Please answer part (b) in great detail, mostly, how to calculate the 3rd vector
ID: 2987173 • Letter: P
Question
Please answer part (b) in great detail, mostly, how to calculate the 3rd vector for the doubly degenerate value of lambda = 2 with a corresponding eigenvector of [1 0 -1]^t
Explanation / Answer
eigen values for X1 and X2 are 1, 2 respectively.
let ? be an eigen value of A.Then by definition,
A X = ?X, where X is an eigen vector.
=> (A-?I)X = O.
=> |A-? I| = 0
=> -?^3+5 ?^2-8 ?+4 = 0.
=> (?-1)(?-2)^2 = 0
=> 1 and 2 are the only eigen values of A. so we cant find an eigen vector which is independent of X1 and X2.