Linear Algebra Worksheet

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Linear Algebra Worksheet
University of California, Berkeley
Scott Alan Carson
Econometrics is much easier if we use linear algebra. You don’t have to be too versed in linear
algebra to be able to use if affectively in introductory econometrics. Below are some of the
operations we will use throughout the semester
1. Matrix dimensions
Given a system of equations
y1   0  1 x1   2 x2  . . .   k xk   1
y2   0  1 x1   2 x2  . . .   k xk   2
.
.
yn   0  1 x1   2 x2  . . .   k xk   n
The system can be represented in matrix and vector form as
 y1  1
  
 y 2  1
 y  = 1
 3 
 .  .
 y  1
 n 
x11
x12
.
x21
x22
.
x31
x32
.
.
.
.
xn1
xn 2
.
x1k 

x2 k 
x3 k 

. 
xnk 
 1    1 
   
 2   2 
 .  + . 
   
 .   . 
    
 k  n
The left hand side is an nX1 vector. The right hand side is an nXk matrix. It has n rows
and k columns. Matrix notation is that each element’s first subscript is the row identifier,
and the second is the column identifier. In the future, we will just identify each matrix
component with its subscript identifying the column.
 x0

 x0
x
 0
 .
x
 0
x1
x2
x1
x2
x1
x2
.
.
x1
x2
. xk 

. xk 
. xk 

. . 
. xk 
Given two matrices with the same dimensions, their elements can be added. Let
2
 a11

 a21
A=  .

 .
a
 n1
a12
a22
.
.
an 2
 a11  b11

 a 21  b21
A B  
.

.

a b
 n1 n1
. . a1m 
 b11 b12


. . a2 m 
 b21 b22
. . .  and B=  .
.


. . . 
.
 .


. . anm 
 bn1 bn 2
a12  b12
a 22  b22
.
.
a n 2  bn 2
. . b1m 

. . b2 m 
. . . 

. . . 
. . bnm 
. . a1m  b1m 

. . a2 m  b 

. .
.

. .
.

. . a nm  bnm 
Example
3 6 2
1 2 0
 4 8 2






A   5 1 3  and B   0 1 2  => A  B   5 2 5 
1 1 1
1 0 1
 2 1 2






Matrix subtraction is just taking the difference between each element.
3. Matrix multiplication
If two matrices are conformable, they can be multiplied. Conformability means the
number of columns in the first matrix is equal to the number of rows in the second
matrix. The dimension of the product matrix is the number of rows in the first matrix by
the number of columns in the second matrix. If A is 2X4 and B is 2X2, the matrices are
not conformable. If A is 2X4 and B is 4X2, the matrices are conformable, and the
dimension of the product is 2X2. Given the matrices
 a11

 a21
A=  .

 .
a
 n1
a12
a22
.
.
an 2
. . a1m 
 b11 b12


. . a2 m 
 b21 b22

. . . and B=  .
.


. . . 
.
 .


. . anm 
 bn1 bn 2
. . b1m 

. . b2 m 
. . . 

. . . 
. . bnm 
Conformable matrices are multiplied with a series of inner products that match the
conformability of the product matrix. For example,
3
 a11

a
A   21
a
 31
a
 41
a12
a13
a 22
a 23
a32
a33
a 42
a 43
a14 

a 24 
a34 

a 44 
and
 b11 b12

b
b
B   21 22
b
b
 31 32
b
 41 b42
b13
b23
b33
b43
b14 

b24 
b34 

b44 
The product matrix is
 d11

d
D   21
d
 31
d
 41
d12
d13
d 22
d 23
d 32
d 33
d 42
d 43
d14 

d 24 
d 34 

d 44 
The subscript on this product matrix tells us how to construct to inner products. The 1X1
subscript for the d11element indicates that the first row of the pre-multiplied matrix A is
multiplied by the post-multiplied matrix B.
d11  a11b11  a12b21  a13b31  a14b41
d12  a11b12  a12b22  a13b32  a14b42
.
.
d 44  a41b14  a42b24  a43b34  a44b44
Note that the dimension of each element is also the dimension of the element in the
product matrix. For example, the dimension for each element in the right-hand side’s
product equals the left hand side.
A quantitative example is
 2 3
1 3 
 . Each matrix is a 2X2. So the matrix is conformable
 and B  
A  
1 2 
 4 1
because the number of columns in the first matrix is equal to the number of rows in the
second matrix. The dimesion of the product matrix is 2X2 because the number of rows in
the first matrix is 2, and the number of columns in the second matrix is 2. Taking the
inner products is
d11  2 X 1  3X 1  2  3  5
d12  2 X 3  3X 2  6  6  12
d 21  4 X 1  1X 1  4  1  5
4
d 22  4 X 3  1X 2  12  2  14
This means that the product matrix D is
d12   5 12 
d
  

D   11
 d 21 d 22   5 14 
For any square matrix A, the following statements are equivalent.
1. A is invertible.
2. A is non-singular.
3. detA is non zero.
4. A is of rank n.
Determinant
So, we need to be able to calculate a determinant. A determinant is a value that can be calculated
from a square matrix. The most important use in econometrics is to calculate the inverse of a
matrix, therefore, the coefficients in a linear system of equations. There are various ways we can
define the determinant of square matrices. For large matrices—matrices for values greater than a
three by three—the typical way of calculating a determinant is a Laplace expansion. We leave
those larger determinants to Laplace expansion and computers, but for smaller matrices,
determinants are calculated by sweeping the matrix. For example, given a two by two matrix.
a12 
a
  a11a22  a12 a22
A  11
 a21 a22 
Given the matrix
 5 12 
 the determinant is 5 14  5 *12  70  60  10
A  
 5 14 
For a three by three matrix, the determinant is calculated in similar fashion.
 a11 a12

A   a21 a22
a
 31 a32
a13 

a23   a11a22 a33  a12 a23a31  a13a21a32  a12 a21a33  a11a23a32  a13a22 a31
a33 
For example,
 1 5 3


A   2 1 3  1 *1 *1  5 * 3 *1  3 * 2 * 2  5 * 2 *1  1 * 3 * 2  3 *1 *1
 1 2 1


 1  15  12  10  6  3  28  19  11
5
Inverse
To estimate the vector of regression coefficients, it is necessary to calculate a matrix inverse.
There are various ways to calculate an inverse, and econometricians and applied mathemiticians
almost never calculate a matrix inverse by hand: there is just to many moving parts and a
computer is more accurate. Nevertheless, it is useful to understand how a matrix inverse in
calculated in its role in estimating the vector of regression coefficients.
1
A
This is illustrated by breaking it down into steps and with an example. The example is the same
as Gujarati and Porter, p. 847. Let the matrix equal
A1 
1 2 3 
A  5 7 4
2 1 3
Calculate the determinant.
a. 1*7*3+2*4*2+3*5*1-2*5*3-1*4*1-3*7*2=21+16+15-30-4-42=-24
b. Obtain the cofactor matrix
 7 4
5 4
5 7 



2 3
2 1 
 1 3
1 3
1 2
 2 3
C  


2 3
2 1
 1 3
1 3
1 2 
 2 3
 7 4 5 4
5 7 

 17  7  9
   3  3 3 
 13 11  3
c. Transpose the cofactor matrix
 17  3  13
  7  3 11 
  9 3
 3 
6
Multiply the cofactor matrix by the inverse of the determinant
d.
 17  3  13
1 
1
A    7  3 11 
24
 9 3  3 
 17
 24
 7

 24
 9
 24
3
24
3
24
3

24
13 
24 
11 
 
24 
3 
24 
4. Transpose
Because of the dimensions in linear algebra, we do not square matrices but multiply by the same
matrix that is symmetric. This symmetric matrix is the matrix transpose. Stated precisely,
aij  a ji
In other words, rows become columns and columns become rows. For example, given the matrix
 a11
A  a 21
 a31
a12
a 22
a32
a13 
a 23 
a33 
 a11
 A  a12
 a13
T
5. Idempotent（幂等）
A=AA
6. Orthogonal（正交）
a 21
a 22
a 23
a31 
a32 
a33 
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