351Hmw8solns[1]

# 351Hmw8solns[1] - ± 0 0 0 1 ² = ± 0 1 0 2 ² we have A =...

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HW 8 solutions April 8, 2011 Section 4.2 43) T ( f ( t )) = f (5) f (7) f (11) Then T ( f ( t ) + g ( t )) = f (5) + g (5) f (7) + g (7) f (11) + g (11) = f (5) f (7) f (11) + g (5) g (7) g (11) and T ( kf ( t )) = kf (5) kf (7) kf (11) = k. f (5) f (7) f (11) Hence T is a linear transformation. Now, T ( f ( t )) = 0 means f (5) = f (7) = f (11) = 0. Since a non-zero polynomial in P 2 has atmost two zeros, f ( t ) = 0. Hence Ker ( T ) = 0. Since dim ( P 2 ) = dim ( R 3 ) = 3, T is an isomorphism. 53) Given T ( f ( t )) = f ”( t )+4 f 0 ( t ). Let f ( t ) = at 2 + bt + c . Then T ( f ( t )) = 8 at +2 a +4 b . So T ( f ( t )) = 0 means 8 a = 0 and 2 a +4 b = 0. So a = b = 0. Hence ker ( T ) = { f P 2 | f ( t ) = c } and nullity ( T ) = dim ( Ker ( T )) = 1. Also Im ( T ) = { 8 at + 2 a + 4 b } = { a 1 t + b 1 } . So Im ( T ) = P 2 and rank ( T ) = dim ( Im ( T )) = 2. 60) A and B are similar means there exists S 6 = 0 such that AS = SB . Hence T ( S ) = AS - SB = 0. So S Ker ( T ) which implies T is not an isomorphism. Section 4.3 44)a) The change of matrix S is given by S = 1 0 1 0 0 1 0 1 - 1 0 2 0 0 - 1 0 2 1

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44)b) Since 1 1 2 2 . 1 0 0 0 = 1 0 2 0 1 1 2 2 . 0 1 0 0 = 0 1 0 2 1 1 2 2 . 0 0 1
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Unformatted text preview: . ± 0 0 0 1 ² = ± 0 1 0 2 ² we have A = 1 0 1 0 0 1 0 1 2 0 2 0 0 2 0 2 Since ± 1 1 2 2 ² . ± 1-1 0 ² = ± 0 0 0 0 ² ± 1 1 2 2 ² . ± 1-1 ² = ± 0 0 0 0 ² ± 1 1 2 2 ² . ± 1 0 2 0 ² = ± 3 0 6 0 ² = 3 . ± 1 0 2 0 ² ± 1 1 2 2 ² . ± 0 1 0 2 ² = ± 0 3 0 6 ² = 3 . ± 0 1 0 2 ² we have B = 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 3 Now SB = AS = 0 0 3 0 0 0 0 3 0 0 6 0 0 0 0 6 60)a) Since 3 3 = 1 2 2 + 2-2 1 we have S B→U = ± 1 1 0 1 ² 2 60)b) S U→B = S-1 B→U = ± 1-1 1 ² 60)c) ³ a 1 a 2 ´ .S B→U = ³ b 1 b 2 ´ 3...
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