Multiple View Geometry: Solution Sheet 1
Prof. Dr. Florian Bernard, Florian Hofherr, Tarun Yenamandra
Computer Vision Group, TU Munich
Exercise: April 21st, 2021
Part I: Theory
1. Show for each of the following sets (1) whether they are linearly independent, (2) whether they
spanR3 and (3) whether they form a basis ofR3:
(a) B1 =
11
1
,
01
1
,
00
1
The set B1 (1) is linearly independent, (2) spansR3, (3) forms a basis ofR3.
This can be shown by building a matrix and calculating the determinant:
det
1 0 01 1 0
1 1 1
= 1 6= 0 .
As the determinant is not zero, we know that the vectors are linearly independent. Three
linearly independent vectors in R3 span R3. A set is a basis ofR3 if it is linearly indepen-
dent and spansR3, so B1 forms a basis.
(b) B2 =
21
0
,
11
0
The set B2 (1) is linearly independent, (2) does not span R3, (3) does not form a basis of
R
3.
Since the two vectors are not parallel, linear independence is given. To span R3, there are
at least three vectors needed. Hence, the set cannot be a basis either.
(c) B3 =
21
0
,
31
0
,
00
1
,
10
1
The set B3 (1) is not linearly independent, (2) spansR3, (3) does not form a basis ofR3.
In R3, there cannot be more than three independent vectors. Using e.g. the determinant,
one finds that any three of the four vectors form a basis of R3 and thus the four together
spanR3. Since they are not linearly independent, they cannot form a basis.
2. Which of the following sets forms a group (with matrix-multiplication)? Prove or disprove!
(a) G1 :=
{
A ∈ Rn×n| det(A) 6= 0 ∧A> = A
}
The set is not closed under multiplication, thus no group. To show this, one counter-
example is enough: choose n = 3 and
A =
1 2 32 0 4
3 4 5
∈ G1 , B =
1 0 00 2 0
0 0 3
∈ G1 : AB =
1 4 92 0 12
3 8 15
/∈ G1
1
Note:
You can also show that if G1 was a group, for any A,B ∈ G1, (AB)> = AB would have
to be true, but is not. This is equivalent to saying BA = AB would have to be true:
(AB)> = B>A> = BA
However, to show that there exist A and B in G1 for which AB 6= BA (which is an im-
portant step in the proof!), the easiest way again is to choose a concrete counter-example.
(b) G2 := {A ∈ Rn×n| det(A) = −1}
The set contains no neutral element, thus no group:
det(Idn) = 1 6= −1 ⇒ Idn /∈ G2
(c) G3 := {A ∈ Rn×n| det(A) > 0}
The set forms a group. The easiest way to show this is to show that G3 is a subgroup of
the general linear group GL(n). We simply need to show that for any two elements A,B
of G3, AB−1 is also in G3: 1 for A,B ∈ G3,
det(AB−1) = det(A)︸ ︷︷ ︸
>0
[det(B)]−1︸ ︷︷ ︸
>0
> 0 ⇒ AB−1 ∈ G3
Thus, G3 is a subgroup of GL(n) and hence a group.
3. Prove or disprove: There exist vectors v1, …,v5 ∈ R3 \ {0}, which are pairwise orthogonal,
i.e.
∀i, j = 1, …, 5 : i 6= j =⇒ 〈vi,vj〉 = 0
Assume there exist five pairwise orthogonal, non-zero vectors v1, …,v5 ∈ R3. In R3, there
are at most three linearly independent vectors. Thus, the vectors are linearly dependent, which
means
∃ ai :
5∑
i=1
aivi = 0 ,
with at least one ai 6= 0. Without loss of generality, assume that a1 = −1, resulting in
v1 = a2v2 + a3v3 + a4v4 + a5v5
As the vectors are assumed to be pairwise orthogonal, we can derive
||v1||2 = 〈v1,v1〉 =
= 〈v1, a2v2 + a3v3 + a4v4 + a5v5〉 =
= a2〈v1,v2〉+ a3〈v1,v3〉+ a4〈v1,v4〉+ a5〈v1,v5〉 =
= 0 + 0 + 0 + 0 = 0
⇒ v1 = 0 ,
which contradicts the assumption of pairwise orthogonal, non-zero vectors.
1See e.g. https://en.wikipedia.org/wiki/Subgroup test for a proof if this is not clear to you.
2