Question

1: 1 131 2 Given matrix A 2 2 2. matrix P and I S set 2. a) Show that matrix P diaqonalizes A and find D(diagonal matnx) that matches. 6) Find the eigen values of A Observe that the columns of P form set S c) orthogonal Set using the inner product standard show that set S is not an Use the Gram- Schmidt process to get an orthonormal set from S using inner product standard

Answer 1

b). The eigenvalues of A are solutions to its characteristic equation det(A- λI₃)= 0 or, λ³-3λ²-9ʎ-5 = 0 or, (λ+1²)(λ-5)= 0. Hence, the eigenvalues of A are λ₁, λ₂ = -1 and λ₃ = 5.

a). Further, the eigenvectors of A corresponding to the eigenvalue -1 are solutions to the equation (A+I₃)X= 0. To solve this equation, we have to reduce A+I₃ to its RREF which is

1	1	1
0	0	0
0	0	0

Now, if X = (x,y,z)^T, then the equation (A+I₃)X= 0 is equivalent to x+y+z = 0 or, x = -y-z. Then, X = (-y-z,y,z)^T = y(-1,1,0)^T+z(-1,0,1)^T.This implies that every solution to the equation (A+I₃)X= 0 is a linear combination of 2 linearly independent vectors (-1,1,0)^T,(-1,0,1)^T. Hence, the eigenvectors of A corresponding to the eigenvalue -1 are v₁ = (-1,1,0)^T and v₁ =           (-1,0,1)^T.

Similarly, the eigenvector of A corresponding to the eigenvalue 5 is solution to the equation (A-5I₃)X= 0. To solve this equation, we have to reduce A-5I₃ to its RREF which is

1	0	-1
0	1	-1
0	0	0

Now, if X = (x,y,z)^T, then the equation (A-5I₃)X= 0 is equivalent to x-z = 0 or, x = z and y -z = 0 or, y = z. Then, X = (z,z,z)^T = z(1,1,1)^T. This implies that every solution to the the equation (A-5I₃)X= 0 is a scalar multiple of the vector (1,1,1)^T. Hence, the eigenvector of A corresponding to the eigenvalue 5 is v₃ = (1,1,1)^T.

Now, let P = [v₁,v₂,v₃] =

-1	-1	1
1	0	1
0	1	1

and D = diag[λ₁ , λ₂ , λ₃] =

-1	0	0
0	-1	0
0	0	5

Then A = PDP^-1. T^he matrix P diagonalizes A.

( c). We have v₁.v₂ = (-1,1,0)^T.(-1,0,1)^T = 1 ≠ 0, v₁.v₃ = (-1,1,0)^T.(1,1,1)^T =-1+1= 0 and v₂.v₃ = (-1,0,1)^T.(1,1,1)^T =-1+1 = 0. Hence v₁ , v₂ are not orthogonal so that S is not an orthogonal set.

(d). Let u₁ = v₁= (-1,1,0)^T, u₂ = v₂- proj_u1(v₂) = v₂-[(v₂.u₁)/(u₁.u₁)]u₁ = v₂-[(1+0+0)/(1+1+0)] u₁ = (-1,0,1)^T –(1/2) (-1,1,0)^T = (-1/2,-1/2,1)^T and u₂ = v₃- proj_u1(v₃)- proj_u2(v₃) = (1,1,1)^T.

Further, let e₁ = u₁/||u₁|] = (-1/√2,1√2,0)^T, e₂ = u₂/||u₂|] = (-1/√6,-1√6,√2/√3)^T and e₃ = u₃/||u₃|] =(1/√3,1/√3,1/√3)^T.

Then {e₁,e₂,e₃} is the required orthonormal set.