Math 417 homework 2 solutions
(Given only for problems that are not straightforward computation; contact
the instructor if you still have questions about the others.)
Section 2.1 problem 1
A transformation T : R3 → R3 is linear only if it satisfies T (~0) = ~0. The
transformation is not linear. For x1 = x2 = x3 = 0 we get y2 = 2, not y0 = 0.
However, the transformation can be written
0 2 0 x1 0
T (~x) = 0 1 0 x2 + 2 .
0 2 0 x3 0
Outside of algebra, adding a constant vector still counts as “linear”.
Section 2.1 problem 1
A linear transformation T : R3 → R3 must satisfy T (α · ~x) = α · ~x for all α ∈ R.
But here
1 −1 −2
2T (0) = 2 1 = 2
1 1 2
which is not the same as
1 2 −2
T (2 0) = T (0) = 4 .
1 2 2
Note: alternatively one could check that another property of linear transfor-
mations,
T (~x + ~y ) = T (~x) + T (~y ),
is not satisfied either.
1
This study source was downloaded by 100000869267694 from CourseHero.com on 07-12-2023 04:32:04 GMT -05:00
https://www.coursehero.com/file/3630/hw2sol/
, Section 2.1 problem 44
The transformation is linear, with matrix
0 −a3 a2
a3 0 −a1 .
−a2 a1 0
Section 2.2 problem 2
Counterclockwise rotation around the origin by an angle α is multiplication by
the matrix
cos α − sin α
sin α cos α
For α = 60◦ we get " √ #
1 3
√2
− 2
3 1
2 2
Section 2.2 problem 10
√ √
4
is a vector on the line. Its length is 42 + 32 = 25 = 5, so a unit vector
3
on the line is
u 4/5
~u = 1 = .
u2 3/5
The projection matrix is
2
u1 u1 u2 16/25 12/25 1 16 12
= = .
u1 u2 u22 12/25 9/25 25 12 9
The reflection matrix is
2
2u1 − 1 2u1 u2 7/25 24/25 1 7 24
= = .
2u1 u2 2u22 − 1 24/25 −7/25 25 24 −7
Section 2.2 problem 26
(a) The second vector is 4 times the first, so
4 0
A= = 4I.
0 4
(b) The line to project on is not given explicitly. However, the second vector
2
is the projection of the first, so it must lie on the line. A corresponding unit
0
vector is
u1 1
=
u2 0
2
This study source was downloaded by 100000869267694 from CourseHero.com on 07-12-2023 04:32:04 GMT -05:00
https://www.coursehero.com/file/3630/hw2sol/
(Given only for problems that are not straightforward computation; contact
the instructor if you still have questions about the others.)
Section 2.1 problem 1
A transformation T : R3 → R3 is linear only if it satisfies T (~0) = ~0. The
transformation is not linear. For x1 = x2 = x3 = 0 we get y2 = 2, not y0 = 0.
However, the transformation can be written
0 2 0 x1 0
T (~x) = 0 1 0 x2 + 2 .
0 2 0 x3 0
Outside of algebra, adding a constant vector still counts as “linear”.
Section 2.1 problem 1
A linear transformation T : R3 → R3 must satisfy T (α · ~x) = α · ~x for all α ∈ R.
But here
1 −1 −2
2T (0) = 2 1 = 2
1 1 2
which is not the same as
1 2 −2
T (2 0) = T (0) = 4 .
1 2 2
Note: alternatively one could check that another property of linear transfor-
mations,
T (~x + ~y ) = T (~x) + T (~y ),
is not satisfied either.
1
This study source was downloaded by 100000869267694 from CourseHero.com on 07-12-2023 04:32:04 GMT -05:00
https://www.coursehero.com/file/3630/hw2sol/
, Section 2.1 problem 44
The transformation is linear, with matrix
0 −a3 a2
a3 0 −a1 .
−a2 a1 0
Section 2.2 problem 2
Counterclockwise rotation around the origin by an angle α is multiplication by
the matrix
cos α − sin α
sin α cos α
For α = 60◦ we get " √ #
1 3
√2
− 2
3 1
2 2
Section 2.2 problem 10
√ √
4
is a vector on the line. Its length is 42 + 32 = 25 = 5, so a unit vector
3
on the line is
u 4/5
~u = 1 = .
u2 3/5
The projection matrix is
2
u1 u1 u2 16/25 12/25 1 16 12
= = .
u1 u2 u22 12/25 9/25 25 12 9
The reflection matrix is
2
2u1 − 1 2u1 u2 7/25 24/25 1 7 24
= = .
2u1 u2 2u22 − 1 24/25 −7/25 25 24 −7
Section 2.2 problem 26
(a) The second vector is 4 times the first, so
4 0
A= = 4I.
0 4
(b) The line to project on is not given explicitly. However, the second vector
2
is the projection of the first, so it must lie on the line. A corresponding unit
0
vector is
u1 1
=
u2 0
2
This study source was downloaded by 100000869267694 from CourseHero.com on 07-12-2023 04:32:04 GMT -05:00
https://www.coursehero.com/file/3630/hw2sol/
