Linear Algebra Solutions 74

# Linear Algebra Solutions 74 - C3 β C 3 β C 2 =(Ξ β 3...

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Unformatted text preview: C3 β C 3 β C 2 = (Ξ» β 3) Ξ» β 5 β2 4 β2 Ξ» β 5 βΞ» + 7 0 1 0 Ξ»β5 4 β2 βΞ» + 7 = β(Ξ» β 3) {(Ξ» β 5)(βΞ» + 7) + 8} = β(Ξ» β 3) = β(Ξ» β 3)(βΞ»2 + 5Ξ» + 7Ξ» β 35 + 8) = β(Ξ» β 3)(βΞ»2 + 12Ξ» β 27) = β(Ξ» β 3)(β1)(Ξ» β 3)(Ξ» β 9) = (Ξ» β 3)2 (Ξ» β 9). We have to ο¬nd bases for each of the eigenspaces N (A β 9I3 ) and N (A β 3I3 ). First we solve (A β 3I3 )X = 0. We have 1 1 β1 2 2 β2 0 . 2 β2 β 0 0 A β 3I3 = 2 00 0 β2 β2 2 Hence the eigenspace consists of vectors X = [x, y, z ]t satisfying x = βy + z , with y and z arbitrary. Hence 1 β1 βy + z = y 1 + z 0 , y X= 1 0 z so X1 = [β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 β 9I3 )X = 0. We have β4 2 β2 101 A β 9I3 = 2 β4 β2 β 0 1 1 . β2 β2 β4 000 Hence the eigenspace consists of vectors X = [x, y, z ]t satisfying x = βz and y = βz , with z arbitrary. Hence βz β1 X = βz = z β1 z 1 and we can take X3 = [β1, β1, 1]t as a basis for the eigenspace corresponding to the eigenvalue 9. 77 ...
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