Solution of Exercise 1
Prof. Dr. Dominik Liebl
1 −2 3 −6 3· 5 7 = 15 21
It holds that
=
The matrices D and E are not defined.
Solution of Exercise 2
Remember:
A =
B =
C =
F =
G =
H =
3 1+−2 −3=1 −2
3 16 7 −3 16 20 −5
Econometrics Solution 1 (Ch. 2)
2 0 −2 −1 4 = 12 14 −6 3 3 27
7 1 9 = 7 63
3 5−3 −3
−4 2 −9 7 −9 = −30 −6 −9 =144
6 3 −3′1 −8 −4 6 41 −8 −4
4 1 2 7 5 2 =3 17 5 2 −3 2
34 −28 −16 10 −19 −10 11 34 16
• The rank of a matrix can be defined as the maximal number of linearly independent columns of the matrix.
• If A is a (n × m) dimensional matrix, then rank(A) = min{m, n}.
• A column vector is said to be linearly dependent if it can be written as a linear combination of the
other column vectors.
• A set of column vectors {v1, . . . , vp} is said to be linearly independent if the vector equation
c1v1 +···+cpvp =0
has only the trivial solution c1 = ··· = cp = 0. Otherwise, the set of column vectors is linearly
dependent.
Solution: One can compute the rank of a matrix using R as following:
1
library(“Matrix”)
A <- rbind(c(1, 0, 3, 2),
c(5, 0,-1,-6),
c(4, 0, 2,-2)) B <- rbind(c(1, -1),
c(-1, 0)) rankMatrix(A)[1]
## [1] 2
rankMatrix(B)[1] ## [1] 2
Explanations: Matrix A has rank 2, since with the zero vector the set of column vectors of A is always linearly dependent. Moreover, the first column of A can be expressed as a linear combination of the third and the fourth column;
A[,1] == A[,3] - A[,4] ## [1] TRUE TRUE TRUE
Matrix B has rank 2 since both column vectors are linearly independent. Solution of Exercise 3
Remember: A (n × n) matrix D−1 is called the inverse of a (n × n) matrix D, if D−1D = DD−1 = In.
In exercise 3 we are asked whether the inverse of AB, i.e. (AB)−1, equals A−1B−1. So, we have to check the following two statements:
1. (B−1A−1)(AB) = In
2. (AB)(B−1A−1) = In To 1.
To 2.
(B−1A−1)(AB) = B AA−1 B−1 = BB−1 = In
=In
(AB)(B−1A−1) = A BB−1 A−1 = AA−1 = In
=In
So, (B−1A−1) is indeed the inverse of (AB), i.e., (AB)−1 = (B−1A−1).
Solution of Exercise 4
a) (AB)′(B−1A−1)′ = B′A′(A−1)′(B−1)′ = B′ A′A′−1 B′−1
=In
=B′B′−1 =In
=In
b) (A(A−1 +B−1)B)(B+A)−1 =((AA−1 +AB−1)B)(B+A)−1 =(B+AB−1B)(B+A)−1 = (B+
A)(B + A)−1 = In
=In =In
2
Solution o Exercise 5
Solution of Exercise 6
X=x1 x2···xn′
(n×m)
(m×1) (m×1) (m×1)
X′X =x1 x2 ··· xn·x1 x2 ··· xn′ (m×m)
x ′1
x′ ·2
=x1 x2 ··· xn ·· ·
x′n =x1x′1 +x2x′2 +···+xnx′n
(m×m) (m×m) n
=xix′i i=1
(m×m)
nn
(xi − x)(yi − y) = i=1
(xiyi−xiy−xyi+xy) i=1
nnn
= xiyi −yxi −xyi +nxy
i=1 i=1 i=1 nn
= xiyi −ynx−xyi +nxy i=1 i=1
nn
= xiyi−xyi
i=1 i=1 n
= (xiyi − xyi) i=1
n
= (xi−x)yi
i=1 Note that by similar arguments one can show that
and that
nn
(xi −x)(yi −y)=(yi −y)xi i=1 i=1
nnn
(xi −x)2 =(xi −x)(xi −x)=(xi −x)xi.
i=1 i=1
i=1
3