Answer:
c) To determine if A is diagonalizable, we need to check if there are enough linearly independent eigenvectors to form a basis for the vector space. In this case, we have two distinct eigenvalues, λ = 2 and λ = 3.
Explanation:
To compute the eigenvalues of matrix A, we need to find the values of λ that satisfy the equation (A - λI)x = 0, where I is the identity matrix.
The matrix A - λI is given by:
A - λI =
2 - λ 0 3
1 3 - λ 0
0 0 0
To find the eigenvalues, we set the determinant of A - λI equal to zero and solve for λ:
det(A - λI) =
(2 - λ)((3 - λ)(0) - (0)(0)) - (0)((1)(0) - (0)(3 - λ))
= (2 - λ)(3 - λ) = 0
Expanding and simplifying the expression, we have:
(2 - λ)(3 - λ) = 0
λ^2 - 5λ + 6 = 0
Factoring the quadratic equation, we get:
(λ - 2)(λ - 3) = 0
Setting each factor equal to zero gives us two eigenvalues:
λ - 2 = 0 --> λ = 2
λ - 3 = 0 --> λ = 3
a) The eigenvalues of matrix A are λ = 2 and λ = 3.
To compute the eigenvectors associated with each eigenvalue, we solve the equations (A - λI)x = 0.
For λ = 2:
(A - 2I)x =
(2 - 2) 0 3
1 3 - 2 0
0 0 0
Simplifying the matrix equation, we have:
0 0 3
1 1 0
0 0 0
From the row reduced echelon form, we can see that x1 and x2 are free variables. Solving the equation, we get:
x1 = -3x3
x2 = x3
Therefore, the eigenvector associated with λ = 2 is:
v1 = [x1, x2, x3] = [-3x3, x3, x3] = x3[-3, 1, 1]
For λ = 3:
(A - 3I)x =
(2 - 3) 0 3
1 3 - 3 0
0 0 0
Simplifying the matrix equation, we have:
-1 0 3
1 0 0
0 0 0
From the row reduced echelon form, we can see that x2 and x3 are free variables. Solving the equation, we get:
x1 = 3x3
x2 = x2
x3 = x3
Therefore, the eigenvector associated with λ = 3 is:
v2 = [x1, x2, x3] = [3x3, x2, x3] = x2[0, 1, 0] + x3[3, 0, 1]
b) A basis for the eigenspace associated with λ = 2 is {[-3, 1, 1]}.
A basis for the eigenspace associated with λ = 3 is {[0, 1, 0], [3, 0, 1]}.
c) To determine if A is diagonalizable, we need to check if there are enough linearly independent eigenvectors to form a basis for the vector