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Invertibility, Determinants and Eigenvalues of 2-by-2
Matrices with Entries from a Finite Field
Wu, Daiyi
Department of Mathematics and Computer Science
The Citadel, Charleston S.C, 29409
[email protected]
1. Introduction
In this paper, I investigate the nature of linear algebra results when the field of real
numbers is replaced by a finite field. We will write GF(q) (known as the Galois Field), or
more simply just Gq to denote the finite field with q elements. Necessarily of course, q is
a power of a prime. When p is prime, Z p and Gq denote the same field. However, if n is
not prime, then Z n is just a ring.
We use the notation M n (R) to denote the set of n-by-n matrices with integers from
a ring R which is always assumed to have the usual matrix addition and multiplication but
using the operation defined on the set R. For example, M 2 (G5 ) is the set of 2-by-2
matrices with entries from G5 .
2. Matrix Theory
As in general linear algebra, we define a function det, the determinant of any square
matrix. However, the det(A), for any square matrix A, is not a real number, it is an
element of the ring R. More formally, if A is in M n (R) , we define det(A) inductively as
follows. For any 1-by-1 matrix, say A = ( ) , we define the determinant of A to be . If
n  2 and A is an n  n matrix, we write det ( Aij ) to mean the determinant of the
(n  1)  (n  1) matrix formed by deleting row i and column j from A. Proceeding
inductively, for n  2 , the determinant of a matrix A is the sum of n terms of the
form  a1 j det A1 j , with plus and minus sign alternating, where the entries a11 , a12 ..., a1n
are from the first row of A. In symbols,
n
det A  a11 det A11  a12 det A12  ...  (1)1 n a1n det A1n   (1)1 j a1 j det A1 j
j 1
a b 
  ad  bc in the case that n=2. In the case that n=3, we have:
For example, det 
c d 
1
a b

det  d e
g h

c

b
f   a  det 
e
i 
c
d
  b  det 
f
g
f
d
  c  det 
i
g
 c(dh  eg )
e
  a(bf  ce)  b(dc  gf )
h 
.
Those are the usual rules for expressing a determinant. But again, in this paper, ad-bc is a
ring element.
Definition: A matrix in M n (G q ) is invertible iff there is a matrix B  M n (Gq ) , so that
A  B  B  A  I n where I n is the familiar identity matrix with the unity element (1) of
Gq on the main diagonal and 0 elsewhere.
a b 
 is invertible iff
In the usual matrix theory, we have the theorem that a matrix A= 
c d 
ad-bc  0. If ad-bc = 0, then A is not invertible. The usual proof works:
Suppose that A  M 2 (G p ). If ad-bc  0, it is invertible in G p . Thus,
 d
 a b   ad  bc
  
AB = 
 c d    c
 ad  bc
 d

B   ad  bc
 c

 ad  bc
b 

ad  bc    1 0 
a   0 1 

ad  bc 
where
b 

 d  b
1
ad  bc  =

 
a 
det( A)   c a 

ad  bc 
The Lemma and the previous comments, prove the following:
Lemma: In a ring R with unity u, if an element has a multiplication inverse, it is unique.
Proof:
Suppose t is an invertible element in a ring R with unity u. We assume that two
elements are inverses of t.
Let t  t 1  t  y , so
t 1 (t  t 1 )  t 1 (t  y )
(t 1  t )  t 1  (t 1  t )  y
u  t 1  u  y
t 1  y
Theorem: Let R be a communicative ring with unity and suppose ad-bc is a unit in R.
2
a b 
 in M 2 ( R) . Then A is invertible and
Suppose matrix A= 
c d 
1  d  b

.
B  A 1 
ad  bc   c a 
3. Scalars and Eigenvalues
In linear algebra, one may multiply a matrix A  (aij ) in M n (R) by a scalar 
from a field so that   A  (  aij ) . Thus, every entry in A is multiplied (in the ring R)
by  , and we do exactly this in the paper. In general,  will be in Gq , and if q  p t for
some t, we may restrict scalar to some subfield of Gq . Later in this paper, the scalar  is
an eigenvalue.
Definition: An eigenvector of a n  n matrix A is a nonzero vector v such that Av  v
for some scalar  . The scalar  is called an eigenvalue of A if there is a nontrivial solution
v of Av  v . Such a v is called an eigenvector corresponding to  .
v
2
Let   be a vector in V  G p  G p  G p .
u 
1 2
 in  5 . In order to find its
Main Example: For our main example, let M  
3 4
eigenvalue  , we solve the characteristic equation det( M  I )  0.
(1   )(4   )  6  4  5  2  6  2  5  2  2  2 .
The equation 2  2  0 is irreducible in the field  5 .
We add
2 to the field, and now we have 25 elements in our new field,  5 ( 2 ) =
GF (25). The elements of  5 ( ) , where  2  2 are { b  c : b, c   5 }.
The vector ( x, y ) in GF (25) 2 is an eigenvector if it solves the equation:
 1 2  x 
 x

      where  2  2 .
 3 4  y 
 y
There are 25 elements which we solved for by inspection:
3
 x  0  1   2   3   4 
    , 
, 
, 
, 

 y   0   3  2     4   4  1  2  3 
      1    2     3     4 

, 
, 
, 
, 

 2  1  3   3     2   4  4 
 2   2  1   2  2   2  3   2  4 

, 
, 
, 
, 

 4  2   2  4   1   3  3    
 3   3  1  3  2   3  3   3  4 

, 
, 
, 
, 

   3   4   2  2   4   3  1 
 4   4  1  4  2   4  3   4  4 

, 
, 
, 
, 

 3  4     1   4  3   2   2 
In all, I got 25 eigenvectors. They are all different vectors. Top entries (first components)
contain all 25 elements and so do bottom entries. They match perfectly without repetition.
Besides, when I add any two of eigenvectors together, the result will still be one of the 25
possible eigenvectors. For example:
 1   3   4   4 

  
  
  

 3  2   4  1  7  3   2  3 
 2    3    5    

  
  
  

   4     2   2  6   2  1
 3   4  3   7  3   2  3 

  
  
  

   3   2   3  3   3  3 
If we multiply either one of those 25 eigenvectors by scalars (in  5 ( ) ), the result will be
one of 25 eigenvectors.
 2  1   4  2   4  2 
  
  

2  
 2  4   4  8   4  3 
   3   3  9   3  4 
  
  

3  
   2   3  6   3  1 
 3   3 2   6   1 
  
   2
  

  

   3     3   2  3   2  3 
By the definition of vector space, [DL 191] we know that our nonempty set V, with
scalars from GF(9), is a vector space. Also, a subspace of a vector space V is a subset H
of V such that H is itself a vector space under the same operations of addition and scalar
4
multiplication that are already defined on V.
Theorem: A subset H of a vector space V is a subspace of V if and only if the following
conditions are all satisfied:
a. The zero vector of V is in H.
b. If u and v are in H, then u+v is in H.
c. If u is in H and c is any scalar, then cu is in H.
By the definition and theorem above, these 25 eigenvectors are a subspace of GF (25) 2 ,
which has 625 elements (25  25) . It is actually the null space of A-I.
In linear algebra, the set of eigenvectors corresponding to an eigenvalue is a vector
subspace of V. The same is true here.
1 2
 in  5 is   , so we solve the equation for
The other eigenvalue of M  
3 4
eigenvectors.
 1 2  x 
 x
 x

       4  
 3 4  y 
 y
 y
where  2  2
There are 25 elements which we found:
 x  0  1   2   3   4 
    , 
, 
, 
, 

 y   0   2  2   4  4     1  3  3 
      1    2    3    4 

, 
, 
, 
, 

 2  4   4  1    3   3   2 
 2   2  1  2  2   2  3   2  4 

, 
, 
, 
, 

 4  3      3  2   4   2  1 
 3   3  1   3  2   3  3   3  4 

, 
, 
, 
, 

   2   3  4   1   2  3   4 
 4   4  1  4  2   4  3   4  4 

, 
, 
, 
, 

 3  1  3   2   2  4     4 
By adding two of eigenvectors together, the result will still be one of 25 possible
eigenvectors. For example:
5
 4    1     5    

  
  
  

 3  3   4  1  7  4   2  4 
 4  4     3   5  7   2 

  
  
  

   4   3   4  4   4  4 
   2   4  3   5  5   0 

  
  
   
   3   4  2   5  5   0 
If we multiply either one of those 25 eigenvectors by scalars (in  5 ( ) ), the result will be
one of 25 eigenvectors.
 2  1  4  2 
  

2  
    2 
 3  3   6  9     4 
  
  

(  1)  
 2  3   5  7   2 
 3   3 2   6   1 
  
   2
  

  

   2     2   2  2   2  2 
 0 4
 in  5 we get
Another example, similar to the previous one, with M  
2
0


2
2
( )  8  0    3  0 .
2
Thus,   3 is the characteristic equation and it is irreducible in  5 . We expand to
 5 ( ) where  2  3 . The eigenvectors are:
 x  0  1   2   3   4 
    ,  ,  ,  ,  
 y   0   4    3   2     
      1     2     3     4
 , 
, 
, 
, 

2
4


2
3


2
2


2


2
 




 2   2  1   2  2   2  3   2  4 
 , 
, 
, 
, 

 4   4  4    4  3   4  2    4   
 3   3  1   3  2   3  3   3  4 
 , 
, 
, 
, 

1
1

4

1

3

1

2

1


 




 4   4  1   4  2   4  3   4  4 
 , 
, 
, 
, 

 3   3  4    3  3   3  2    3   
In this example, I got a different set of eigenvectors from the previous example.
However, there are still 25 eigenvectors without elements repetition either on top or
6
bottom entries.
When I add two of eigenvectors together, the result will still be one of the 25
possible eigenvectors. Theses eigenvectors are again closed under vector addition and
scalar multiplication as was checked.
 1   2   3   3 
          
 4    3   7    2  
   2   2   4   3  6   3  1 

  
  
  

3


2
4


4


6
4


1

 
 
 

 3  4   4   4   7   8   2   3 

  
  
  

 1     3     2  4   2  4 
If we multiply either one of those 25 eigenvectors by scalars (in  5 ( ) ), the result
will be one of 25 eigenvectors.
 2   8   3 
   
4     
 3  1 2   2  
   2   2 2  4   4  1 
  
   2

2   

 3  2   6   4    4   3 
 4   4  1 2  1 2  2  2 
  
  

3  
 3     3  9   3  4 
The other eigenvalue is   , and we solve for its eigenvectors
 1 2  x 
 x
 x

        4  
 3 4  y 
 y
 y
where  2  3
There are 25 elements which we found:
 x  0  1   2   3   4 
    , 
, 
, 
, 

 y   0   2   2   4   4     1   3  3 
      1     2     3    4 

, 
, 
, 
, 

 2   4   4   1    3   3   2 
 2    2   1  2   2   2   3   2   4 

, 
, 
, 
, 

 4   3      3  2   4   2   1 
 3   3  1   3  2   3  3   3  4 

, 
, 
, 
, 

   2   3  4   1   2   3   4  
 4    4   1  4   2   4   3   4   4 

, 
, 
, 
, 
.
 3  1  3   2    4   2     4 
7
4. Some Different Examples
Now, we try a couple of a little bit different examples, and see what happens.
 1 3
 in  5 we get
Using M  
 2 2
(1   )( 2   )  6  0  2  2    2  6  2  3  4  (  4)(  1)
  4,   1  4
This time,  is actually an integer. Solving the characteristic equation we get
 x   0  1  2   3   4 
    ,  ,  ,  ,  
 y   0  1  2   3   4 
2
This also is a subspace of scalar V  G p  G p  G p with p = 5.
 0 4
 . The calculation is:
The second example in this section: M  
 2 3
( )(3   )  8  0  3  2  8  2  2  3  (  1)(  3)
  3  2,   1
The  ’s are different integers.
 0 4  x 
 x
   2  . There are only 5 eigenvectors.
Solving for   2, 
 2 3  y 
 y
 x   0 1  2   3   4 
    ,  ,  ,  ,  
 y   0   3  1   4   2 
Solve
  1,
 0 4  x   x 

   1 
 2 3  y   y 
 x   0  1  2  3  4
    ,  ,  ,  ,  
 y   0  4  3  2  1
1 4
 .
Another example: M  
 0 3
(1   )(3   )  0
  3,   1
8
 1 4  x   x 
   3  and get eigenvectors:
For   3, we solve 
 0 3  y   y 
 x   0  1  2  3  4
    ,  ,  ,  ,  .
 y   0   3  1   0   2 
 1 4  x   x 
   1  and get eigenvectors
For   1, we solve 
 0 3  y   y 
 x   0  1  2  3  4
    ,  ,  ,  ,  .
 y   0  0  0  0  0
5. Diagonalization
1 2
 in  5 .
We return to the first example in the paper here, M  
3 4
There are 25 eigenvectors for each of its two eigenvalues   ,4 , and we pick one
vector (not zero) randomly from each of these two sets of eigenvalues.
(4 )  V j j 1
  U i i251
U i  0,
25
Vj  0
However, we observe that U i  V j  0
Therefore, P  U i
for any i = 2,…,25,
j = 2,…,25.
V j  cannot be orthogonal matrix. P T  P 1
By the definition of diagonalization from a linear algebra book ([DL}, p.288-289),
for 2-by-2 matrix,
0

A22 [U i V j ]  [ AU i AV j ]  [1U i  2V j ]  U i V j  1

 0 2 


Vj   1
0
0

 P 1 AP   1

 0 2 
U
Vj
 AU

D 1
0
0
 2 
i
1
i

0
 2 
9

A  PDP 1
When   
  4
  3

U  
  2 
 3 

V  
   1
3 
  3

 , D  
p  
   2   1
0
So,
PDP 1 
1  2  3

  1  2  2
0 

4 
P 1 
3 
1   1


  1     2   3 
2    1
3 
1    4 3  2   1 2 

 


 =A
3  4     2   3    1  3  2 4  1   3 4 
Since the condition holds, we can conclude that
1 2
  0  1

  P
 P
and P  U i
3 4
 0 4 
25
(4 )  V j j 1 .
  U i i251
V j ,
I also did a couple more examples to show if the matrix with entries in finite field is
diagonalization or not (not in this paper), and the results all come out nicely. One reason I
believe it is diagonalizable is because when P  U i V j  , PP 1  P 1 P  I (since the
invertibility theory for matrix with entries in a finite field is the same as regular matrix
theory).
Therefore, the matrix with entries in a finite field and which has two eigenvalues is
diagonalizable, and one may also apply the formula A  PDP 1
References
[AJ] Arthur Jones, Sidney A. Morris, Kenneth R. Pearson, Abstract Algebra and Famous
Impossibilities, 1994, Springer-Verlag New York, Inc. New York.
[DL] David C. lay, Linear Algebra, instructor’s edition, 1994, Addison-Wesley
Publishing Company, Inc. Maryland.
[JR] Joseph J. Rotman, Abstract Algebra With Applications, 3rd edition, 2006. Pearson
Education, Inc. New Jersey.
10