Question

4. Let E) 6 3 0 [8 Marks] 3 6 0 A = 0 0 11 a) Find the eigenvalues of A b) For each eigenvalue of A, find a basis for the corresponding eigenspace. c) Consider the linear transformation T : R3 -> R3 defined by T(x) = Ax for every xE R3. Find a basis of R3 in which the matrix representing T is diagonal

Answer 1

4. a). The eigenvalues of A are solutions to its characteristic equation det(A- λI₃)= 0 or, λ³-23λ²+159ʎ-297 = 0 or, (ʎ-3)(λ-9)(λ-11)= 0. Hence, the eigenvalues of A are λ₁ =3 , λ₂ = 9 and λ₃ = 11.

b). Further, the eigenvector of A corresponding to the eigenvalue 3 is solution to the equation (A-3I₃)X= 0. To solve this equation, we have to reduce A-3I₃ to its RREF which is

1	1	0
0	0	1
0	0	0

Now, if X = (x,y,z)^T, then the equation (A-3I₃)X = 0 is equivalent to x+y = 0 or, x = -y and z = 0. Then, X = (-y,y,0)^T = y,,. This implies that every solution to the equation (A-3I₃)X = 0 is a scalar multiple of the vector (-1,1,0)^T. Hence, the eigenvector of A corresponding to the eigenvalue 3 is v₁ = (-1,1,0)^T. The set {(-1,1,0)^T } is a basis for the eigenspace E₃ of A corresponding to the eigenvalue 3.

Similarly, the eigenvector of A corresponding to the eigenvalue 9 is solution to the equation (A-9I₃)X= 0. To solve this equation, we have to reduce A-9I₃ to its RREF which is

1	-1	0
0	0	1
0	0	0

Now, if X = (x,y,z)^T, then the equation (A-9I₃)X= 0 is equivalent to x-y = 0 or, x = y and z= 0 . Then, X =(y,y,0)^T = y(1,1,0)^T. This implies that every solution to the the equation (A-9I₃)X= 0 is a scalar multiple of the vector (1,1,0)^T. Hence, the eigenvector of A corresponding to the eigenvalue 9 is v₂ = (1,1,0)^T. The set {(1,1,0)^T } is a basis for the eigenspace E₉ of A corresponding to the eigenvalue 9.

Also, the eigenvector of A corresponding to the eigenvalue 11 is solution to the equation (A-11I₃)X= 0. To solve this equation, we have to reduce A-11I₃ to its RREF which is

1	0	0
0	1	0
0	0	0

Now, if X = (x,y,z)^T, then the equation (A-11I₃)X= 0 is equivalent to x = 0 and y = 0. Then, X = (0,0,z)^T = z(0,0,1)^T. This implies that every solution to the the equation (A-11I₃)X= 0 is a scalar multiple of the vector (0,0,1)^T. Hence, the eigenvector of A corresponding to the eigenvalue 11 is v₃ = (0,0,1)^T. The set {(0,0,1)^T } is a basis for the eigenspace E₁₁ of A corresponding to the eigenvalue 11.

Now, let B = {v₁,v₂,v₃} = {(-1,1,0)^T , (1,1,0)^T , (0,0,1)^T }. Then the matrix representing T is D =

diag[λ₁ , λ₂ , λ₃] =

3	0	0
0	9	0
0	0	11

which is a diagonal matrix.