42_hwc1SolnsODDA

42_hwc1SolnsODDA - n ) . Now just observe that the right...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
SOLUTION Since f is a linear transformation from IR n to IR , Theorem 2 says that f ( x ) = A f x ( * ) where A f is the 1 × n matrix A f = [ f ( e 1 ) , f ( e 2 ) , . . . , f ( e n )] . The single row of this matrix – written in vertical form, as is our convention with vectors – is a f = f ( e 1 ) f ( e 2 ) . . . f ( e n ) . Now, V.I.F. 2 says that A f x = a f · x . ( ** ) Putting ( * ) and ( ** ) together, we have shown that f ( x ) = a f · x . ( * * * ) Remark: It is worth remembering that although Theorem 2 is fundamentally important, it is also a very simple consequence of the notion of linearity. In fact, one can deduce (***) directly from the linearity of f : Just write x = x 1 e 1 + x 2 e 2 + ··· + x n e n . Then, by the linearity of f , f ( x ) = x 1 f ( e 1 ) + x 2 f ( e 2 ) + ··· + x n f ( e
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n ) . Now just observe that the right hand side is x a f . This line of reasoning is just what we used to prove Theorem 2 in the rst place. 5.25 For any vector a b c , dene the polynomial p ( x ) = a + bx + cx 2 . In turn, dene the number R 1 p ( x )d x . Putting the pieces together, we get a function f from IR 3 to IR we dene f a b c = Z 1 a + bx + cx 2 d x . (a) Show that f is a linear functional on IR 3 ; i.e., a linear transformation from IR 3 to IR . 21 /september/ 2005; 22:09 43...
View Full Document

Ask a homework question - tutors are online