--- //[[honeypot@handbasket.org|David Wagner]] 2009/09/20 01:07//

====== √EGR 6013 Homework Set 2 (38-49) ======

//Section 1.10.//

==== √Problem 38. ====
Determine the dimensions of the vector spaces generated by each of the following sets of vectors:

**(a)** {1, 1, 0}, {1, 0, 1}, {0, 1, 1}.

dimension = <m>rank delim{[}{ matrix{3}{3}{ 
1 1 0
0 1 1
1 0 1
 } }{]}</m>
=
<m>rank delim{[}{ matrix{3}{3}{ 
1 1 0
0 1 1
0 {-1} 1
 } }{]}</m>
=
<m>rank delim{[}{ matrix{3}{3}{ 
1 1 0
0 1 1
0 0 2
 } }{]}</m> = 3

**(b)** {1, 0, 0}, {0, 1, 0}, {0, 0, 1}, {1, 1, 1}.

dimension = <m>rank delim{[}{ matrix{3}{3}{ 
1 0 0 1
0 1 0 1
0 0 1 1
 } }{]}</m>= 3


**(c)** {1, 1, 1}, {1, 0, 1}, {1, 2, 1}.

dimension = <m>rank delim{[}{ matrix{3}{3}{ 
1 1 {1~}
1 0 {1~}
1 2 {1~}
 } }{]}</m>
=
<m>rank delim{[}{ matrix{3}{3}{ 
1 1 1
0 {-1} 0
0 1 0
 } }{]}</m>
=
<m>rank delim{[}{ matrix{3}{3}{
1 1 1
0 1 0
0 0 0
 } }{]}</m> = 2


==== √Problem 39. ====
Determine whether the vector {6, 1, -6, 2} is in the vector space generated by the vectors
 {1, 1, -1, 1}, {-1, 0, 1, 1}, and {1, -1, -1, 0}.

<m>c_1 delim{lbrace}{ matrix{4}{1}{ 1  1 {-1} 1 } }{rbrace}
+ c_2 delim{lbrace}{ matrix{4}{1}{ {-1} 0 1 1 } }{rbrace}
+ c_3 delim{lbrace}{ matrix{4}{1}{ 1 {-1} {-1} 0 } }{rbrace}
= delim{lbrace}{ matrix{4}{1}{ 6 1 {-6} 2 } }{rbrace}</m>

<m>delim{[}{ matrix{4}{3}{
{1} {-1} {1~}
{1} {0} {-1~}
{-1} {1} {-1~}
{1} {1} {-1~}
} }{]}
delim{lbrace}{ matrix{3}{1}{ c_1 c_2 c_3 } }{rbrace}
= delim{lbrace}{ matrix{4}{1}{ 6 1 {-6} 2 } }{rbrace}</m>
=>
<m>delim{[}{ matrix{4}{3}{
{1} {-1} {1}
{0} {1} {-2}
{0} {0} {0}
{0} {2} {-2}
} }{]}
delim{lbrace}{ matrix{3}{1}{ c_1 c_2 c_3 } }{rbrace}
= delim{lbrace}{ matrix{4}{1}{ 6 {-5} 0 {-4} } }{rbrace}</m>
=>
<m>delim{[}{ matrix{4}{3}{
{1} {-1} {1~}
{0} {1} {-2~}
{0} {0} {0~}
{0} {0} {2~}
} }{]}
delim{lbrace}{ matrix{3}{1}{ c_1 c_2 c_3 } }{rbrace}
= delim{lbrace}{ matrix{4}{1}{ 6 {-5} 0 6 } }{rbrace}</m>

  * <m>c_3 = -2</m>
  * <m>c_2 = -4</m>
  * <m>c_1 = 2</m>

Three vectors are not sufficient to define a space with dimension four, so the vector of dimension four being considered cannot be in the smaller space defined.

However, the vector in question is a linear combination of the vectors defining the space and  unconstrained along its third axis,
 implying that this vector may be projected along this axis and into the vector space defined.



==== √Problem 40 (a) ====
Determine the angle θ between the vectors
  * {u} = {1, 1, 1, 1},
  * {v} = {1, 0, 0, 1}.

<m>|u| = sqrt{u · u} = sqrt{4}=2</m>

<m>|v| = sqrt{v · v} = sqrt{2}</m>

<m>theta = arccos({u · v}/{|u||v|})
= arccos( {2}/{2sqrt{2}} )
= arccos(sqrt{2}/2) = pi/4</m>


//Section 1.11.//

==== √Problem 46. ====
Show that the set of equations
  * <m>x_1 + x_2 + x_3 = 3</m>,
  * <m>x_1 + x_2 - x_3 = 1</m>,
  * <m>3 x_1 + 3 x_2 -5 x_3 = 1</m>
possesses a one-parameter family of solutions,
and verify directly that the vector {c} whose elements comprise the right-hand members is orthogonal to all vector solutions of the transposed homogeneous set of equations.

<m>[A] lbrace x rbrace = lbrace c rbrace =
delim{[}{ matrix{3}{3}{ 
1 1 1
1 1 {-1}
3 3 {-5}
 } }{]}
delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 3 1 1 } }{rbrace}</m>
=>
<m>delim{[}{ matrix{3}{3}{ 
1 1 1
0 0 {-2}
0 0 {-8}
 } }{]}
delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 3 {-2} {-8} } }{rbrace}</m>
=>
<m>delim{[}{ matrix{3}{3}{ 
1 1 {1~}
0 0 {1~}
0 0 {1~}
 } }{]}
delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 3 {1} {1} } }{rbrace}</m>.

The matrix has order 3, rank 2, so one additional parameter is needed.

  * <m>x_3 = 1</m>.
  * Assume <m>x_2 = b_2</m>,
  * <m>x_1 + x_2 + x_3 = 3 = x_1 + b_2 + 1</m> => <m>x_1 = 2 - b_2</m>.

<m>x = delim{lbrace}{ matrix{3}{1}{ {2 - b_2} {b_2} {1} } }{rbrace}</m>.


<m>[A]^T x prime = delim{[}{ matrix{3}{3}{
1 {-1} {-5}
1 1 {3}
1 1 {3}
 } }{]}
delim{lbrace}{ matrix{3}{1}{ {x_1 prime} {x_2 prime} {x_3 prime} } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 0 0 0 } }{rbrace}</m>
=>
<m>[A]^T x prime = delim{[}{ matrix{3}{3}{
1 {-1} {-5}
0 2 {8}
0 2 {8}
 } }{]}
delim{lbrace}{ matrix{3}{1}{ {x_1 prime} {x_2 prime} {x_3 prime} } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 0 0 0 } }{rbrace}</m>
=>
<m>[A]^T x prime = delim{[}{ matrix{3}{3}{
1 {-1} {-5}
0 1 {4}
0 0 {0}
 } }{]}
delim{lbrace}{ matrix{3}{1}{ {x_1 prime} {x_2 prime} {x_3 prime} } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 0 0 0 } }{rbrace}</m>.

  * <m>x_3 prime = b_3 prime</m>,
  * <m>x_2 prime =- 4 b_3 prime</m>,
  * <m>x_1 prime =x_2 prime +5x_3 prime = - 4 b_3 prime+5b_3 prime=b_3 prime</m>.

<m>x prime · c = 3b_3 prime - 4 b_3 prime +b_3 prime=0</m>.

Thus, {c} is perpendicular to all {x'}.

//Section 1.12.//

==== √Problem 49. ====
Show that the problem
  * <m>x_1 - 2x_2 = lambda x_1</m>,
  * <m>x_1 -  x_2 = lambda x_2</m>
does not possess real nontrivial solutions for any values of λ.

  - <m>x_2 = 0 right x_1=0</m>, a trivial solution.
  - <m>x_1 = lambda x_2+x_2</m>
  - <m>x_1 - 2x_2 = lambda x_1</m> = <m>lambda x_2+x_2 - 2x_2 = lambda (lambda x_2+x_2) = lambda x_2 - x_2 = lambda^2 x_2 +lambda x_2</m>
  - <m>lambda^2 x_2 +lambda x_2 -lambda x_2 + x_2=(lambda^2 +1)x_2  =0</m>
  - <m>x_2 ne 0 right lambda^2 +1=0 right lambda^2 = -1</m>.

Thus, no real value of λ satisfies the characteristic equation.