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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 i251 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 A22 [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 AU 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 i251 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