Q7rs - ( a-b b a ) . 2 Solution: The characteristic...

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Quiz VII 1. The matrix A = 2 0 1 0 1 0 1 0 2 is symmetric. Find a basis for R 3 consisting of eigenvectors for A . Solution: Since A is symmetric, there will be an orthonormal basis consisting of eigenvectors of A , but you do not need to use that to solve the problem. First find the eigenvalues det 2 - λ 0 1 0 1 - λ 0 1 0 2 - λ = (2 - λ ) 2 (1 - λ ) - (1 - λ ) = (1 - λ )(3 - 4 λ + λ 2 ) . So the eigenvalues are 3 and 1. Solving for the eigenspaces, ⃗v 1 = 1 0 1 is an eigenvector for the eigenvalue 3. For the eigenvalue 1, A - I = 1 0 1 0 0 0 1 0 1 , and the reduced row echelon form of that is just 1 0 1 0 0 0 0 0 0 So a basis for the eigenspace E 1 is { ⃗v 2 = 0 1 0 ,⃗v 3 = - 1 0 1 } . The basis B = { ⃗v 1 ,⃗v 2 ,⃗v 3 } is a basis of eigenvectors. Note that, if you divide ⃗v 1 and ⃗v 3 by their lengths (both lengths are 2), this becomes an orthonormal basis. 2. Find a basis for R 2 such that the matrix of T ( ⃗x ) = ( 3 2 - 1 5 ) ⃗x with respect to this basis has the form
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Unformatted text preview: ( a-b b a ) . 2 Solution: The characteristic polynomial det ( 3-λ 2-1 5-λ ) = λ 2-8 λ + 17 has the roots λ = 4 ± i , and ( 3-(4 + i ) 2-1 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 ⃗u-⃗w + 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 1-1 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.

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Q7rs - ( a-b b a ) . 2 Solution: The characteristic...

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