Ec402 Problem Set 2 Linear Algebra

Ec402
Problem Set 2
Linear Algebra
Alex Moore
LSE
October 2014
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
1 / 13
Question 1
A0 A = 0 if every element is zero
A is nxm so A0 A is an mxm square matrix
Consider the diagonals:
0 n
∑ a2
B k =1 k 1
B
..
B
.
B
A0 A = B
..
B
.
B
@
1
n
2
∑ akn
k =1
2 + a2 + ... + a2 = 0
Take the …rst diagonal: a11
21
n1
Hence ak 1 = 0 for all k
C
C
C
C
C
C
C
A
The same applies for all diagonal elements
Hence aki = 0 for all k and all i
Hence A = 0
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
2 / 13
Question 2(a)
(a)
Recall de…nition of positive de…nite:
I¤ A is positive de…nite, x 0 Ax > 0 for any x 6= 0
Use proof by contradiction.
Suppose A is both positive de…nite and singular...
This means x 0 Ax > 0 (positive de…nite) and Ax = 0 (singular) for
some x 6= 0
As Ax = 0, pre-multiplying by x 0 gives x 0 Ax = x 0 0 = 0
Contradicts claim that A is p.d.
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
3 / 13
Question 2(b)
?
(b) "If X is full column rank =) X’X p.d.?"
Full column rank: The columns are linearly independent (they are not
linear combinations of each other - e.g. dummy variable trap)
Positive de…nite: X 0 X is positive de…nite if Z 0 (X 0 X )Z > 0 for Z 6= 0
Z 0 (X 0 X )Z
= (Z 0 X 0 )XZ
= (XZ )0 XZ
= a0 a where a is nx1 vector
= ∑n ai2 0
(Because a is nx1, a0 a is a scalar)
Hence X 0 X is positive semi-de…nite...
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
4 / 13
Question 2(b)
But since X has full column rank, XZ = 0 only if Z = 0
because XZ is a linear combination of the columns of X
So if XZ = 0 and Z 6= 0 then the columns of X are zeroes...so X
would not be full column rank!
So we know that XZ 6= 0 hence (XZ )0 XZ 6= 0
Hence (XZ )0 XZ > 0 (already shown that it is
Equivalently,
and so
Alex Moore
Z 0 (X 0 X )Z
X 0X
(LSE)
0)
>0
is positive de…nite
Ec402 Problem Set 2 Linear Algebra
October 2014
5 / 13
Question 3
Both parts make use of identity matrices:
(a) Start with AB (AB ) 1 = I
(b) Start with AA 1 = I
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
6 / 13
Question 4
0 1
x1
@: A
xn
11
1 0 2
σ1 σ12
µ1
CC
B
B
..
N @@ : A , @ :
AA
.
2
µn
σn
00
a0 x is a linear combination of the xi ...it is a linear combination of
jointly Normally distributed random variables
Recall: Any linear combination of jointly Normally distributed random
variables is also Normal
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
7 / 13
Question 4
Univariate case: E (ax ) = aµ, Var (ax ) = a2 µ2
In our case, E (a0 x ) = a0 µ, Var (a0 x ) = a0 Σa
a0 x
N (a0 µ, a0 Σa)
In more general case, where A is kxn matrix of constants:
Ax
Alex Moore
(LSE)
N (Aµ, AΣA0 )
Ec402 Problem Set 2 Linear Algebra
October 2014
8 / 13
Question 5
See class.
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
9 / 13
Question 6(a)
n
tr (A) = ∑ aii , i.e. sum of all diagonal elements.
i =1
(a)
In transpose, element aij becomes element aji .
Hence aii becomes...aii : diagonal elements are the same. Hence:
n
tr (A) = ∑ aii = tr (A0 ) .
i =1
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
10 / 13
Question 6(b)
(b) tr (A + B ) = tr (A) + tr (B )
0
a11 + b11
..
B
a
+
b22
22
B
A+B = B
@
:
..
1
.
ann + bnn
C
C
C
A
Hence tr (A + B ) = ∑i (aii + bii ) = ∑i aii + ∑i bii = tr (A) + tr (B ).
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
11 / 13
Question 6(c)
(b) tr (AB ) = tr (BA)
AB
C =
a11 b11 + a12 b21
..
..
a21 b12 + a22 b22
n
Hence cii = ∑ aij bji
j =1
And: tr (C ) = ∑i cii = ∑i (∑j aij bji ) = ∑i ∑j bji aij = ∑j ∑i bji aij (try this
to convince yourself) = ∑j djj
where djj is the diagonal element of BA [i.e. BA = D where djj = ∑j bji aij ]
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
12 / 13
Questions 7&8
See class.
Alex Moore
(LSE)
Ec402 Problem Set 2 Linear Algebra
October 2014
13 / 13

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