Question

(8 points) [102] The matrix A= 0 3 0 (205 has a single real eigenvalue = 3 with algebraic multiplicity three (a) Find a basis for the associated eigenspace. Basis = { (b) is the matrix A defective? A. A is not defective because the eigenvectors are linearly independent O B. A is defective because the geometric multiplicity of the eigenvalue is less than the algebraic multiplicity c. A is defective because it has only one eigenvalue D. A is not defective because the eigenvalue has algebraic multiplicity three

Answer 1

a)- Giren with 2 os 3 iu Single Yew eigenvalue of wish - Characteristics poly of A D - CACW) = (2-3) 3 - .. Algebruic muliplicity of 3 = Now Considio A-31 - 00 By apply rou operation we reduce it A-BIN R, (-) Rilioi Rza - 13 oool R2 - (t) R2 lol ~ R2 - Rz-R1 0 1 . rank of it = 1 and dim of eigen space of eigen value 30 is given by - (no. of variable) - rank orot. - 3-1-2. :: Algebra multiplicity 0+3 12. and we find out Busis of eigenspure associated with eigen-setre 3 is hy putting 24+31320 g uz is also tree variable = 2y=-213 ne is free variable. Put n3=1 =) pu=-1)

eigen Basis value 3 of eigenspace associated with is given by B = {2,,*2} closely = = { 2,=(-1,0,1), Te-.(0,1,0)) G.m & A.m. since 3 +2. A is not diagonilizable b) me lsnow that matrix A is detective it it is not diagonalizahle. or Matrix A is defective is am G.m. Here also a m=2<G.m=3. : A is detective because Amc Gm it G.me A.m. Matrix Here also - A is so option A is defective G.m=2<A.m=3 . defective because B is correcta G.m A.m.