Exam2MA262fall2011

# Solution first we need to find c 1 c 2 and c 3 such

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Solution. First we need to find c 1 , c 2 and c 3 such that x 2 + x +1 = c 1 ( x 2 - 1)+ c 2 (2 x +3)+ c 3 ( x 2 - 2 x ) = ( c 1 + c 3 ) x 2 +(2 c 2 - 2 c 3 ) x +3 c 2 - c 1 . Then we have c 1 + c 3 = 1 2 c 2 - 2 c 3 = 1 - c 1 + 3 c 2 = 1 You can either apply Gaussian elimination and solve the system above or, since it is an easy system, adding the first and third equation gives 2 c 2 - 2 c 3 = 1 3 c 2 + c 3 = 2 c 2 = 5 8 , c 3 = 1 8 . So c 1 = 7 8 , and then T ( x 2 + x +1) = 7 8 T ( x 2 - 1)+ 5 8 T (2 x +3)+ 1 8 T ( x 2 - 2 x ) = 1 8 7( x 2 + x ) + 5(2) + ( x + 1) T ( x 2 + x + 1) = 1 8 (7 x 2 + 8 x + 11) . 9. Find all eigenvalues and a basis for the respective eigenspaces of the (15 points) following matrix and determine if it is defective or nondefective A = 1 - 2 0 2 - 3 0 2 - 2 - 1 . To find the eigenvalues, we need to find all λ such that 1 - λ - 2 0 2 - 3 - λ 0 2 - 2 - 1 - λ = 0 . Expanding the third column we get 1 - λ - 2 0 2 - 3 - λ 0 2 - 2 - 1 - λ = ( - 1 - λ ) 1 - λ - 2 2 - 3 - λ = - (1+ λ )[(1 - λ )( - 3 - λ )+4] = 0 ⇒ - (1+ λ )[(1 - λ )( - 3 - λ )+4] = - (1+ λ )( λ 2 +2 λ +1) = - (1+ λ )(1+ λ ) 2 = 0 . Hence (1 + λ ) 3 = 0, λ = - 1 with multiplicity 3. To find the eigenspace we plug λ = - 1 and solve 1 - ( - 1) - 2 0 2 - 3 - ( - 1) 0 2 - 2 - 1 - ( - 1) x y z = 0 0 0 Applying Gaussian elimination on the system above we get 2 - 2 0 0 0 0 0 0 0 x y z = 0 0 0 So there are two free variables, say y and z and the eigenspace is { ( y, y, z ) } with a basis given by { (1 , 1 , 0) , (0 , 0 , 1) } .
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