Linear Algebra Solutions 74

Linear Algebra Solutions 74 - C3 β†’ C 3 βˆ’ C 2 =(Ξ βˆ’ 3...

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Unformatted text preview: C3 β†’ C 3 βˆ’ C 2 = (Ξ» βˆ’ 3) Ξ» βˆ’ 5 βˆ’2 4 βˆ’2 Ξ» βˆ’ 5 βˆ’Ξ» + 7 0 1 0 Ξ»βˆ’5 4 βˆ’2 βˆ’Ξ» + 7 = βˆ’(Ξ» βˆ’ 3) {(Ξ» βˆ’ 5)(βˆ’Ξ» + 7) + 8} = βˆ’(Ξ» βˆ’ 3) = βˆ’(Ξ» βˆ’ 3)(βˆ’Ξ»2 + 5Ξ» + 7Ξ» βˆ’ 35 + 8) = βˆ’(Ξ» βˆ’ 3)(βˆ’Ξ»2 + 12Ξ» βˆ’ 27) = βˆ’(Ξ» βˆ’ 3)(βˆ’1)(Ξ» βˆ’ 3)(Ξ» βˆ’ 9) = (Ξ» βˆ’ 3)2 (Ξ» βˆ’ 9). We have to find bases for each of the eigenspaces N (A βˆ’ 9I3 ) and N (A βˆ’ 3I3 ). First we solve (A βˆ’ 3I3 )X = 0. We have 1 1 βˆ’1 2 2 βˆ’2 0 . 2 βˆ’2 β†’ 0 0 A βˆ’ 3I3 = 2 00 0 βˆ’2 βˆ’2 2 Hence the eigenspace consists of vectors X = [x, y, z ]t satisfying x = βˆ’y + z , with y and z arbitrary. Hence 1 βˆ’1 βˆ’y + z = y 1 + z 0 , y X= 1 0 z so X1 = [βˆ’1, 1, 0]t and X2 = [1, 0, 1]t form a basis for the eigenspace corresponding to the eigenvalue 3. Next we solve (A βˆ’ 9I3 )X = 0. We have βˆ’4 2 βˆ’2 101 A βˆ’ 9I3 = 2 βˆ’4 βˆ’2 β†’ 0 1 1 . βˆ’2 βˆ’2 βˆ’4 000 Hence the eigenspace consists of vectors X = [x, y, z ]t satisfying x = βˆ’z and y = βˆ’z , with z arbitrary. Hence βˆ’z βˆ’1 X = βˆ’z = z βˆ’1 z 1 and we can take X3 = [βˆ’1, βˆ’1, 1]t as a basis for the eigenspace corresponding to the eigenvalue 9. 77 ...
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