厦门大学数理经济学midterm2011wise-solutions

Define a positive number
B
=
min{
1 2
y
,
y1 , . . . , yN1 }, then we have
yn
≥B
for all n.
Similar, there exist N2 > 0 such that
xn − x
<
1 2
Bε,
and N3
such
that
yn − y
A11 = (−1)2M11 =
21 12
8 −4
= −180
A12
= (−1)3M21
=−
6 12
5 −4
= 84
A13 = (−1)4M31 =
6 21
5 8
= −57
A21
= (−1)3M12
=−
6 4
8 −4
= 56
A22 = (−1)4M22 =
2 4
5 −4
= −28
A23
= (−1)5M23
And the inverse of A is
A−1 =
1
1 ⎛−180
adj A = − ⎜ 56
det A
84 ⎝ −12
84 −28
0
15
−57⎞ ⎛ 7
14
⎟
=
⎜⎜−
2 3
6 ⎠ ⎝1
7
−1
1 3
0
19
28 ⎞
1
−6
⎟ ⎟
−
1 14
⎠
Problem 2
(20 marks) For an m × n matrix A and a p × q matrix B, define A ⊗ B as
Answer:
First we examine the whether v1, v2 and v3 are linearly independent. Start with the equation c1v1 +
c2v2 + c3v3 = 0, that is
⎡⎢7 1 −5⎤⎥ ⎡⎢c1⎤⎥ ⎡⎢0⎤⎥
⎝4 12 −4⎠ exists.
Answer:
The (i, j)-th element of the adjoint matrix of A, say Aij, can be written as Aij = Cji = (−1)i+jMji, where Mji is the (j, i)th minor of A. Therefore,
⎢⎢3 ⎢
9
15
⎥ ⎥ ⎥
⎢⎢⎢c2⎥⎥⎥
=
⎢⎢0⎥⎥ ⎢⎥
⎢⎣5 0 −5⎥⎦ ⎢⎣c3⎥⎦ ⎢⎣0⎥⎦
so
⎡⎢7 1 −5⎤⎥
⎡⎢7 1 −5⎤⎥
⎡⎢1 3 5⎤⎥
rank ⎢⎢3 9
15⎥⎥ = rank ⎢⎢1 3
5
⎥ ⎥
=
rank
⎢⎢0
1
2⎥⎥ = 2
⎢
⎥
⎢⎣5 0 −5⎥⎦
⎢
⎥
⎢⎣1 0 −1⎥⎦
(b) This is to show that (int E)c = cl(Ec), i.e. int E = (cl (Ec))c. ((int E)c ⊂ cl(Ec)) If x ∈ (int E)c, we have x ∉ int E; then for any open sets B contained in E, x ∉ B. That is, x ∈ Bc for all closed set Bc ⊃ Ec. Since cl (Ec) is the intersection of all Bc, it must be that x ∈ cl(Ec). (cl(Ec) ⊂ (int E)c) If x ∈ cl (Ec), then x belongs to all closed set T that contains Ec, thus x ∉ T c for all closed set T ⊃ Ec. Since T c are open sets that satisfy T c ⊂ E, and int E is the union of all open sets contained in E, x ∉ int E, that is, x ∈ (int E)c.
Problem 5
(15 marks) For any positive real number a, prove that the limit of {a1 n}∞ n=1 is 1.
Proof:
We can see that
(1) When a = 1, a1 n is always 1.
(2) When 0 < a < 1 (ln a < 0), take N =
⎝ −12 0 6 ⎠
the determinant of A can be calculated directly or as follows
det A = a11A11 + a12A21 + a13A31 = 2 ⋅ (−180) + 6 ⋅ 56 + 5 ⋅ (−12) = a21A12 + a22A22 + a23A32 = 6 ⋅ 84 + 21 ⋅ (−28) + 6 ⋅ 0 = a31A13 + a32A23 + a33A33 = 4 ⋅ (−57) + 12 ⋅ 14 + (−4) ⋅ 6 = −84
1 n
ln
a
− 1 < exp
1 n
(−
ln
a)
− 1 < exp
−
ln(1+ε) ln a
⋅
(− ln a)
−1
= exp ln(1 + ε) − 1 = 1 + ε − 1 = ε
that
is,
when
n
>
−
ln a ln(1+ε)
,
a1 n − 1
< ε.
(3)
When
a
>
1,
take
N
=
[
ln a ln(1+ε)
[w1, w2] = [v1, v2, v3]A, rank(A) = 2
where A is a 3 × 2 matrix of constants, and its columns does not take the form e1, e2 or e3. Examples of w1 and w2 can be v1 + v2, v2 − v3, etc.
2
Proof:
Since lim yn = y, there exists N1 > 0 such that when n > N1,
yn − y
<
1 2
y
,
then
for
n
≥
N.
Since
y
=
yn + y − yn
≤
yn
+ yn − y
<
yn
+
1 2
y
,
we
have
yn
>
1 2
y
for all n ≥ N1.
=−
2 6
5 8
= 14
A31 = (−1)4M12 =
6 4
21 12
= −12
A32
= (−1)5M23
=−
2 4
6 12
=0
A33 = (−1)6M33 =
2 6
6 21
=6
So the adjoint matrix of A is
⎛−180 84 −57⎞ adjA = ⎜ 56 −28 14 ⎟
ln a
− ln(1+ε)
+ 1,
then
when
n > N,
1 n
<
−
ln(1+ε) ln a
,
so
a1 n − 1 = exp
1 n
ln
a
− 1 > exp
−
ln(1+ε) ln a
⋅
ln
a
− 1 = exp
− ln(1 + ε)
−1
1
ε
=
− 1 = − > −ε
1+ε
1+ε
and
a1 n − 1 = exp
3
Proof:
By definition, intts contained in E; cl E is the intersection of all closed sets containing E.
(a) (E is open ⇒ int E = E) For any x ∈ E where E is an open set, we can find an ε > 0 such that the open set Bε(x) centered at x is completely contained in E thus by definition Bε(x) ⊂ int E. So E ⊂ int E. On the other hand, for any x ∈ int E, we can find an open set B ⊂ int E such that x ∈ B. Since int E contains all open subsets of E, B ⊂ E, we obtain x ∈ E, so int E ⊂ E. Combining these results, for an open set E, int E = E. (E is open ⇐ int E = E) Since int E = E, any x ∈ E also satisfies x ∈ int E. Since x ∈ int E, we can find an open set B ⊂ int E such that x ∈ B ⊂ E. Since B is open, we can find the open set Bε(x) centered at x with radius ε is completely contained in B and Bε(x) ⊂ E. Since for all x ∈ E such ε exists, E is an open set.
1+ε
1+ε
and
a1 n − 1 = exp
1 n
ln
a
− 1 < exp
ln(1+ε) ln a
⋅
ln
a
−1
= exp ln(1 + ε) − 1 = 1 + ε − 1 = ε
that
is,
when
n
>
ln a ln(1+ε)
,
a1 n − 1
< ε.
Problem 6
Let int E be the interior of a set E. Prove that (a) (15 marks) E is open if and only if int E = E. (b) (15 marks) The complement of int E is the closure of the complement of E.
Mathematical Economics
Answer Key to Mid-Term
Nov 13, 2011 † Alternative answers are correct as well.
Problem 1
⎛2 6 5 ⎞ (20 marks) A = ⎜6 21 8 ⎟. Calculate the adjoint matrix of A and the inverse matrix of A if it
⎢
⎥
⎢⎣0 0 0⎥⎦
therefore v1, v2 and v3 are linearly dependent, with c1 = c3, c2 = −2c3, so
v1 − 2v2 + v3 = 0
So we can select two linear combinations of v1, v2, v3, say w1 and w2, as a basis for W such that
Alternative answer: e1, e2, e3 can also form a basis (but less efficient than the one formed by w1 and w2).
Problem 4
(20 points) If {xn} has a limit of x and {yn} has a limit of y, where yn and y are nonzero, prove that {xn yn} has a limit of x y.
⎛a11BT ⋯
AT ⊗ BT = ⎜ ⋮
⋮ ⎟ ⊗ BT = ⎜ ⋮
⎝a1n ⋯ amn⎠
⎝a1nBT ⋯
⎛ a11B
⋯
a1n
B
T
⎞
=⎜ ⋮
⋮ ⎟ = (A ⊗ B)T
⎝am1B ⋯ amnB⎠
am1BT ⎞
⋮⎟ amnBT ⎠
Problem 3
(20 points) In R3 space, let v1 = (7, 3, 5)T , v2 = (1, 9, 0)T , and v3 = (−5, 15, −5)T . And W = L[v1, v2, v3] denotes the set spanned by v1, v2, v3. Please find a basis for the set W, which contains no v1, v2 or v3.
]+1
(or
take
0
<
ε
<
1,
N
=
[
ln a ln(1−ε)
]
+
1),
then
when
n
>
N,
1 n
<
ln(1+ε) ln a
,
so
a1 n − 1 = exp
1 n
ln
a
− 1 > exp
−
1 n
ln a
− 1 = exp
ln(1+ε) ln a
⋅
(−
ln
a)
−1
1
ε
= exp − ln(1 + ε) − 1 = − 1 = − > −ε
<
y 2x
Bε,
then
when
n
>
max{N1,
N2,
N3}
xn x xny − ynx (xn − x) ⋅ y + x ⋅ (y − yn) xn − x x yn − y
−=
=
=
+⋅
yn y
yn ⋅ y