues of a matrix, following steps are to followed:

Step 1: Make sure the given matrix A is a square matrix. Also, determine the identity matrix I of the same order.

Step 2: Estimate the matrix AλIA – \lambda I, where λ\lambda is a scalar quantity.

Step 3: Find the determinant of matrix AλIA – \lambda I and equate it to zero.

Step 4: From the equation thus obtained, calculate all the possible values of λ\lambda which are the required eigenvalues of matrix A.

Decomposition of Eigenvalues

The computation of eigenvalues and eigenvectors for a square matrix is known as eigenvalue decomposition. When we process a square matrix and estimate its eigenvalue equation and by the use of it, the estimation of eigenvalues is done, this process is formally termed as eigenvalue decomposition of the matrix.

Eigenvalues so obtained are usually denoted by λ1\lambda_{1}, λ2\lambda_{2}, …. or e1,e2,e_{1}, e_{2}, ….

Properties on Eigenvalues

Let A be a matrix with eigenvalues λ1,,λn{\displaystyle \lambda _{1},…,\lambda _{n}}

The following are the properties of eigenvalues.

1. The trace of A, defined as the sum of its diagonal elements, is also the sum of all eigenvalues,

tr(A)=i=1naii=i=1nλi=λ1+λ2++λn.{\displaystyle {tr} (A)=\sum _{i=1}^{n}a_{ii}=\sum _{i=1}^{n}\lambda _{i}=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{n}.}

2. The determinant of A is the product of all its eigenvalues, det(A)=i=1nλi=λ1λ2λn.{\displaystyle \det(A)=\prod _{i=1}^{n}\lambda _{i}=\lambda _{1}\lambda _{2}\cdots \lambda _{n}.}

3. The eigenvalues of the kthk^{th} power of A; that is the eigenvalues of AkA^{k}, for any positive integer k, are λ1k,,λnk.{\displaystyle \lambda _{1}^{k},…,\lambda _{n}^{k}}..

4. Matrix A is invertible if and only if every eigenvalue is nonzero.

5. If A is invertible, then the eigenvalues of A1A^{-1} are 1λ1,,1λn{\displaystyle {\frac {1}{\lambda _{1}}},…,{\frac {1}{\lambda _{n}}}} and each eigenvalue’s geometric multiplicity coincides. The characteristic polynomial of the inverse is the reciprocal polynomial of the original, the eigenvalues share the same algebraic multiplicity.

6. If A is equal to its conjugate transpose, or equivalently if A is Hermitian, then every eigenvalue is real. The same is true of any symmetric real matrix.

7. If A is not only Hermitian but also positive-definite, positive-semidefinite, negative-definite, or negative-semidefinite, then every eigenvalue is positive, non-negative, negative, or non-positive, respectively.

8. If A is unitary, every eigenvalue has absolute value λi=1{\displaystyle |\lambda _{i}|=1}.

9. If A is a n×n{\displaystyle n\times n} matrix and {λ1,,λk}{\displaystyle \{\lambda _{1},\ldots ,\lambda _{k}\}} are its eigenvalues, then the eigenvalues of matrix I + A (where I is the identity matrix) are {λ1+1,,λk+1}{\displaystyle \{\lambda _{1}+1,\ldots ,\lambda _{k}+1\}}.

Also read

Types of matrices

Solving linear equations using matrix

Matrix operations

Adjoint and inverse of a matrix

Solved Problems on Eigenvalues

Sample problems based on eigenvalue are given below:

Example 1: Find the eigenvalues for the following matrix?

A = [2145]\begin{bmatrix} 2 & 1\\ 4 & 5 \end{bmatrix}

Solution:

Given A = [2145]\begin{bmatrix} 2 & 1\\ 4 & 5 \end{bmatrix}

A-λI = [2λ145λ]\begin{bmatrix} 2-\lambda & 1\\ 4 & 5-\lambda \end{bmatrix}

AλI\left | A-\lambda I \right | = 0

 ⇒2λ145λ=0\begin{vmatrix} 2-\lambda &1\\ 4& 5-\lambda \end{vmatrix} = 0

(2-λ)(5-λ) – 4 = 0

 ⇒ 10- 5λ – 2λ +λ2-4 = 0

⇒ λ2-7λ +6 = 0

⇒( λ-6)(λ-1) = 0

⇒λ = 6 or λ= 1

Hence the required eigenvalues are 6 and 1.

Example 2: Find the eigenvalues of a

2 [2011]\begin{bmatrix}2 & 0\\-1 & 1\end{bmatrix}.

Solution –

Let A = [2011]\begin{bmatrix}2 & 0\\-1 & 1\end{bmatrix}

Then,

AλI=[2λ011λ]A – \lambda I=\begin{bmatrix}2 – \lambda & 0\\-1 & 1- \lambda\end{bmatrix}

|A – λ\lambda I| = 0

(2 – λ\lambda) (1 – λ\lambda) – 0 = 0

(2 – λ\lambda) (1 – λ\lambda) = 0

λ\lambda = 1, 2

These are required eigenvalues.

Example 3: Calculate the eigenvalue equation and eigenvalues for the following matrix –

[100012200]\begin{bmatrix}1 & 0 & 0\\0 & -1 & 2\\2 & 0 & 0\end{bmatrix}

Solution –

Let us consider, A = [100012200]\begin{bmatrix}1 & 0 & 0\\0 & -1 & 2\\2 & 0 & 0\end{bmatrix} AλI=[1λ0001λ2200λ]A – \lambda I = \begin{bmatrix}1-\lambda & 0 & 0\\0 & -1-\lambda & 2\\2 & 0 & 0 – \lambda \end{bmatrix}

We can calculate eigenvalues from the following equation:

|A – λ\lambda I| = 0

(1 – λ\lambda) [(- 1 – λ\lambda)(- λ\lambda) – 0] – 0 + 0 = 0

λ\lambda (1 – λ\lambda) (1 + λ\lambda) = 0

Which is the required eigenvalue equation.

From this equation, we are able to estimate eigenvalues which are –

λ\lambda = 0, 1, -1.

Example 4: Find the eigenvalues for the following matrix?

A = [6345]\begin{bmatrix} -6 & 3\\ 4 & 5 \end{bmatrix}

Solution:

Given A = [6345]\begin{bmatrix} -6 & 3\\ 4 & 5 \end{bmatrix}

A-λI = [6λ345λ]\begin{bmatrix} -6-\lambda & 3\\ 4 & 5-\lambda \end{bmatrix}

AλI\left | A-\lambda I \right | = 0

 6λ345λ=0\begin{vmatrix} -6-\lambda &3\\ 4& 5-\lambda \end{vmatrix} = 0

(-6-λ)(5-λ) – 3×4 = 0

 ⇒ -30- 5λ+6λ +λ2-12 = 0

⇒ λ2+λ -42 = 0

⇒( λ-6)(λ+7) = 0

⇒λ = 6 or λ= -7

Hence the required eigenvalues are 6 and -7.

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