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5.2(2d,3b,7,8,12) - Linear Algebra-115 Solutions to...

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Linear Algebra -115 Solutions to Eleventh Homework Problem 2( d ) (Section 5 . 2) det ( A - λI ) = 7 - λ - 4 0 8 - 5 - λ 0 6 - 6 3 - λ = (3 - λ ) 7 - λ - 4 8 - 5 - λ = (3 - λ )[(7 - λ )( - 5 - λ ) + 32] = (3 - λ )( λ 2 - 2 λ - 3) = (3 - λ )( λ - 3)( λ + 1) . So, - 1 , 3 , 3 are the eigenvalues. We need to find the eigenspaces, too. A + 1 I = 8 - 4 0 8 - 4 0 6 - 6 4 2 - 1 0 0 0 0 3 - 3 2 . So, ( x, y, z ) E - 1 iff 2 - 1 0 0 0 0 3 - 3 2 x y z = 0 0 0 iff 2 x - y = 0 and 3 x - 3 y + 2 z = 0 iff y = 2 x and - 3 x + 2 z = 0 iff y = 2 x and z = 3 2 x, i.e. E - 1 = { x (2 , 4 , 3) } = span (2 , 4 , 3). A - 3 I = 4 - 4 0 8 - 8 0 6 - 6 0 4 - 4 0 0 0 0 0 0 0 . So, ( x, y, z ) E 3 iff 4 - 4 0 0 0 0 0 0 0 x y z = 0 0 0 iff 4 x - 4 y = 0 iff x = y i.e. E 3 = { x (1 , 1 , 0) + z (0 , 0 , 1) } = span { (1 , 1 , 0) , (0 , 0 , 1) } . 1
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In total, the matrices we are looking for are Q = 2 1 0 4 1 0 3 0 1 , D = - 1 0 0 0 3 0 0 0 3 . Problem 3( b ) (Section 5 . 2) Let γ = { 1 , x, x 2 } the standard basis. We will compute [ T ] γ . Since T (1) = x 2 , T ( x ) = x , T ( x 2 ) = 1, [ T (1)] γ = (0 , 0 , 1) , [ T ( x )] γ = (0 , 1 , 0) , [ T ( x 2 )] γ = (1 , 0 , 0) , or A = [ T ] γ = 0 0 1 0 1 0 1 0 0 .
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