Assignment 5 solution

Assignment 5 solution - MATH 223, Linear Algebra Winter,...

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MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 5 1. Let V = P 3 ( X ), the space of real polynomials with degree at most 3. Define T : V -→ V by T ( p ( x )) = ( x 2 - 2 x ) p 00 ( x ) + (2 x + 2) p 0 ( x ) - 12 p ( x ) . (a) Verify that T is a linear operator on V . Solution: For any p 1 ( x ) ,p 2 ( x ) V , T [ p 1 ( x ) + p 2 ( x )] = ( x 2 - 2 x )( p 1 + p 2 ) 00 ( x ) + (2 x + 2)( p 1 + p 2 ) 0 ( x ) - 12( p 1 + p 2 )( x ) = ( x 2 - 2 x )( p 00 1 ( x ) + p 00 2 ( x )) + (2 x + 2)( p 0 1 ( x ) + p 0 2 ( x )) - 12( p 1 ( x ) + p 2 ( x )) = [( x 2 - 2 x ) p 00 1 ( x ) + (2 x + 2) p 0 1 ( x ) - 12 p 1 ( x )]+ [( x 2 - 2 x ) p 00 2 ( x ) + (2 x + 2) p 0 2 ( x ) - 12 p 2 ( x )] = T ( p 1 ( x )) + T ( p 2 ( x )) . Also, for any real r and p ( x ) V , T ( rp ( x )) = ( x 2 - 2 x )( rp ) 00 ( x ) + (2 x + 2)( rp ) 0 ( x ) - 12( rp ( x )) = ( x 2 - 2 x )( rp 00 ( x )) + (2 x + 2)( rp 0 ( x )) - 12( rp ( x )) = r [( x 2 - 2 x ) p 00 ( x ) + (2 x + 2) p 0 ( x ) - 12 p ( x )] = rT ( p ( x )) . That does it. (b) Find [ T ] B where B = (1 ,x,x 2 ,x 3 ) is the standard ordered basis for V . Solution: T 1 = ( x 2 - 2 x )0 + (2 x + 2)0 - 12(1) = - 12 = - 12(1) + 0 x + 0 x 2 + 0 x 3 , so the first column of [ T ] B is - 12 0 0 0 . Tx = ( x 2 - 2 x )0 + (2 x + 2)1 - 12 x = 2 - 10 x and the second column is 2 - 10 0 0 . Tx 2 = ( x 2 - 2 x )(2) + (2 x + 2)(2 x ) - 12 x 2 = - 6 x 2 and the third column is 0 0 - 6 0 . Tx 3 = ( x 2 - 2 x )(6 x )+(2 x +2)(3 x 2 ) - 12 x 3 = - 6 x 2 so the fourth column is the same as the third. The matrix [ T ] B is - 12 2 0 0 0 10 0 0 0 0 - 6 - 6 0 0 0 0 . (c) Find [ T ] B 0 where B 0 = (1 ,x +1 ,x 3 - x 2 ,x 2 ) is a nonstandard ordered basis for V . Solution: The first column here is the same as above. T ( x + 1) = 2 - 10 x - 12 = 0(1) - 10( x +1)+0( x 3 - x 2 )+0( x 2 ) so the second column 1
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is 0 - 10 0 0 . [It is not - 10 - 10 0 0 . ] T
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Assignment 5 solution - MATH 223, Linear Algebra Winter,...

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