# Sol2 - SOLUTION TO ASSIGNMENT 2 MATH 251 WINTER 2007 1 Let...

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: SOLUTION TO ASSIGNMENT 2 - MATH 251, WINTER 2007 1. Let B = { (1 , 1) , (1 , 5) } and C = { (2 , 1) , (1 ,- 1) } be bases of R 2 . Find the change of basis matrices B M C and C M B between the bases B and C . Let v = ( 8 28 ) with respect to the standard basis. Find [ v ] B and [ v ] C . Proof. We have St M B = 1 1 1 5 , St M C = 2 1 1- 1 . Thus, B M St = 1 4 5- 1- 1 1 . Therefore, B M C = 1 4 5- 1- 1 1 2 1 1- 1 = 1 4 9 6- 1- 2 , and C M B = ( B M C )- 1 =- 1 3- 2- 6 1 9 . We have 8 28 B = B M St 8 28 = 1 4 5- 1- 1 1 8 28 = 3 5 and 8 28 C = C M B 8 28 B =- 1 3- 2- 6 1 9 3 5 = 12- 16 . 2. Which of the following functions is a linear map? (provide proof): (1) T : R 2 → R 2 , T ( x,y ) = (3 x- 2 y,x + y ). (2) T : R 2 → R 2 , T ( x,y ) = ( x 2- y,x + y + 1). (3) T : R [ x ] 3 → R 2 , T ( f ( x )) = ( f (1) ,f (1)). (4) T : R [ x ] 3 → R [ x ] 4 , T ( f ( x )) = xf ( x ) + f (1) . In each case where T is a linear map, find its kernel. Proof. (1) T is a linear map. Its kernel are the vectors ( x,y ) s.t. 3 x = 2 y,x =- y , namely, just (0 , 0). (2) T is not a linear map. T (2 , 0) = (4 , 3), while 2 T (1 , 0) = 2(1 , 2) = (2 , 4) 6 = T (2 , 0). (3) T is a linear map. Its kernel are the polynomials that vanish at 1 and whose derivatives vanish at 1. Writing a polynomial as a 3 x 3 + a 2 x 2 + a 1 x + a , the conditions are a 3 + a 2 + a 1 + a = 0 , 3 a 3 + 2 a 2 + a 1 = 0 . (4) Again, T is a linear map. Its kernel is just 0, because if f is not zero then xf ( x ) is a non-zero polynomial with no constant term, so for every scalar r , xf ( x ) + r is not zero, in particular, xf ( x ) + f (1) is not zero. 1 2 SOLUTION TO ASSIGNMENT 2 - MATH 251, WINTER 2007 3. Let T : V → W be a surjective linear map between vector spaces of finite dimension. Prove that if dim( V ) = dim( W ) then T is an isomorphism. Proof. We have dim(Ker( T )) = dim( V )- dim(Im( T )) = dim( W )- dim( W ) = 0 and so Ker( T ) can only be the zero subspace. Thus, T is injective. 4. Prove the following Proposition. Proposition 0.1. Let V and W be vector spaces over F . Let B = { b 1 ,...,b n } be a basis for V and let t 1 ,...,t n be any elements of W . There is a unique linear map T : V → W, such that T ( b i ) = t i , i = 1 ,...,n. Proof. Define T ( n X i =1 α i b i ) = n X i =1 α i t i . Since any vector in V can be written uniquely as ∑ n i =1 α i b i this gives a well defined function T : V → W , which clearly satisfies T ( b i ) = t i . It only remains to check that T is linear: T ( α n X i =1 α i b i ) = T ( n X i =1 αα i b i ) = n X i =1 αα i t i = α n X i =1 α i t i = αT ( n X i =1 α i b i ) , and T ( n X i =1 α i b i + n X i =1 β i b i ) = T ( n X i =1 ( α i + β i ) b i ) = n X i =1 ( α i + β i ) t i = n X i =1 α i t i + n X i =1 β i t i = T ( n X i =1 α i b i )+ T ( n X i =1 β i b i ) ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

Sol2 - SOLUTION TO ASSIGNMENT 2 MATH 251 WINTER 2007 1 Let...

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

View Full Document
Ask a homework question - tutors are online