Linear Algebra Exercises-n-Answers.pdf

72 Linear Algebra, by Hefferon
-a
T
D
a
R
a
+
b
T
D
b
R
-b
+
c
T
D
-c
R
c
=
-a+b+c
3 (a) A two-voter election can have a majority cycle in two ways. First, the two voters could be
opposites, resulting after cancellation in the trivial election (with the majority cycle of all zeroes).
Second, the two voters could have the same spin but come from different rows, as here.
−1 voter
1 ·
T
D
R
1 voter
1 voter
1 voter
+ 1 ·
T
D
R
1 voter
−1 voter
1 voter
+ 0 ·
T
D
R
T
−1 voter
D
R
a+b-c
1 voter
a-b+c
=
0 voters
T
D
R
2 voters
0 voters
(b) There are two cases. An even number of voters can split half and half into opposites, e.g., half
the voters are D > R > T and half are T > R > D. Then cancellation gives the trivial election. If
the number of voters is greater than one and odd (of the form 2k + 1 with k > 0) then using the
cycle diagram from the proof,
-a
T
D
a
R
a
+
b
T
D
b
R
-b
+
c
T
D
-c
R
c
=
-a+b+c
T
D
R
a+b-c
a-b+c
we can take a = k and b = k and c = 1. Because k > 0, this is a majority cycle.
4 This is one example that yields a non-rational preference order for a single voter.
character experience policies
Democrat most middle least
Republican middle least most
Third least most middle
The Democrat is preferred to the Republican for character and experience. The Republican is preferred
to the Third for character and policies. And, the Third is preferred to the Democrat for experience
and policies.
5 First, compare the D > R > T decomposition that was done out in the Topic with the decomposition
of the opposite T > R > D voter.
⎛
⎝ −1 ⎞ ⎛
1 ⎠ = 1 3 · ⎝ 1 ⎞ ⎛
1⎠ + 2 3 · ⎝ −1 ⎞ ⎛
1 ⎠ + 2 3 · ⎝ −1 ⎞ ⎛
0 ⎠ and ⎝ 1 ⎞ ⎛
−1⎠ = d 1 · ⎝ 1 ⎞ ⎛
1⎠ + d 2 · ⎝ −1 ⎞ ⎛
1 ⎠ + d 3 · ⎝ −1 ⎞
0 ⎠
1 1 0
1
−1 1
0
1
Obviously, the second is the negative of the first, and so d 1 = −1/3, d 2 = −2/3, and d 3 = −2/3. This
principle holds for any pair of opposite voters, and so we need only do the computation for a voter
from the second row, and a voter from the third row. For a positive spin voter in the second row,
c 1 − c 2 − c 3 = 1
c 1 + c 2 = 1
c 1 + c 3 = −1
−ρ 1+ρ 2 (−1/2)ρ 2+ρ 3
−→
−ρ 1+ρ 3
−→ 2c 2 + c 3 = 0
c 1 − c 2 − c 3 = 1
(3/2)c 3 = −2
gives c 3 = −4/3, c 2 = 2/3, and c 1 = 1/3. For a positive spin voter in the third row,
c 1 − c 2 − c 3 = 1
c 1 + c 2 = −1
c 1 + c 3 = 1
gives c 3 = 2/3, c 2 = −4/3, and c 1 = 1/3.
−ρ 1 +ρ 2 (−1/2)ρ 2 +ρ 3
−→
−ρ 1+ρ 3
−→ 2c 2 + c 3 = −2
c 1 − c 2 − c 3 = 1
(3/2)c 3 = 1
6 It is nonempty because it contains the zero vector. To see that it is closed under linear combinations
of two of its members, suppose that ⃗v 1 and ⃗v 2 are in U ⊥ and consider c 1 ⃗v 1 + c 2 ⃗v 2 . For any ⃗u ∈ U,
and so c 1 ⃗v 1 + c 2 ⃗v 2 ∈ U ⊥ .
(c 1 ⃗v 1 + c 2 ⃗v 2 ) ⃗u = c 1 (⃗v 1 ⃗u) + c 2 (⃗v 2 ⃗u) = c 1 · 0 + c 2 · 0 = 0
Topic: Dimensional Analysis
1 (a) This relationship
(L 1 M 0 T 0 ) p 1
(L 1 M 0 T 0 ) p 2
(L 1 M 0 T −1 ) p 3
(L 0 M 0 T 0 ) p 4
(L 1 M 0 T −2 ) p 5
(L 0 M 0 T 1 ) p 6
= L 0 M 0 T 0