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Unformatted text preview: ( ab b a ) . 2 Solution: The characteristic polynomial det ( 3λ 21 5λ ) = λ 28 λ + 17 has the roots λ = 4 ± i , and ( 3(4 + i ) 21 1(4 + i ) ) ⃗v = ⃗ has the solution ⃗v = ( 2 1 + i ) . [It actually has the solution ⃗v = ( 2 z (1 + i ) z ) , but z = 1 is the simplest choice.] Writing ⃗v = ⃗u + i⃗w where ⃗u and ⃗w have real components, we have T ( ⃗u ) + iT ( ⃗w ) = T ( ⃗v ) = (4 + i ) ⃗v = (4 + i )( ⃗u + i⃗w ) = 4 ⃗u⃗w + i (4 ⃗w + ⃗v ) . Equating real and imaginary parts on the far left and far right in that string of equalities, the matrix of T with respect to the basis B = { ⃗u = ( 2 1 ) , ⃗w = ( 1 ) } is ( 4 11 4 ) ....
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This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.
 Fall '08
 lee
 Eigenvectors, Vectors

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