Math 2270 Spring 2004
Homework 5 Solutions
Section 2.2) 1, 2, 6, 7, 8, 10, 11, 17, 24ab, 27, 43, 49
(1) The image is a parallelogram with vertices (0, 0), (3, 1), (4, 3) and (1, 2).
(2) We can use the formula derived in class (Fact 2.2.2).
"
cos α
sin α
− sin α
cos α
#
=
"
#
√
− 3/2
1/2
√1/2
3/2
(6) First, we must find a unit vector on the line L. To do this we determine the length of
the given vector.


2 √
√


22 + 12 + 22 = 4 + 1 + 4 = 3
 1  =
2 Then, we define the unit vector, ~u.
~u =


2


 1 
2
1
3
Now, we calculate the orthogonal projection of the given vector, ~v , onto L.

  


 
2/3
1
2/3
  
 
 
2
projL~v =   1/3  ·  1    1/3 
 = 3+
2/3
1
2/3
1
3
+
2
3


1
9


2/3


 1/3  =
2/3


10


 5 
10
(7) We use the result from number (6).






11/9
20/9 − 1
1
10/9








ref L~v = 2 (projL~v) − ~v = 2  5/9  −  1  =  10/9 − 1  =  1/9 
11/9
20/9 − 1
1
10/9
(8) The transformation leaves the x1 -axis in place, suggesting that it is a shear parallel to
the x1 -axis. All vectors on the x1 -axis can be written as scalar multiples of the standard
vector ~e1 , or as k~e1 where k is a scalar. We must test the two properties of shears (Def.
2.2.4). We see that (a) holds since T (k~e1 ) = k~e1 . For part (b) we see that
T (~x) − ~x =
"
1 −1
0 1
#
"
x1
x2
#
−
"
x1
x2
1
#
=
"
x1 − x2
x2
#
−
"
x1
x2
#
=
"
−x2
0
#
= −x2
"
1
0
#
Math 2270 Spring 2004
Homework 5 Solutions
The last vector is a scalar multiple of ~e1 and is therefore parallel to the line spanned by the
vector ~e1 . Therefore, this is a shear parallel to the x1 − axis.
(10) Here, we can either use the formula derived in class, or we can calculate the matrix
directly. In either case, we must
first find a unit vector on the line L. We see that the given
√
2
vector has a magnitude of 3 + 42 = 5. Thus, we can define the unit vector, ~u to be 1/5
times the given vector. Then, u1 = 4/5 and u2 = 3/5.
"
projL~v =
=



4/5
3/5
#
16
v
25 1
+
12
v
25 1
+
·
"
v1
v2
# ! "


12
v
25 2
9
v
25 2
=




#
4/5
3/5
=
4
v
5 1
16/25 12/25
12/25
9/25



+
3
v
5 2
"
v1
v2
"
4/5
3/5
#
#
(11) We can use the result from number (10).

ref L~v = 2 

16
v
25 1
12
v
25 1
+
+
12
v
25 2
9
v
25 2





− 

v1

= 


v2

7
v
25 1
24
v
25 1
+
−
24
v
25 2
7
v
25 2




= 

7/25
24/25
24/25 −7/25
 
 
 
v1
v2



(17) The unit square is transformed into the segment of the line x1 = x2 from the origin to
the point (2, 2). The transformation projects all vectors onto the line x1 = x2 . Using √
the
formula√
derived in class and the unit vector on the line x1 = x2 with components u1 = 1/ 2,
u2 = 1/ 2, the projection matrix is


1
2


1
2
1
2
1
2


If we multiply this matrix by 2 we get the matrix associated with the transformation.

2 

1
2
1
2
1
2
1
2




= 

1 1
1 1



= T (~x)
Therefore, the original transformation can be interpreted as a projection onto the line x1 = x2
followed by a dilation by a factor 2.
2
Math 2270 Spring 2004
Homework 5 Solutions
(24) The given transformation is a shear parallel to the x2 axis. The second column of the
given matrix is T (~e2 ) indicating that all scalar multiples of ~e2 will map to themselves. It
remains to check part (b) of Def. 2.2.4.
T (~x) − ~x =
"
# "
1 0
2 1
x1
x2
#
−
"
x1
x2
#
=
"
x1
2x1 + x2
#
−
"
x1
x2
#
=
"
0
2x1
#
= 2x1
"
0
1
#
The last vector is a scalar multiple of ~e2 and is therefore parallel to the line spanned by the
vector ~e2 . Therefore, this is a shear parallel to the x2 − axis.
(27) We must check the two properties of linear transformations given in Fact 2.2.1. Let F
be the composite transformation defined as F (~x) = L(T (~x)). First,
F (~v + w)
~ = L(T (~v + w))
~ = L(T (~v ) + T (w))
~
by the fact that T is a linear transformation. Then, since L is a linear transformation
L(T (~v) + T (w))
~ = L(T (~v)) + L(T (w))
~ = F (~v) + F (w)
~
Second, if k is a scalar,
F (k~v ) = L(T (k~v )) = L(kT (~v ))
since T is linear. Then, since L is linear,
L(kT (~v )) = kL(T (~v )) = kF (~v )
Therefore, the composite of T and L is a linear transformation as well.
(43) The inverse of a rotation by an angle α is just another rotation by the same angle, but
in the opposite direction, or by −α. Using the formula for rotation matrices we have
A
−1
=
"
cos (−α) − sin (−α)
sin (−α) cos (−α)
#
=
"
cos α sin α
− sin α cos α
#
(49) The transformation stretches the unit circle so that it becomes an ellipse with a horizontal width of 10 units and a vertical height of 4 units.
3