Linear Algebra Solutions 66

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

This preview shows page 1. Sign up to view the full content.

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 defined 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 fact that P is non–singular is easily proved by showing the columns of
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern