√EGR 6013 Homework Set 2 (38-49)

— 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 = rank delim{[}{ matrix{3}{3}{
1 1 0
0 1 1
1 0 1
} }{]} = = = 3

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

dimension = rank delim{[}{ matrix{3}{3}{
1 0 0 1
0 1 0 1
0 0 1 1
} }{]} = 3

dimension = rank delim{[}{ matrix{3}{3}{
1 1 {1~}
1 0 {1~}
1 2 {1~}
} }{]} = = = 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}.

$c_1 delim{lbrace}{ matrix{4}{1}{ 1 1 {-1} 1 } }{rbrace} + c_2 delim{lbrace}{ matrix{4}{1}{ {-1} 0 1 1 } }{rbrace} + c_3 delim{lbrace}{ matrix{4}{1}{ 1 {-1} {-1} 0 } }{rbrace} = delim{lbrace}{ matrix{4}{1}{ 6 1 {-6} 2 } }{rbrace}$

$delim{[}{ matrix{4}{3}{ {1} {-1} {1~} {1} {0} {-1~} {-1} {1} {-1~} {1} {1} {-1~} } }{]} delim{lbrace}{ matrix{3}{1}{ c_1 c_2 c_3 } }{rbrace} = delim{lbrace}{ matrix{4}{1}{ 6 1 {-6} 2 } }{rbrace}$ ⇒ $delim{[}{ matrix{4}{3}{ {1} {-1} {1} {0} {1} {-2} {0} {0} {0} {0} {2} {-2} } }{]} delim{lbrace}{ matrix{3}{1}{ c_1 c_2 c_3 } }{rbrace} = delim{lbrace}{ matrix{4}{1}{ 6 {-5} 0 {-4} } }{rbrace}$ ⇒ $delim{[}{ matrix{4}{3}{ {1} {-1} {1~} {0} {1} {-2~} {0} {0} {0~} {0} {0} {2~} } }{]} delim{lbrace}{ matrix{3}{1}{ c_1 c_2 c_3 } }{rbrace} = delim{lbrace}{ matrix{4}{1}{ 6 {-5} 0 6 } }{rbrace}$

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}.

|u| = sqrt{u · u} = sqrt{4}=2

|v| = sqrt{v · v} = sqrt{2}

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

Section 1.11.

√Problem 46.

Show that the set of equations

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.

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

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

.
Assume ,
⇒ .

$x = delim{lbrace}{ matrix{3}{1}{ {2 - b_2} {b_2} {1} } }{rbrace}$ .

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