Test: Eigenvalues & Eigenvectors - 1 - Mechanical Engineering

According to the property of the Eigenvalues, the sum of the Eigenvalues of a matrix is its trace that is the sum of the elements of the principal diagonal. 
Therefore, the sum of the Eigenvalues = 3 + 4 + 1 = 8.

A - λI = 0

Now, After taking the determinant:

(4 - λ)² - 1 = 0

16 + λ² - 8λ - 1 = 0

λ² - 8λ + 15 = 0

(λ - 3) (λ - 5) = 0

λ = 3, 5

One eigen value is zero

Let the matrix be A.  We know, Trace (A) = sum of eigen values. 

Let the given matrix be A.  A is singular. 

We know If λ be the eigen value of A ⇒λ² is an eigen value of A² .

For S matrix, if eigen values are λ₁, λ₂, λ₃,... then for S² matrix, the eigen values will be λ²₁ λ²₂ λ²₃......
For S matrix, if eigen values are 1 and 5 then for S² matrix, the eigen values are 1 and 25

Number of linear independent vectors is equal to the sum of Geometric Multiplicity of eigen values. Here only eigen value is 2.

To find Geometric multiplicity find n-r of (matrix-2I), where n is order and r is rank.

Rank of obtained matrix is 1 and n = 2 so n-r = 1. Therefore the no of linearly independent eigen vectors is 1

The eigen vectors of A are given by  AX= λ X  

So we can check by multiplication.

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix.

The determinant of this matrix equated to zero

i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Characteristic equation will be :λ² -4 =0 thus root of characteristic equation will be +2 and - 2.

Concept: 

(i) A complex matrix A is said to be self-adjoint if A = A* where A* = A^-T

(ii) Eigenvalues of a self-adjoint matrix are real

Explanation:

A is an n × n complex matrix and A is self-adjoint so

A = A*

B denotes the inverse of (A + iIn).

So B = (A + iIn)^-1

Let λ be an eigenvalue of A

then λ + i is an eigenvalue of A + iI_n

and (λ + i)^-1 is an eigenvalue of B

Also λ - i is an eigenvalue of A - iI_n

Hence eigenvalue of (A - iIn)B

= (λ - i)(λ + i)-1 

Rank of a matrix is equal to the No. of linearly independent row or no. of linearly  independent column vector. 

Eigen Value (λ ) are 1,− 2.

 be the eigen  of A. Corresponding 

To λ then. 

be the eigen vector corrosponding to λ = 1