Answers Exercise Set I.2 2007 PDF

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Answers to Exercise Set I. Drills    1 1 −1 −1 0 1. (a)  1 −1 0  (b)  0 2 0 1 1 0 0   1/2 1/2 2. 1/2 1/2   0 cos θ 0  (c)  0 1 − sin θ 0 1 0   sin θ 0 0  (d)  3 cos θ −2 −3 0 1  2 −1  0 3. Notice that the x–axis is turned onto the y–axis, the y–axis is turned onto the z–axis, and t...

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    

     cos θ −1 0 0 1 1 −1      1. (a) 1 −1 0 0 2 0 (c) 0 (b) 0 0 1 0 1 1 − sin θ   1/2 1/2 2. 1/2 1/2 

  0 sin θ 0   1 0 3 (d) −2 0 cos θ

 −3 2 0 −1  1 0

3. Notice that the x–axis is turned onto the y–axis, the y–axis is turned onto the z–axis, and the z–axis is turned onto the x–axis. So the required matrix is 

 0 0 1 1 0 0. 0 1 0 4. (a)



1 −1 1 −1



6. (a) 1, x, x , x

(b)



2 −1 2 −1



(c)

(b) (0, 1), (1, 0)



cos θ sin θ

(1 − cos  θ)/ sin θ − cos θ



(c) (1, 0, 0), (0, 1, 0), (0, 0, 1)

(d) (1, 0), (x, 0), (0, 1), (0, x), (0, x)         0 1 0 0 0 0 1 0 , , , . (e) 0 0 0 0 1 0 0 1 7. (a) For ker T : {1}. For T (V ): {1, 2x}. case is the empty set ∅). For T (V ): {x, T (V ): {(1, 1)}.

(d)  0 (e) For ker T : 1  1 (f) For ker T : 0  1 (g) For ker T : 1

8. (a)



−1 3



  1 (b) i

(b) ker T = {0} (basis for ker T in this x

, x}.

(c) For ker T : {(−1, 1)}. For

For ker T = {(0, 0)} (basis for ker T = ∅). For T (V ): {1, x}.        0 0 0 1 0 1 0 , , . . For T (V ): 0 0 1 1 0 0 1        −1 0 0 0 0 1 0 , , . . For T (V ): 0 0 0 1 −1 1 0        1 0 0 0 −1 1 0 , , . . For T (V ): 0 0 1 0 1 1 0

(c)



5/2 −1/2





 1/ 2 (d)  1/2  1/ 2

  3 2 (e)   2 0

9. [(The missing part of this qestion in the course notes) In each of the following cases, find the representing matrix of T : V → W relative to the basis B in V and C in W : (a) V = W = P, B = C = the standard basis in P and T (p(x)) = p ′(x). (b) V , W , B and C are the same as those in (a), T (p(x)) = p(x − 1) + p′(x). (c) V = P , W = P, B and C are standard bases in P and P respectively, and T (p(x)) = x p(x).]     0 0 1 0 0 1 (a)  1 0 0  (b)  1 0 0  0 1 0 0 1 0 

0 0 (e)  1 0

−1 1 0 1

 0 0  0 −1

0 0 0 −1



0 0  (c)  1  0 0

0 0 0 1 0

  0 1 0 0 0 0   0  (d)  0 1   0 0 0 1 0 0

 0 0  0  0 1

 1. The matrix A = [a a     an ] in M ,n corresponds to MA in L (n , ) given by MA [x x    xn ]⊤ = a x + ax  +    + an xn . The matrix A = [a a    an]⊤ in Mn, corresponds to MA in L (,  n ) given by MA x = [a x a x    an x]⊤ = Ax. 2. Clearly a rotation has the required form. Conversely, suppose that A has the given form. Let θ be the angle between the vectors (a, b) and (1, 0) in  . Then a = cos θ and b = sin θ. So



a A= b

  cos θ −b = a sin θ

 − sin θ . cos θ

To show that A− = A⊤ , we compute         1 0 a b a + (−b)  ab − ba a −b ⊤ = AA = =I = ba + a(−b) b + a 0 1 b a −b a in view of a + b = 1. In the same way we can check that A⊤ A = I     a b  a  b 3. Let B = , B = (a − b = a − b  = 1, a, a  > 0) be boosts. b  a b a Then     a b a a  + b b a b + b a = B B ‘ = b a b  a + a b b  b + a a

where a = aa  + b b  and b = a b + b a  satisfy a − b  = (a  a + bb ) − (a b  + ba ) = (a  a + bb  + 2a a bb  ) − (ab + b a  + 2ab ba ) = aa  − a b + b b − b  a = (a − b )(a  − b ) = 1. Finally we check a > 0. Notice that, from a  − b = 1 and a > 0 we deduce that a  = |a | =



a 

>



b = |b |

Similarly, we have a > |b|. Thus a = a a  + bb ≥ a a  − |b||b  | = a (a − |b  |) + (a − |b |)|b  | > 0 4. The map P is given by P =

 . 

Applying P on the basis vectors  = (1, 0) and   = (0, 1), we have        1 a a a , =  P  = =    a +b b a + b ab         b 1 a ab P  = =  =  .   b  a +b b a +b So the required matrix for P is 1 [P ] =  a + b



 a ab . ab b 

5. Let α = (cos α, sin α) and let Pα be the projection of  onto Lα . Then T α = 2Pα − I. (Indeed, this can be checked as follows. If  is parallel to L α , we have Pα =  and hence (2Pα − I) = . If  is perpendicular to L α, then T α = − and Pα =  which gives (2P α − I) = −.) The last exercise tells us the matrix [P α ] for a = cos α and b = sin α. So    cos 2α 2 cos  α − 1 2 cos α sin α = [T α] = 2[P α] − I =  sin 2α 2 cos α sin α 2 sin α − 1

 sin 2α . − cos 2α

 cos θ − sin θ , where θ = 2(α + β). Direct computation shows [Tα ][T β] == sin θ cos θ 

6. (a) Let S =  − 2(  ). When  is lying on H, then we have T  =  and    = 0, which gives S =  − 2(  ) = . In the normal direction, we have T  = − and S =  − 2(  ) =  − 2 = −. Hence T = S. (b) We have T  =  − 2(  ) = (1, 0, 0) − 2n  (n, n, n  ) = (1 − 2n , −nn , −n n). We can find T  , T   in the same fashion. So [T ] = [T   T 



1 − 2n T  ] =  −n n  −n n 

 −n n  −n n 1 − 2n −n n  . −n n  1 − 2n 

(c) Since  is perpendicular to both  and , we have T  =  and T   = . Hence T T  = T  = . 7. Consider the triangle ∆ABC with vertices A(3, 0, 0), B(0, 3, 0), C = (0, 0, 3). The −−→ center of this trangle is M (1, 1, 1). The vector  = OM = (1, 1, 1) is fixed by the −−→ rotation, that is, T  = . After the 60o of rotation, vector  = M A = (2, −1, −1) −−→ becomes − = −M C = (1, 1, −2), or T  = −. Similarly T  = −, where −−→ −→ −−→ −−→  = M B = (−1, 2, −1). Now the trivial identity OA = OM + M A becomes 3 =  +  (here  = (1, 0, 0)) and hence 3T  = T  + T  =  −  = (1, 1, 1) + (1, 1, −2) = (2, 2, −1) 3T  = T  + T  =  −  = (1, 1, 1) + (−2, 1, 1) = (−1, 2, 2) 3T  = T  + T  =  −  = (1, 1, 1) + (1, −2, 1) = (2, −1, 2) So the matrix representing T is 

2 1 2 [T ] = 3 −1

−1 2 2

 2 −1  . 2

A direct computation shows 

 0 0 1 [T ]  =  1 0 0  , 0 1 0 which is the matrix representing the 120o rotation described in Drill #3 above....