One of the Eigen vectors of the matrix A = 2 1 1 3 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ is [ME:...
Solution:
Given:
Matrix A = 2 1
1 3
To find:
One of the eigen vectors of matrix A.
Explanation:
To find the eigen vector of a matrix, we need to solve the equation (A - λI)v = 0, where A is the given matrix, λ is the eigen value, I is the identity matrix, and v is the eigen vector.
Let's assume the eigen vector v = [x, y], where x and y are constants.
Step 1:
Calculate the determinant of the matrix (A - λI).
(A - λI) = (2 - λ) 1
1 (3 - λ)
Determinant of (A - λI) = (2 - λ)(3 - λ) - 1*1
= λ^2 - 5λ + 5
Step 2:
Set the determinant to zero and solve for λ.
λ^2 - 5λ + 5 = 0
Using the quadratic formula, we get:
λ = (5 ± √5i)/2
Since the eigen values can be complex, we will consider both the values of λ.
For λ = (5 + √5i)/2:
Substitute this value of λ in the equation (A - λI)v = 0 and solve for v:
(2 - (5 + √5i)/2) * x + y = 0
(3 - (5 + √5i)/2) * y + x = 0
Simplifying the above equations, we get:
(4 - 5 - √5i) * x + 2y = 0
(6 - 5 - √5i) * y + x = 0
(-1 - √5i) * x + 2y = 0
(1 - √5i) * y + x = 0
We can solve these equations to find the values of x and y. However, since the question asks for one of the eigen vectors, we can choose any non-zero value for x and solve for y using the above equations.
Let's assume x = 1:
Substituting x = 1 in the above equations, we get:
(-1 - √5i) + 2y = 0
(1 - √5i) * y + 1 = 0
Solving these equations, we get:
y = (1 + √5i)/2
Therefore, one of the eigen vectors corresponding to λ = (5 + √5i)/2 is [1, (1 + √5i)/2].
For λ = (5 - √5i)/2:
Similarly, substituting this value of λ in the equation (A - λI)v = 0, we get:
(-1 + √5i) * x + 2y = 0
(1 + √5i) * y + x = 0
Again, we can choose any non-zero value for x and solve for y using the above equations.
Let's assume x = 1: