PLEASE HELP ME Let A 1 2 1 1 1 5 0 1 0 2 1 4 0 0 1 0 4 3 0

PLEASE HELP ME

Let A = (1 2 1 ?1 ?1 5)( 0 1 0 2 ?1 4)( 0 0 1 0 ?4 3)( 0 0 0 1 ?2 2)( 0 0 0 0 1 1)( 0 0 0 0 0 1)

A) Here’s just a larger version of the previous matrix where ? = 1 now has multiplicity 6. Find the dimension of its eigenspace and the defect and write out all possibilities for chain structures.

B) Use the dimension of the appropriate solution space to determine which chain structure it has. Indicate the solution space you used and its dimension.

12 1 1 -1 5 0 1 0 2 1 4 2. Let A-0 0 1 0 4 3 0 0 0 1 22 A) Here\'s just a larger version of the previous matrix where = 1 now has multiplicity 6. Find the dimension of its eigenspace and the defect and write out all possibilities for chain structures. B) Use the dimension of the appropriate solution space to determine which chain structure it has. Indicate the solution space you used and its dimension. C) Now, use the Backward method to find the longest chain and whatever method you want to find all other chains.

Solution

(a) As this is an upper triangular matrix, all the eigenvalues are 1 (as given,) with multiplicity 6.

Let [x y z u v w]\' be an eigen vector. (\' denoting the transpose). Then

A [x y z u v w]\' = [x y z u v w]\'

gives the system of equations

x+2y+z-u-v+5w=x

   y      +2u-v+4w=y

       z        -4v+3w=z

             u-2v+2w=u

                   v+w=v

                       w=w

Clearly theae imply

w=0, v=0, u=0   and 2y+z=0

So the solution (eigenspace for 1) space has dimension two , spanned by

      [0 1 -2 0 0 0]\' and [1 0 0 0 0 0]\'.

By definition, defect of an eigenvalue = algebraic multiplicity - geometric multiplicity

In this case, defect of the eigenvalue 1 = 6-2 =4