Solutions to Problems – 3
MAT 2355.
16. a) Express the translation of R2 defined by f (x, y) = (x 3, y + 4) as a product
of two reflections, and sketch your solution (with each line labelled) on a set of labelled
axes.
Solution: Since f (v) = v + ( 3, 4), if we set
L = {(x, y) |
3x + 4y = 0}
and
K = {(x, y) |
3x + 4y =
25
}
2
we know that f = RK RL . (You can sketch these yourselves.)
b) Express the rotation ⇢ about (1, 1) by ⇡2 in the positive sense as a product of two
reflections in lines, and sketch your solution (with each line labelled) on a set of
labelled axes.
Solution: If we set
L = {(x, y) | x
y = 0}
and
K = {(x, y) | x = 1}
we know that f = RK RL . (You can sketch these yourselves.)
c) Let P = (0, 0), Q = (0, 1), R = (1, 0) and S = (1, 1)).
(i) Find an orientation preserving isometry f such that f (P ) = R and f (P Q) = RS.
Solution: Make a sketch of these segments. Then note that if we reflect in the line
L = {(x, y) | y = 0}, RL (P ) = P , and RL (Q) = (0, 1). If we then follow this by a
reflection in K = {(x, y) | x = 12 }, RK (P ) = R and RK (0, 1) = (1, 1) = S. Hence if we
define f = RK RL , f (P ) = R and f (Q) = S. Hence, by a previous exercise, g(P Q) = RS.
Note that since f is a product of two reflections, f is is orientation preserving.
(ii) Find an orientation reversing isometry g such that g(P ) = R and g(P Q) = RS.
Solution: Make a sketch of these segments. Then note that if we reflect in the line
L = {(x, y) | y = 0}, RL (P ) = P , and RL (Q) = (0, 1).
If we then follow this by the translation by t(v) = v +(1, 0), t(P ) = R and t(0, 1) = S.
Hence if we define g = tRL , i.e. g(v) = RL (v)+(1, 0), g(P ) = R and g(Q) = S. Hence, by
a previous exercise, g(P Q) = RS. Note that g is a glide reflection (since (1, 0) is parallel
to L), g is orientation reversing.
17. Let f : R2 ! R2 be defined by f (x, y) = ( y, x + 4).
a) Find a 2 ⇥ 2 matrix A and b 2 R2 , and write the formula for f as f (v) = Av + b.
Use this to show that f is an isometry.
b) Is f orientation preserving or orientation reversing?
c) Is f a translation, a rotation, a reflection or a glide reflection?
d) Give details for your response to (c).





x
0
1
x
0
0
Solution: (a) Note that f (
)=
+
. Since A =
y
1 0
y
4
1
t
2
satisfies A A = A = I2 , A is orthogonal and so f is an isometry
b) Since det A =
1
0
1, f reverses orientation.




0
0
4
0
c) We first compute f (f (v)). But f (f (v)) = A(Av+
)+
= A2 v+
+
=
4
4
0
4

4
v+
.
4


4
2
1
2
Since f 6= id, f is a glide reflection with glide vector v0 = 2
=
.
4
2
To find the line of reflection involved,
we set g(v) = f (v) v0 and find L = {v 2

2
2
R 2 | g(v) = v}: But g(v) = Av +
, so we have the system (A I2 )v =
. The
2
2

1 1 | 2
augmented matrix of this system reduces to
, and so
0 0 | 0
L = {(x, y) | x + y = 2}.
(Note that L is indeed parallel to v0 .)
p
2
18. Let ↵ =
, and f : R2 ! R2 be defined by f (x, y) = (↵x ↵y 2↵+1, ↵x+↵y 1).
2
Note that 2↵2 = 1.
a) Find a 2 ⇥ 2 matrix A and b 2 R2 , and write the formula for f as f (v) = Av + b.
Use this to show that f is an isometry.
b) Is f orientation preserving or orientation reversing?
c) Is f a translation, a rotation, a reflection or a glide reflection?
d) Give details for your response to (c).





x
↵
↵
x
1 2↵
↵
Solution: (a) Note that f (
)=
+
. Since A =
y
↵ ↵
y
1
↵
 2
2↵
0
satisfies At A =
= I2 , A is orthogonal and so f is an isometry.
0
2↵2
↵
↵
b) Since det A = 2↵2 = 1, f preserves orientation.
c) Since A 6= I2 , f is not a translation and so is a rotation. Solving cos ✓ = ↵, sin ✓ = ↵
for ✓ 2 [0, 2⇡)), we obtain ✓ = ⇡4 .
To find the center c of the rotation, we find c such that f (c) = c: this equation is


1 2↵
2↵ 1
(A I)c =
=
.
1
1
Then

2↵ 1
c = (A I) 1
1

1
↵ 1
↵
2↵ 1
=
↵
↵ 1
1


1
1
a b
d
b
Since
=
, and det(A I) = (↵ 1)2 + ↵2 = 2↵2 2↵ + 1 =
c d
c
a
ad bc
2 2↵, we have
c=
1
2
2↵
1


↵
1
↵
↵ ↵ 1

2↵
1
(↵ 1)(2↵ 1) + ↵
↵(2↵ 1) + ↵ 1
2 2↵

1
2↵2 2↵ + 1
=
2↵2 + 2↵ 1
2 2↵

1
2 2↵
=
2 2↵ 2↵ 2

1
=
.
1
=
Hence, f is the rotation about (1, 1) by
positive mathematical sense).
⇡
4
1
in the anticlockwise sense (i.e., the
2
0 1
21. Let A = 4 1 0
0 0
3
0
0 5 and consider the map f : R3 ! R3 defined by f (v) = Av.
1
a) Show that f is an isometry, but is not a reflection. (You may not use part (d)).
b) Find the set F = {v 2 R3 | f (v) = v} of fixed points of f , and show that F is a line
through the origin.
c) Show that e3 = (0, 0, 1) is perpendicular to every vector in F .
d) Indeed, f is a rotation about the line F . Find the angle of rotation by considering
the vectors e3 and f (e3 ).
Solution: (a) A short computation shows that At A = I3 , so A is orthogonal. (Or
simply note that the columns form an orthonormal basis is R3 .) Thus f is an isometry.
Since det A = 1, f preserves orientation and so f is not a reflection. (See Q. 24.)
b)
2
1
4
But A I =
1
0
indeed a line through
F = {v 2 R3 | f (v) = v}
= {v 2 R3 | Av = v}
= {v 2 R3 | (A I)v = 0}
3
2
3
1
0
1
1 0
1 0 5 ⇠ 4 0 0 1 5, so F = span{(1, 1, 0)}, which is
0
2
0 0 0
the origin (as 6= (0, 0, 0)).
c) Clearly, e3 · (t, t, 0) = (0, 0, 1) · (t, t, 0) = 0 for all t 2 R, so e3 is perpendicular to
every vector in F .
d) Since f is a linear isometry, f preserves the lengths of vectors, so kf (e3 )k = ke3 k = 1.
Thus the angle between e3 and f (e3 )) is arccos(e3 · f (e3 ) = arccos( 1) = ⇡. So f is
the rotation about F by ⇡. (The sense is irrelevant since the angle of rotation is ⇡.)
Remark: Looking at the RRE form of A I above, we see that F is the intersection
of the planes H and K with equations x y = 0 (say, ) and z = 0 respectively. These
planes are perpendicular (since their normals are). If you compute the formula for RK RH ,
you’ll find it is f : i.e. the product of the two reflections is a rotation about their line
of intersection through twice the angle between the planes! This is true in general when
the planes intersect, and is thus seen to be a generalization of the same result in two
dimensions.
22. Let T1 be the triangle whose vertices are
(1,
1), (1, 4) and (4, 4) and T2 the triangle
p
p
p
whose vertices are (0, 0), ( 3 2, 0) and ( 3 2 2 , 3 2 2 ). Find an isometry f of R2 such that
f (T1 ) = T2 .
Solution: First, we move A = (1, 1) to D = (0, 0) by reflection RL (x, y) = (1
y, 1 x) in the line L = {(x, y) | x + y = 1}.
p
Then, C 0 := RL (C) = RL (4, 4) = ( 3, 3), which we wish to move to E = ( 3 2, 0),
while keeping the origin D fixed. (Draw yourself a picture to see why.) This we can do
by reflection in the line K that is equidistant from C 0 and E, since D is also equidistant
from C 0 and E. So we know that RK (C 0 ) = RK (E) and RK (D) = D.
But a short calculation shows that F and B 0 := RL (B) = ( 3, 0) are also equidistant
from C 0 and E, so RK (B 0 ) = F .
Thus, if we set f = RK RL , then f (A) = RK RL (A) = RK (D) = d, f (B) =
RK RL (B) = RK (B 0 ) = F and f (C) = RK RL (C) = RK (C 0 ) = E.
Since f is a product of 2 reflections in lines that do intersect, f is a rotation in this
0
case. (How do we know L and K intersect? We know that 0 2 K and C 2+E 2 K, so the
p
0
vector C 2+E ) = 32 (1 + 2, 1) is actually a direction vector for K, and this is clearly not
parallel to (1, 1), a direction vector for L.)
23. If H and K are perpendicular planes in R3 , show that RH (K) = K, where RH
denotes the reflection in H.
Solution: Suppose aH , aK are (unit) normals to H and K respectively and suppose
bH , bK 2 R so that H = {v 2 R3 | v · aH = bH } and K = {v 2 R3 | v · aK = bK }. Let
H 0 = {v 2 R3 | v · aH = 0}. Then, 0 2 H 0 , so RH 0 (v) · RH 0 (w) = v · w for all v, w 2 R3 ,
since RH 0 is a linear isometry.
Moreover, RH (v) = v + 2(bH aH · v)aH = RH 0 (v) + 2bH aH . In particlular, since
aH · aK = 0, RH 0 (aK ) = aK . Hence, if k 2 K,
RH (k) · aK = (RH 0 (k) 2bH aH ) · aK
= RH 0 (k) · aK
= RH 0 (k) · RH 0 (aK )
= k · aK
= bK ,
so RH (k) 2 K. Thus RH (K) ✓ K, and one can see that equality holds by applying RH
to both sides of this set equation.
24. Show that the reflection RH in any hyperplane H ⇢ Rn reverses orientation.
Solution: It suffices to show this for a hyperplane through the origin, so let 0 6=
a 2 Rn be a normal vector for H so that H = {v 2 Rn | a · v = 0}. Note that H is
a subspace of Rn of dimension n 1, and that if {v1 , . . . , vn 1 } is a basis for H, then
{a, v1 , . . . , vn 1 } is a basis for Rn . Note that RH (a) = a and that RH fixes each of
v1 , . . . , vn 1 . Hence we may compute:
det [ RH a RH v1
...
RH vn
1
] = det [ a v1
and so the reflection RH reverses orientation.
...
vn
1
]=
det [ a v1
...
vn
1
],