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Consider the following matrix 2 0 3 A= 1 3 0 0 0 4 a) Compute the eigenvalues of A b) Compute a basis for each eigenspace associated to each distinct eigenvalue of A c) Determine if A is diagonalizable, if it is, compute a diagonalization of A

User Bilcker
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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

User Victor DeMatos
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