Multiple View Geometry: Solution Sheet 3
Prof. Dr. Florian Bernard, Florian Hofherr, Tarun Yenamandra
Computer Vision Group, TU Munich
Link Zoom Room , Password: 307238
Exercise: May 12th, 2021
Part I: Theory
1. (a) M =
(
I T
0 1
)
=
1 0 0 tx
0 1 0 ty
0 0 1 tz
0 0 0 1
(b) M =
(
R 0
0 1
)
=
r11 r12 r13 0
r21 r22 r23 0
r31 r32 r33 0
0 0 0 1
(c) M =
(
I T
0 1
)(
R 0
0 1
)
=
(
R T
0 1
)
=
r11 r12 r13 tx
r21 r22 r23 ty
r31 r32 r33 tz
0 0 0 1
(d) M =
(
R 0
0 1
)(
I T
0 1
)
=
(
R RT
0 1
)
=
r11 r12 r13 r1T
r21 r22 r23 r2T
r31 r32 r33 r3T
0 0 0 1
,
where r1, r2, r3 are the row vectors of R: R =
— r1 —— r2 —
— r3 —
.
2. Let M := (M1 −M2) =:
m11 m12 m13m21 m22 m23
m31 m32 m33
.
”⇒”:
We show that M is skew-symmetric by distinguishing diagonal and off-diagonal elements of
M :
(a) ∀i : 0 = e>i Mei = mii where ei = i-th unit vector
(b) ∀i 6= j : 0 = (ei + ej)>M(ei + ej) where ej = j-th unit vector
= mii +mjj +mij +mji ⇒ mij = −mji
hence, mii = 0 and mij = −mji, i.e. M is skew-symmetric.
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https://tum-conf.zoom.us/s/62772800235?pwd=SUpZN2QrV0JpeXJyR2R1TWx5cHEwdz09
”⇐”:
using M = −M>, we directly calculate
∀x : x>Mx = (x>Mx)> = x>M>x = −(x>Mx)
⇒ x>Mx = 0
Alternative for ”⇐”:
∀x : x>Mx = x>(M̌ × x) = 0
Because M is skew-symmetric, Mx can be interpreted as a cross product. The result of any
cross product with x is orthogonal to x.
3. We know: ω = (ω1 ω2 ω3)> with ||ω|| = 1 and ω̂ =
0 −ω3 ω2ω3 0 −ω1
−ω2 ω1 0
.
(a)
ω̂2 =
−(ω22 + ω23) ω1ω2 ω1ω3ω1ω2 −(ω21 + ω23) ω2ω3
ω1ω3 ω2ω3 −(ω21 + ω
2
2)
=
ω21 − (ω
2
1 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
ω1ω2 ω1ω3
ω1ω2 ω
2
2 − (ω
2
2 + ω
2
1 + ω
2
3)︸ ︷︷ ︸
1
ω2ω3
ω1ω3 ω2ω3 ω
2
3 − (ω
2
3 + ω
2
1 + ω
2
2)︸ ︷︷ ︸
1
=
ω21 ω1ω2 ω1ω3ω1ω2 ω22 ω2ω3
ω1ω3 ω2ω3 ω
2
3
−
1 0 00 1 0
0 0 1
= ωω> − I
ω̂3 = ω̂ ω̂2
= ω̂ (ωω> − I)
= ω̂ ω (ω>)− ω̂I
= (ω × ω) ω> − ω̂
= −ω̂ (as ω × ω = 0)
2
Alternative solution for ω̂3:
ω̂3 =
−(ω22 + ω23) ω1ω2 ω1ω3ω1ω2 −(ω21 + ω22) ω2ω3
ω1ω3 ω2ω3 −(ω21 + ω
2
2)
·
0 −ω3 ω2ω3 0 −ω1
−ω2 ω1 0
=
0 ω3 · (ω21 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
−ω2 · (ω21 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
−ω3 · (ω21 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
0 ω1 · (ω21 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
ω2 · (ω21 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
−ω1 · (ω21 + ω
2
2 + ω
2
3)︸ ︷︷ ︸
1
0
= −ω̂
(b) The formulas for n even and odd can be found by writing down the solutions for n =
1, . . . , 6:
ω̂
ω̂2
ω̂3 = −ω̂
ω̂4 = −ω̂2 as: ω̂4 = ω̂3ω̂ = −ω̂ω̂ = −ω̂2
ω̂5 = ω̂ as: ω̂5 = ω̂4ω̂ = −ω̂2ω̂ = −ω̂3 = −(−ω̂) = ω̂
ω̂6 = ω̂2 as: ω̂6 = ω̂5ω̂ = ω̂ω̂ = ω̂2
For even numbers: ω̂2
ω̂4 = −ω̂2
ω̂6 = ω̂2
For odd numbers: ω̂
ω̂3 = −ω̂
ω̂5 = ω̂
even: ω̂2n = (−1)n+1 ω̂2 for n ≥ 1
odd: ω̂2n+1 = (−1)n ω̂ for n ≥ 0
Proof via complete induction:
i. For even numbers 2n where n ≥ 1:
– n = 1 : ω̂2 = (−1)2ω̂2
– Induction step n→ n+ 1 :
ω̂2(n+1) = ω̂2n · ω̂2
= (−1)n+1 · ω̂2 · ω̂2 (assumption)
= (−1)n+1 · ω̂3 · ω̂
(a)
= (−1)(n+1)+1 · ω̂2
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ii. For odd numbers 2n+ 1 where n ≥ 0:
– n = 0 : ω̂1 = (−1)0ω̂
– Induction step n→ n+ 1 :
ω̂2(n+1)+1 = ω̂2n+1 · ω̂2
= (−1)n · ω̂ · ω̂2 (assumption)
= (−1)n · ω̂3
(a)
= (−1)n+1 · ω̂
(c) We know: ω ∈ R3. Let ν = ω‖ω‖ and t = ‖ω‖. Hence, w = νt, ω̂ = ν̂t.
eω̂ = eν̂t
=
∞∑
n=0
(ν̂t)n
n!
= I +
∞∑
n=1
t2n
(2n)!
ν̂2n +
∞∑
n=0
t2n+1
(2n+ 1)!
ν̂2n+1
(b)
= I +
∞∑
n=1
(−1)n+1
t2n
(2n)!︸ ︷︷ ︸
1−cos(t)
ν̂2 +
∞∑
n=0
(−1)n
t2n+1
(2n+ 1)!︸ ︷︷ ︸
sin(t)
ν̂
(def.)
= I +
ω̂2
‖ω‖2
(1− cos(‖ω‖)) +
ω̂
‖ω‖
sin(‖ω‖)
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