MA2940F09SamplePrelim2Sol

# MA2940F09SamplePrelim2Sol - Practice Prelim 2 Solutions 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Practice Prelim 2 Solutions 1. Note that P 4 is a 5-dimensional linear space with basis B = { 1 ,x,x 2 ,x 3 ,x 4 } . (a) In terms of the basis B , T a b c = a- 2 b + 3 c 3 a + 2 b + c a + 2 b- c a + c B = 1- 2 3 3 2 1 1 2- 1 1 1 · a b c To find the image (and kernel) of T , row-reduce the matrix of T : 1- 2 3 3 2 1 1 2- 1 1 1 - 3 R 1- R 1- R 1-→ 1- 2 3 8- 8 4- 4 2- 2 ÷ 8 ÷ 4 ÷ 2-→ 1- 2 3 1- 1 1- 1 1- 1 - R 2- R 2-→ 1- 2 3 1- 1 The pivots in the first and second columns indicate these columns are independent in the original matrix, so im( T ) = span 1 3 1 1 B , - 2 2 2 B = span 1 + 3 x + x 2 + x 4 ,- 2 + 2 x + 2 x 2 A basis for im( T ) is given by 1 + 3 x + x 2 + x 4 ,- 2 + 2 x + 2 x 2 . (b) Set p 1 ( x ) = 1 + 3 x + x 2 + x 4 and p 2 ( x ) =- 2 + 2 x + 2 x 2 . Then p ( x ) = ( a + c ) p 1 ( x ) + ( b- c ) p 2 ( x ). So the coordinates with respect to the basis 1 + 3 x + x 2 + x 4 ,- 2 + 2 x + 2 x 2 are a + c b- c . (c) For the kernel, the row-reduced matrix gives us the relations x 1- 2 x 2 + 3 x 3 = 0 and x 2- x 3 = 0. x 3 is a free variable, so set x 3 = t . Then x 2 = t and x 1 = 2 t- 3 t =- t . Thus, ker( T ) consists of all vectors - t t t , where t is any real number. So ker( T ) = span - 1 1 1 . Alternatively, this problem can be solved by inspection: if we notice the relationship among the columns: ~v 1- ~v 2 = ~v 3 , where ~v i is the i th column of the matrix for T , we obtain the same answers. (Since the rank of the matrix is 2, the rank-nullity theorem tells us the nullity must be 1, so we have found eveything in the kernel) 2. (a) W is not a subspace of V . It is not closed under scalar multiplication: I = 1 0 0 1 is in W since I 2 = I . But 2 I is not in W since (2 I ) 2 = 4 I 6 = 2 I . (b) W is a subspace of V : • 0 (ie. the function f ( t ) = 0) is in W : f ( t ) = 0 for all t . In particular, f (3) = 0. • W is closed under scalar multiplication: If f ( t ) is in W and k is a scalar, then d ( kf ) dt = k df dt for all t . So d ( kf ) dt t =3 = k df dt t =3 = k · 0 = 0. Hence, kf ( t ) is in W ....
View Full Document

## This note was uploaded on 12/25/2009 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell.

### Page1 / 9

MA2940F09SamplePrelim2Sol - Practice Prelim 2 Solutions 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online