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Linear Algebra and Finite Fields

Thomas Pasko

Dr. Hurd

Math 495

Abstract. The subject of linear algebra mainly focuses on vector spaces and operations on matrices to advance the study of the subject. Specific topics of importance include determinants, inverses, eigenvalues, eigenvectors, transposes, orthogonality and permutations, and linear maps. The purpose of this paper is to connect the Modern Algebra topics like rings and fields, to matrices and this paper studies the 2-by-2 and 3-by-3 matrices with entries from a finite field. In doing this the inverse, determinants, eigenvalues, and eigenvectors will be looked at.

Part I. Introduction

What can one say about the linear algebra of 2-by-2 and 3-by-3 matrices when the usual numbers are replaced with entries from a finite field? This simple question is enough to open up seemingly endless doors. In order to begin it may help to look back on the history of some of these topics. The first thing to examine would be the subject of linear algebra. Then once you think of matrices you think of some other areas such as the inverses and determinants of matrices. As surprising as it may sound though, the chicken did come before the egg in this case. Determinants have been around longer than matrices. It was the discovery of determinants and Cramer’s Rule for solving systems of linear equations that began use of matrices. Galois and Gauss used finite fields before the study of field theory itself.[5]

The following terms will be discussed:

Ring: A set R together with binary operators +,* defined on R  the following axioms are satisfied: - is an abelian group. - Multiplication is associative but not necessarily commutative. -  a,b,c R, the left distributive law a*(b+c), and the right distributive law (a+b)*c = (ac)+(bc) both hold.

Field: A ring F is said to be a field provided that the set F – {0} is a commutative group under the multiplication of F (the identity of this group, the unity element, will be written as 1) [2 and 4].

1 A matrix is a rectangular array of elements taken from a set R, usually enclosed in parentheses or square brackets [1]. In this paper R will always be a commutative ring with unity. In the case of a square matrix the number of rows matches the number of columns, and this number n is the order of the matrix. These are usually referred to as n x n matrices, and we will write Mn(R) for the set of square matrices of order n I with entries from the ring R. We write GF(q) for the finite field with q elements and M2 (GF(5)) means that the set of 2-by-2 matrices with entries in GF(5). We use the notation aij to denote the (i,j) entry (row i, column j) of a matrix D, and we write D = (aij ). Given a matrix D, Dij denotes the (n-1)-by-(n-1) formed from D by deleting the row i and column j. When p is prime, Zp and GF(p) denote the same field, and in any case q will always denote a power of a prime.

The determinant of a square matrix with entries from a ring R is a function from the set of all n-by-n matrices Mn(R) into R. Proceeding with induction, the determinant of a 1-by-1 matrix (a) is a . For n > 1 we define, for any matrix D of order n, n j1 det(D) =  aij *det(Dij )(1) . j1 In this paper we will restrict n ≤ 3. The definition implies a b det   = ad – bc c d and

a b c    e f  d f  d e det d e f  = a * det   - b * det   + c* det   h i  g i  g h g h i 

Suppose u is the unity element of a ring R. The matrix with u on the main diagonal and zeroes elsewhere is called the identity matrix and will be denoted In. A 2-by-2 or 3-by-3 matrix is invertible if there exists a matrix C such that C*A=A*C=In. We will follow custom and denote the unity of GF(q) with 1. The inverse of A is denoted by A-1. If A is not invertible, it is singular[1].

2 5 3 2 Ex: Let A =   , and C =   with entries in Z7. 1 3 6 2

Then

2 5 3 2 36 14 1 0 AC =             I2. 1 3 6 2 21 8  0 1

We have used the facts that 36 ≡ 8 ≡ 1 (mod 7) and 21 ≡ 14 ≡0 (mod 7).

2 Let A be an n x n matrix. A scalar λ is an eigenvalue of A if there is a nonzero column vector v in Rn such that Av = λ. The vector v is then an eigenvector of A corresponding to λ. The characteristic value is given as the equation det (Av – λI) = 0.

2 21 4 1 Ex.        4  3 11 4 1 1 2 2 Thus the vector   is an eigenvector of the matrix   corresponding to the 1 3 1 eigenvalue 4.(1)

An n x n matrix A is diagonalizable if there exists an invertible matrix C such that C-1AC = D, a diagonal matrix. The matrix C is said to diagonalize the matrix A.

Part II Method

Now we look at some 2-by-2 matrices with entries from GF(2) or GF(3):

1 0 0 1   det N= 0, N-1 = DNE   det N= 0, N-1 = DNE 0 0 0 0

0 0 0 0   det N= 0, N-1 = DNE   det N= 0, N-1 = DNE 1 0 0 1

1 1 1 0   det N= 0, N-1 = DNE   det N= 0, N-1 = DNE 0 0 1 0 1 0 1 0 0 1   det N= 1, N-1 =     det N= 0, N-1 = DNE 0 1 0 1 0 1

0 0 0 1 0 1   det N= 0, N-1 = DNE   det N= -1, N-1 =   1 1 1 0 1 0

1 1 1 1 0 1 1 1   det N= 1, N-1 =     det N= -1, N-1 =   0 1 0 1  1 1 1 0 

1 0  1 0 1 1 0 1    det N= 1, N-1 =     det N= -1, N-1 =   1 1 1 1 1 0 1 1

1 1 0 0   det = 0, N-1 = DNE   det = 0, N-1 = DNE 1 1 0 0

3 In GF(2) we note -1 = +1, and in GF(3) we note -1 = 2. But the determinant calculations show that, just as in real-valued matrices, the usual theorem on inverses is true in this setting (if det(A) = 0 then A-1 does not exist).

-1 Theorem: For any N in M 2 (Z 2 ) if det N = 0, then N does not exist. Conversely, if det(N) = 1, then N-1 exists.

a b Let matrix A =    M 2 (Z p ) c d Then det A = 0 if and only if ad ≡ bc (mod p)

If ad ≡ bc (mod p), then in the field Zp, ad = bc or ad-bc = 0. Thus we can leave out the congruence relation whenever possible.

Theorem: A matrix A  M 2 (G q ) is invertible if and only if det(A) ≠ 0.

Proof: 1 Suppose det(A) ≠ 0. Then ad-bc is a non-zero element in GF(q) and thus = r  ad  bc  rd  rb GF(q) for some element r since GF(q) is a field. But then   is the inverse  rc ra  a b of   . c d We know that det A = ad-bc. Now if (a  d)  (b  c) contains a zero then you would be left with (a * 0)-(b * 0) or some similar combination. This of course would leave you with just det A = 0, in which case A-1 does not exist.

1 3 Example: Let A =   with entries from Z4. 1 2 Then det A = (2*1) – (1*3) = -1 = 3

1 3  2 3  2 3 and A-1 = -1   =   or   1 2  1 1 1 3 1 3 2 3 1 0 so   *   =   1 2 1 3 0 1

Thus Z4 has divisors of zero, yet A  M 2 (Z 4 ) has an inverse. So the previous theorem can be made true for M2 (R) even if R is only a commutative ring with unity provided we replace “det R ≠ 0” with “det R is a unit.”.

4 Now we can take a look at 3-by-3 matrices. Since there are 29 or 512 different combinations of 3-by-3 matrices, only about 72 different combinations were observed for this. They include, but are not limited to, examples of the following patterns:

1 0 0 1 0 0 1 0 0   -1   -1   0 0 0 det N= 0, N = DNE 0 1 0 det = 1, N = 0 1 0 0 0 0 0 0 1 0 0 1

1 1 1 0 1 0  1 1 0   -1     -1 1 0 0 det N = -1, N = 1 1 1 1 1 0 det N= 0, N = DNE 0 0 1 0 0 1  0 0 1

a b c  d e f   Theorem: Let matrix B =   M3(Gq). g h i  In order for det B ≠ 0 and B-1 to exist the following axioms must be satisfied:  e f  d f  d e   1.) a *det   b *det   c *det  ≠ 0  h i  g i  g h d e e f  d f  2.) If c* det   = 0, then a * det   ≠ b * det   g h h i  g i  and e f  d e d f  If a * det   = 0, then c * det   ≠ b * det   h i  g h g i  and d f  e f  d e If b * det   = 0, then a * det   ≠ c * det   g i  h i  g h

a b c    Proof: Let matrix B = d e f  g h i  From the definition above, we know that e f  d f  d e det B = a * det   - b * det   + c * det   h i  g i  g h

First check axiom 1. Assume the contradiction:  e f  d f  d e   a *det   b *det   c *det  = 0  h i  g i  g h

5 then e f  d f  d e a *det   - b*det   + c*det   = 0 and B-1 Does not exist h i  g i  g h

Now look at axiom 2. Again assume the contradiction: d e e f  d f  A.) If c*det   = 0, then a *det   = b*det   g h h i  g i  e f  d f  d e Then a *det   - b*det   = 0 and since c*det   = 0, det B h i  g i  g h = 0. Thus B-1 Does not exist. e f  d e d f  B.) If a *det   = 0, then c*det   = b*det   h i  g h g i  d f  d e e f  Then - b*det   + c*det   = 0 and since a *det   = 0, g i  g h h i  1 det B = 0. If det B = 0 then B-1 = Does not exist since det B = 0. det B Hence there is no inverse

1 1 1 1 0 0 Example: Let B =   with entries from Z2 0 0 1

Then det B = 1[(0*1)-(0*0)] – 1[(1*1)-(0*0)] + 1[(1*0)-(0*0)] = 1[0] – 1[1] + 1[0] Thus det B = -1 Now let us calculate B-1. 0 1 0  -1   B  1 1 1 0 0 1 

Next we will examine the eigenvalues and eigenvectors of a finite field. To do this we will first find the eigenvalues of matrix A containing integers in Z3.

1 1 Let matrix A =   2 1 Using the characteristic equation defined above we see that

1 1  0 1  1  A – λI =      =   2 1 0   2 1 

6 det (A – λI) = (1 – λ)2 – 2(1) = (1 – λ)2 – 2 1 – 2λ + λ2 – 2 = λ2 – 2λ – 1

2 2 λ – 2λ – 1 is irreducible in Z3. So let α be a root of λ – 2λ – 1. We therefore add α to Z3. This gives us Z3(α) or GF(9) which is a finite field with 9 elements.

If we extend Z3 by α then we get the field GF(9) which shows the following characteristics  0 1 2 α α+1 α +2 2α 2α +1 2α+2 0 0 1 2 α α+1 α +2 2α 2α +1 2α+2 1 1 2 0 α+1 α +2 α 2α +1 2α+2 2α 2 2 0 1 α +2 α α+1 2α+2 2α 2α +1 α α α+1 α +2 2α 2α +1 2α+2 0 1 2 α+1 α+1 α +2 α 2α +1 2α+2 2α 1 2 0 α +2 α +2 α α+1 2α+2 2α 2α +1 2 0 1 2α 2α 2α +1 2α+2 0 1 2 α α+1 α +2 2α +1 2α +1 2α+2 2α 1 2 0 α+1 α +2 α 2α+2 2α+2 2α 2α +1 2 0 1 α +2 α α+1

And  0 1 2 α α+1 α +2 2α 2α +1 2α+2 0 0 0 0 0 0 0 0 0 0 1 0 1 2 α α+1 α +2 2α 2α +1 2α+2 2 0 2 1 2α 2α+2 2α +1 α α +2 α+1 α 0 α 2α 2α +1 2α +2 2α α +2 2α+2 2 α+1 0 α+1 2α+2 2α +2 α +2 2α 2 α 2α +1 α +2 0 α +2 2α +1 2α 2α 2 2α+2 1 α +2 2α 0 2α α α +2 2 2α+2 2α +1 α+1 1 2α +1 0 2α +1 α +2 2α+2 α 1 α+1 2 2α 2α+2 0 2α+2 α+1 2 2α +1 α +2 1 2α α +2

Here we observe that GF(9) has a multiplicative inverse. Hence GF(9) is a field

To find any eigenvalues and eigenvectors we need to multiply out

x x x A      where x,yGF(9). The results of A   are as follows: y y y

1 1 0 0 1 1 0 1               2 1 0 0 2 1 1 1

1 1 0 2 1 1 0                2 1 2 2 2 1   

7 1 1  0   1 1 1  0    2               2 1  1  1 2 1   2   2

1 1  0  2  1 1  0  2 1               2 1 2  2  2 1 2 1 2 1

1 1  0  2  2 1 1 1 1               2 1 2  2 2  2 2 1 0 2

1 1 1 2 1 1 1 0               2 1 1 0 2 1 2 1

1 1 1  1 1 1  1    2               2 1     2 2 1  1   

1 1  1     1 1  1  2 1               2 1   2  1 2 1 2  2  2

1 1  1  2  2 1 1  1   2                2 1 2 1  2  2 1 2  2 2 1

1 1 2 2 1 1 2 0               2 1 0 1 2 1 1 2

1 1 2 1 1 1 2   2               2 1 2 0 2 1    1

1 1  2     1 1  2   1               2 1  1   2 2 1   2   

1 1  2  2  2 1 1  2   2                2 1 2  2 1 2 1 2 1 2  2

1 1  2  2 1 1 1                    2 1 2  2  2  2 1 0 2 

1 1     1  1 1      2                2 1 1 2 1 2 1 2 2  2

8 1 1   2  1 1    2 1               2 1    0  2 1  1  1 

1 1    2  2 1 1    0               2 1   2  2  2 1 2   

1 1     1  1 1     2                2 1 2 1  1 2 1 2  2   2

1 1  1   1  1 1  1   2               2 1  0  2  2 2 1  1   2 

1 1  1    1 1  1 2 1               2 1  2  2 1 2 1     2 

1 1  1 2  2 1 1  1 2                2 1  1  0  2 1   2  1 

1 1  1  1  1 1   1  2               2 1  2    2 2 1 2 1  

1 1   1   0  1 1   2   2                2 1 2  2  1 2 1  0  2 1

1 1   2    1 1   2  1               2 1  1  2  2 2 1  2   2 

1 1   2 2  2 1 1   2 2                2 1     1  2 1  1  2 

1 1   2 2 1 1 1   2  2                2 1   2  0  2 1  2   1

1 1   2   0  1 1    2  1               2 1 2 1   2 2 1 2  2  

1 1 2  2  1 1 2  2 1               2 1  0     2 1  1    1 

9 1 1 2  2  2 1 1 2   0                2 1  2     2  2 1    2 

1 1  2   1  1 1  2   2                2 1  1 2 1 2 1   2 2  2

1 1 2    1 1  2   1               2 1 2  0 2 1 2 1  1 

1 1  2    2 1 1 2 1 2 1               2 1 2  2  2  2 1  0    2 

1 1 2 1 2  2 1 1 2 1  2                2 1  1     2 1  2   1

1 1 2 1  1  1 1 2 1  2                2 1    2  2 2 1   1  2 

1 1 2 1  0  1 1 2 1  1               2 1   2  2 1 2 1  2   2 

1 1 2 1   2 1 1 2 1                 2 1 2 1  0  2 1 2  2 1

1 1 2  2 2  2 1 1 2  2  2                2 1  0    1  2 1  1    2

1 1 2  2 2 1 1 1 2  2  2                2 1  2     2 1    2 1

1 1 2  2  0  1 1 2  2  1                2 1   1  2  2 2 1    2  2 

1 1 2  2   2 1 1 2  2                 2 1  2   1  2 1 2 1 2

1 1 2  2  1        2 1 2  2  0 

10 After finding these results the next thing to do is to find any eigenvalues and eigenvectors. One solution found is illustrated by:

1 1  2   2  x x        . Now we show that λ   = A   . 2 1 2 1 2  2 y y  2   2   2   2  Notice α     2     , for our eigenvalue α and   is an 2 1 2    2  2 2 1 eigenvector. For this calculation, recall  2  2 1. After looking at the other possibilities we see there is only one eigenvalue; α. This means that matrix A cannot be diagonalized in the field GF(9). In order to make this possible we must extend GF(9). To do this we look at 2  2 1. Applying the quadratic formula the solution is

2  4  4(1)(1) 2  8 λ =   1 2 2 2

So the two eigenvalues of GF(9) are

λ1= 1 + 2

λ2 = 1 – 2 and the two needed corresponding eigenvectors which we found are

 1 1    .  2 2

To test this we look at

1 1 1  1 2        . 2 1 2 2  2

So now we look at the eigenvalues to test them

 1  1 2  1 2     .  2 2  2 And

11 1  2 1 2  2 1 2        .  2  2  2 1 2 

So now the diagonalization is given by conjugation by a matrix whose column are eigenvectors from distinct eigenvalues. We illustrate with matrix A. We claim that

 1 2   1 1   -1 -1 2 4 A = CDC , where the matrix C =   and C =   and the diagonal  2 2 1 2   2 4  1 2 0  matrix of eigenvalues D =   . When we expand CDC-1 we get  0 1 2  1 2     1 11 2 0  2 4 1 1        = A.  2 2 0 1 21 2  2 1  2 4  Therefore the matrix A in GF(9) is diagonalizable, as it is with the linear algebra with matrices of real numbers.

12 Sources:

1) Fraleigh Beauregard, Linear Algebra, 3rd Edition, Addison-Wesley, New York, 1995 2) David Burton, Abstract and Linear Algebra, Addison Wesley, Philippines, 1972 3) Cyrus Colton Mac Duffee, An Introduction to Abstract Algebra, John Wiley & Sons, New York 1940 4) David Burton, Introduction to Modern Abstract Algebra, Addison-Wesley, Massachusetts, 1967 5) David Burton, The History of Mathematics: An Introduction, Allyn & Bacon, Boston, 1985 6) A. Jones, S. Morris, K. Pearson, Abstract Algebra and Famous Impossibilities, Springer-Verlag, New York, 1991

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