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Homework 07 Solution
Note the following class changes:
Prof. Gennert will be away the week of Nov 4. To make up for lost class time, we will meet as
follows:
Mon 28
Oct.
4:30-7:20pm with breaks. HW #7 due.
Wed 30
Oct.
4:30-5:50pm as usual.
Mon 4 Nov.
No class. HW #8 due.
Wed 6 Nov.
No class.
Mon 11
Nov.
4:30-7:20pm with breaks. HW #9 due.
Wed 13
Nov.
Resume usual schedule.
⃗⃗
1. Singular Value Decomposition (4 pts): This is an important tool that allows us to solve
systems of equations
⃗ ≈ 0,
| ⃗|
= 1. This is useful in camera calibration, where we get
such a system of equations and want a non-zero parameter vector ⃗. See Szeliski for an
introduction to SVD.
You can think of the SVD as follows:
⃗ in the
direction, scales it by
1. It takes the component of input
and
1
direction.
1
1
outputs it in the
,
2. Repeat for all
, and .
3. The output vector is the sum of all the contributions
= [ |
] =
1
[1 |
1
0
]=[2 0], =[
] =
1
[
1
1
]
], =[1
1
a. Suppose that 2×2 matrix M has singular value decomposition
1
2
1 −1
0
0 1 1
2
1 −1
√2
1√2
2
Without computing M, predict the value of ⃗ for ⃗ = [1] and ⃗ = [−1] based on
1
⃗, and finally, so does ⃗. Now compute M and use it to verify [1] and
the SVD properties. Hint: Note that in both cases
⃗ has a simple form, so does
1
[−1].
b. Let matrix have singular value decomposition =. Show that matrix
has SVD= 2
.
2. Camera Calibration (10 pts): One restriction in camera calibration is that the world points
chosen must not lie in a single plane, that is, they cannot be ⃗co-,planar,1≤ ≤otherwise calibration
will fail. To see this, suppose that there are K world points. We know that we
a single plane represented by ̃ = ( , , , ), defined by ̃
∙ ̃ = 0 (Szeliski, eqn.
2.7).
need at least 6 points for calibration, ≥ 6. Consider what happens if all world points lie in
Image Plane
Plane containing
·
1
1
·
·
·
⃗⃗
The calibration equation is given by ⃗⃗⃗ = 0, written out as
1
0
0
0
0
0
1 1
1 1
1 1
1
1
1
1
1
1
2
0
0
0
1
1
⋮
]
0
⋮
0
⋯
1
1
1 1
⋮1 1
1 1
1 [
0
12
⃗⃗
⃗⃗
⃗⃗⃗ that we
Ideally, there should be a single vector ⃗⃗⃗ such that ⃗⃗⃗ = 0 (or ≈ 0). This is the
hope to find through the singular value decomposition of . The key concept here is the rank
columns. For this
⃗⃗⃗
to exist and be unique,
must have rank = 11 (or 12 with a very small
of a matrix, which is the number of independent rows, also the number of independent
value for the smallest singular value 12. Show that if all world points are co-planar, then
⃗⃗⃗cannot= 0⃗⃗have rank greater than 9 by finding 3 independent non-zero vectors ⃗⃗⃗ such that
. (These independent vectors establish that the nullspace of has rank at least 3;
hence ⃗⃗⃗’s=rank0⃗⃗ cannot exceed 12-3=9.) In this case, it is not possible to find a unique ⃗⃗⃗ such
that and the calibration ⃗⃗⃗ procedure fails.
Hint: The non ̃-zero∙ ̃ =vectors0 are mostly 0s. Use the fact that if all world points are co-planar, then .
3. Focus of Expansion (8 pts): Suppose that the viewer (camera) is moving. We can model this
⃗
⃗
| |
in the imaging equations as
⃗
⃗
⃗⃗
⃗
=
2
=
+ and
⃗
⃗
⃗⃗
⃗⃗
⃗⃗
∙
by letting R and depend on time, = ( ), = ( ). Assume that the camera is
⃗⃗
⃗⃗
translating with velocity
≡
( ) and that there is no rotation,
( ) = 0 . As we know
from experience, points in the image will seem to move. As we saw in class, image points
⃗⃗
⃗
⃗
⃗⃗
⃗
≡
( ) =
1
((
×
)× ).
will appear to move with velocity
⃗
⃗
∙
⃗⃗
The Focus of Expansion (FOE) is the point in the image toward which the camera appears to be
moving, given by the projection of
⃗
⃗⃗
| |
2
into the image. That is,
⃗
FOE =
⃗
⃗⃗
.
⃗
∙
⃗⃗
⃗
⃗
⃗
Show that for any world point
, its image pointwill appear to move with velocity
= (
−
FOE),
proving that image points appear to move toward or away from the FOE. What is k?
