Question

2 (5 points) Recalled that null space of a matrix A € Mnxn is defined as N(A) = {r € R” : Ar =0}. Now, the eigenspace of A corresponding to the eigenvalue 1 (denoted by Ex(A)) is defined as the nullspace of A-XI, that is, EX(A) = N(A – XI) = {v ER”: (A – XI)v = 0}. You should have three distinct eigenvalues in Problem 1 above. Let say there are li, 12, and 13. (i) Find the eigenspaces Ex, (A), E^2(A), and Eaz(A). (Hint: You don't have to redo any computations, everything you need can be found as part of your solution in Problem 1.) (ii) Find the dimension of the eigenspaces, that is find dim (Ex1(A)), dim (E12 (A)), and dim (Exz (A)). We can also define eigenvalues and eigenvectors for linear operator on general vector spaces. (See Page 299) Definition 1. IfT:V + V is a linear operator on a vector space V, then a nonzero vector r in V is called an eigenvector of T if T(x) is a scalar multiple of r; that is, T(x) = lx for some scalar 1. The scalar 1 is called eigenvalue of T, and x is said to be an eigenvector corresponding to l.

Answer 1

Solution: :) Enica Can be calleulated by Solving Ax= x ie (A-2,5x=0 lie. null Space of A-21I). EA 2 (A) Solme A(X) = Dz (*) =) CA-121) X0 lie. null space of (A - P2I) EDB (A) Sohne An=13& (A-93I) a so lie. null Space of (A-23I) i As Di, D2, D3 are distinct dim(tm) - dim ( 82 ) = dim (Eng) = 1

ñ - X220 ,X3=0 ñ Enz(A) = E2CA) (A-2I) X =0 - 001K ooo 0 0 1 =) -X150, X370 x2 (4) Stans Similarly Es(A) = span { (8) P. dim ( Ei (4) ) = clim (E2 143) = dim (El) - I