How to determine Eigenvalues of a Matrix?

Explanation:

Eigenvalue is defined as a scalar associated with a given linear transformation of a vector space and having the property that there is some non-zero vector which when multiplied by the scalar is equal to the vector obtained by letting the transformation operate on the vector. The roots of the linear equation matrix system are also called eigenvalues. Consider a square matrix A of k x k, v is vector and λ is the scalar quantity, which can be represented as,

Av = λv

Av – λv = 0 ⇢ [λ – Eigen value]

A – λI  = 0 ⇢ [I – Identity matrix]

|A – λI| = 0

Consider a square matrix A with eigenvalues λ₁, λ₂ … λ_n

• The determination of A is a product of all its eigenvalue. [det(A) = λ₁ × λ₂….λ_n]
• Matrix A is invertible if and only if every eigenvalue is non-zero.
• Eigenvalue of real symmetric and Hermitian matrices are equal.
• Eigenvalue of real skew-symmetric and skew Hermitian matrices are either pure or zero.
• Eigenvalue of unitary and orthogonal matrices are of unit modulus |λ| = 1.
• Eigen value of A^-1= 1/λ₁,1/λ₂,… 1/λ_n.
• Eigen value of A^k = λ^k₁, λ^k₂, …. λ^k_n
• If A and B are two matrices of the same order then the eigenvalue of AB = Eigenvalue of BA.
• If a square matrix A is a lower/upper triangular matrix, then its eigenvalue will be the diagonal elements of the matrix.

Below are the steps that are to be followed to find the value of a matrix,

Step 1: Check whether the given matrix is a square matrix or not. If “yes” then, follow step 2.

Step 2: Determine identity matrix(I)

Step 3: Estimate the matrix A – λI.

Step 4: Find the determinant of A – λI.

Step 5: Equate the determinant of A-λI to zero. {|A – λI| = 0}

Step 6: Calculate all the possible values of λ.

Related Articles:

• Eigenvalues & Eigenvectors: Definition, Formula, Examples
• Eigenspace and Eigenspectrum Values in a Matrix

Question 1: Find the eigen value of matrix [Tex]A= \begin{bmatrix}   1 & 4 \\ 3 & 2  \\ \end{bmatrix}[/Tex].

Solution:

[Tex]A-\lambda I= \begin{bmatrix}   1-\lambda & 4 \\ 3 & 2-\lambda  \\ \end{bmatrix}[/Tex]

|A – λI|= 0

(1 – λ)(2 – λ) – 12 = 0

2 – λ – 2λ + λ² – 12 = 0

λ² – 3λ – 10 = 0

λ² – 5λ + 2λ – 10 = 0

(λ + 2)(λ – 5) = 0

λ = -2, 5

Therefore, eigen value will be (-2, 5)

Question 2: Find the eigen value of matrix [Tex]A= \begin{bmatrix}   1 & 0 & 0 \\ 0 & 1 & 2  \\ 0 & 0 & 0 \end{bmatrix}[/Tex]

Solution:

[Tex]A-\lambda I= \begin{bmatrix}   1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 2  \\ 0 & 0 & 0-\lambda   \\ \end{bmatrix}[/Tex]

|A – λI| = 0

(1 – λ)[(1 – λ)(0 – λ) – 2] = 0

(1 – λ)(λ² – λ – 2) = 0

-λ³ + 2λ + λ – 2 = 0

λ = 1, 0

Therefore, the eigen value will be 1, 0.

Question 3: Find the eigen value of matrix [Tex]A= \begin{bmatrix}   4 & 1  \\ 1 & 4  \\ \end{bmatrix}[/Tex]

Solution:

[Tex]A-\lambda I = \begin{bmatrix}   4-\lambda & 1  \\ 1 & 4-\lambda   \\ \end{bmatrix}[/Tex]

[(4 – λ)(4 – λ)] – 1 = 0

16 – 4λ – 4λ + λ² – 1 = 0

λ² – 8λ + 15 = 0

λ² – 3λ – 5λ + 15 = 0

λ(λ – 3) – 5(λ – 3) = 0

(λ – 5)(λ – 3) = 0

λ = 5, 3

Therefore, the eigenvalue will be 5, 3

Question 4: Find the eigen value of the given matrix [Tex]A= \begin{bmatrix}   1 & 4 & 3  \\ 0 & 3 & 8 \\   0 & 0 & 2 \end{bmatrix}[/Tex]

Solution:

As mentioned above in the properties of eigen value i.e If a square matrix A is lower/upper triangular matrix, then its eigenvalue will be the diagonal elements of the matrix.

As the given matrix A is a lower triangular matrix so, its eigenvalue will be 1, 3, 2.

Question 5: Find the eigen value of the matrix [Tex]A= \begin{bmatrix}   2 & 2  \\ 5 & -1 \\   \end{bmatrix}[/Tex]

Solution:

[Tex]A-\lambda I= \begin{bmatrix}   2-\lambda & 2  \\ 5 & -1-\lambda\\   \end{bmatrix}[/Tex]

[(2 – λ)(-1 – λ)] – 10 = 0

-2 – 2λ + λ + λ² – 10 = 0

λ² – λ – 12 = 0

λ² – 4λ + 3λ – 12 = 0

λ(λ – 4) + 3(λ – 4) = 0

(λ – 4)(λ + 3) = 0

λ = 4, -3

Therefore, the eigenvalue will be 4, -3

Question 6: Find the eigenvalue of matrix [Tex]A= \begin{bmatrix}   -1& 8  \\  0 & -1\\   \end{bmatrix}[/Tex]

Solution:

[Tex]A-\lambda I= \begin{bmatrix}   2-\lambda & 2  \\ 5 & -1-\lambda\\   \end{bmatrix}[/Tex]

|A – λI| = 0

(-1 – λ)² – 0 = 0

(λ + 1)² = 0

λ = -1

Therefore, the eigenvalue will be -1