Linear Algebra Solutions 66

Linear Algebra Solutions 66 - ab be a real matrix with...

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6. Let A = a b c d be a real matrix with non–real eigenvalues λ = a + ib and λ = a - ib , with corresponding eigenvectors X = U + iV and X = U - iV , where U and V are real vectors. Also let P be the real matrix defned by P = [ U | V ]. Finally let a + ib = re , where r > 0 and θ is real. (a) As X is an eigenvector corresponding to the eigenvalue λ , we have AX = λX and hence A ( U + iV ) = ( a + ib )( U + iV ) AU + iAV = aU - bV + i ( bU + aV ) . Equating real and imaginary parts then gives AU = aU - bV AV = bU + aV. (b) AP = A [ U | V ] = [ AU | AV ] = [ aU - bV | bU + aV ] = [ U | V ] a b - b a = P a b - b a . Hence, as P can be shown to be non–singular, P - 1 AP = a b - b a . (The ±act that P is non–singular is easily proved by showing the columns o±
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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