Let v1 vector 0 2 2 v2 vector 2 0 1 be eigenvectors of the

Let v_1 vector = [0 2 2], v_2 vector = [-2 0 1] be eigenvectors of the matrix A which correspond to the eigenvalues lambda_1 = - 1, lambda_2 = 1, and lambda_3 = 3, respectively, and let x vector = [6 -2 1]. Express x vector as a linear combination of v_1 vector, v_2 vector, and v_3 vector, and find Ax vector. x vector = v_1 vector + v_2 vector + v_3 vector. A x vector = []

Solution

Let B =

0

2

-2

6

2

-2

0

-2

2

0

1

1

We will reduce B to its RREF as under:

Interchange the 1st row and the 2nd row

Multiply the 1st row by ½

Add -2 times the 1st row to the 3rd row

Multiply the 2nd row by 1/2       

Add -2 times the 2nd row to the 3rd row

Multiply the 3rd row by 1/3

Add 1 times the 3rd row to the 2nd row

Add 1 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

1

0

1

0

2

0

0

1

-1

It is now apparent that x = v₁+2v₂-v₃. Now, Ax = A(v₁+2v₂-v₃) = Av₁+2Av₂-Av₃=-v₁+2v₂-3v₃ = -(0,2,2)^T+2(2,-2,0)^T-3(6,-2,1)^T=(0,-2,-2)^T+(4,-4,0)^T+(-18,6,-3)^T = (-14,0,-5)^T.

0	2	-2	6
2	-2	0	-2
2	0	1	1