2.1_2 - Linear Algebra -115 Solutions to Sixth Homework...

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Unformatted text preview: Linear Algebra -115 Solutions to Sixth Homework Problem 5 (Section 2 . 1) We compute: N ( T ) = { f ( x ) P 2 (R) | T ( f ( x )) = 0 } = { f ( x ) P 2 (R) | xf ( x ) + f ( x ) = 0 } = { f ( x ) = ax 2 + bx + c P 2 (R) | x ( ax 2 + bx + c ) + (2 ax + b ) = 0 } = { f ( x ) = ax 2 + bx + c P 2 (R) | ax 3 + bx 2 + ( c + 2 a ) x + b = 0 } = { f ( x ) = ax 2 + bx + c P 2 (R) | a = 0& b = 0& c + 2 a = 0& b = 0 } = { } So, a basis for N ( T ) is the empty set (not the zero vector!). We also conclude that T is one-to-one. By Theorem 2 . 2, R ( T ) is spanned by T ( ), where is a basis for P 2 (R). If we choose = { 1 , x, x 2 } , we get that T ( ) = { x, x 2 + 1 , x 3 + 2 x } is a basis for R ( T ) (see also problem 14( c ) for this). Verifying the dimension theorem, we have null ( T ) + rank ( T ) = 0 + 3 = 3 = dim ( P 2 (R)) . We mentioned already that T is 1- 1. Since rank ( T ) = 3 < 4 = dim ( P 3 (R)) we conclude that T cant be onto....
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2.1_2 - Linear Algebra -115 Solutions to Sixth Homework...

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