Eigenvalues and Eigenvectors of Matrices over a Finite Field by. Kristopher R. Lewis

Transcription

1 Eigenvalues and Eigenvectors of Matrices over a Finite Field by Kristopher R Lewis Abstract: This paper is a study of eigenvalues and their corresponding eigenvectors of x matrices A with entries in Z p over Z p or the expansion of Z p,z p We will examine some common properties of eigenvalues in relation to matrix A and properties of eigenvectors under certain conditions We will first identify them and prove them as such in, and then explore how strong these properties are when we restrict them to finite fields Introduction: The equation Ax x occurs in many applications such that A is an nxn matrix, x is an nx vector and is a scalar in C If the equation has a nonzero solution x, then is said to be an eigenvalue of A, andx is said to be an eigenvector corresponding to These eigenvalues and eigenvectors are a common part of our everyday life These values and vectors have been used to decipher and understand the how and why of our universe by many of the greatest mathematicians Eigenvalues are often introduced in the context of matrix theory Historically, however, they arose in the study of quadratic forms and differential equations In the first half of the 8th century, the Bernoulli s, d Alembert and Euler encountered eigenvalue problems when studying the motion of a rope, which they considered to be a weightless string loaded with a number of masses Laplace and Lagrange continued their work in the second half of the century They realized that the eigenvalues are related to the stability of the motion They also used eigenvalue methods in their study of the solar system Euler had also studied the rotational motion of a rigid body and discovered the importance of the principal axes As Lagrange realized, the principal axes are the eigenvectors of the inertia matrix In the early 9th century, Cauchy saw how their work could be used to classify the quadric surfaces, and generalized it to arbitrary dimensions Cauchy also coined the term "racine caractéristique" [](characteristic root) for what is now called eigenvalue; his term survives in characteristic equation Fourier used the work of Laplace and Lagrange to solve the heat equation by separation of variables in his famous 8 book Théorie analytique de la chaleur Sturm developed Fourier s ideas further and he brought them to the attention of Cauchy, who combined them with his own ideas and arrived at the fact that symmetric matrices have real eigenvalues This was extended by Hermite in 855 to what are now called Hermitian matrices Around the same time, Brioschi proved that the eigenvalues of orthogonal matrices lie on the unit circle, and Clebsch found the corresponding result for skew-symmetric matrices Finally, Weierstrass clarified an important aspect in the stability theory started by Laplace by realizing that defective matrices can cause instability In the meantime, Liouville had studied similar eigenvalue problems as Sturm; the discipline that grew out of their work is now called Sturm-Liouville theory Schwarz studied the first eigenvalue of Laplace s equation on general domains towards the end of the 9th century, while Poincaré studied Poisson s equation a few years later At the start of the 0th century, Hilbert studied the eigenvalues of integral operators by considering them to be infinite matrices He was the first to use the German word eigen to denote eigenvalues and eigenvectors in 904, though he may have been following a related usage by Helmholtz "Eigen" can be translated as "own", "peculiar to", "characteristic" or "individual" emphasizing how important eigenvalues are to defining the unique nature of a specific transformation[4] For some time, the standard term in English was "proper value", but the more distinctive term "eigenvalue" is standard today These are just a few of many

2 applications and studies into the properties of eigenvalues and how they describe many functions of our universe Lets briefly look at a real world problem in which eigenvalues have changed how this problem can be solved Example : Let us assume, 30% of the sailors in the United States Navy leave the service and 0% of the public sector will join the Navy to serve their country each year There are 80,000 members in the Navy, 00,000 members of the public sector and the total population remains constant Let us investigate the long-range prospects if these percentages continue indefinitely into the future To find the number of Navy sailors after year, we multiply the vector w 0 80k,00k T by A The number of sailors and civilians after year is given by w Aw To determine the number of sailors to civilians after years, we compute w Aw A w 0 and in general for n years we must compute w n A n w 0 When we compute w 0,w 0,w 30 in this way and round the entries or each to the nearest integer w , w , w After a certain point we seem to always get the same answer In fact, w 6 000,68000 T and since Aw It follows that all the succeeding vectors in the sequence remain unchanged after rounding The vector 000,68000 T is the long range prospectus Now, suppose we had different initial proportions of sailors to civilians If, for example, we had 80,000 sailors and 0 civilians, then w ,0 T, and we compute w n as before In this case it turns out that w 9 000,68000 T, getting the same long range prospectus vector This result is not by coincidence as we explore eigenvectors and eigenvalues we will find that for matrix A this vector will result despite the initial values of w 0 Definitions and Properties Definition of Eigenvectors and Eigenvalues

3 Let A be an n n matrix A scalar is said to be an eigenvalue or a characteristic value of A if there exists a nonzero vector x such that Ax x The vector x is said to be an eigenvector or a characteristic vector corresponding to Example : LetA 4 and x Since Ax x, x is an eigenvector of A corresponding to eigenvalue 3 Actually, any nonzero multiple of x will be an eigenvector, since Ax Ax x x Thus, for example, 4, T is also an eigenvector corresponding to 3, Computational Procedure For any given n n matrix A, an eigenvalue and its eigenvector can be computed as follows (i) Solve deta I 0for, wherei is the n n identity matrix (ii) Solve Ax x or A I 0 for a nontrivial solution x for each obtained in (i) Let us revisit Example and this time we will use eigenvalues and eigenvectors to find the long range prospectus of sailors and civilians Let A (i) Compute eigenvalues of A deta I , and 05 are solutions of the equation (ii) Compute an eigenvector for each of and 05 : A Ix x x 0or x x 0 By one elementary row operation: Row Row Row, we have 3

4 x x 0 Solving the equation 03x 0x 0or03x 0x,weletx andthenx 3 Vector is an eigenvector corresponding to 3 05 : Following the same steps above, an eigenvector corresponding to 05 is Now if we look closely at the results in Example, we found A x x We unknowingly found an eigenvector for A without finding the eigenvalue We can quickly 000 see that the eigenvalue is Note that Properties of Eigenvalues and Eigenvectors: Theorem Every n n matrix A has n eigenvalues in C n Proof Let A a ij i,j Then the characteristic polynomial equation deta I 0 can be expressed as (*) n a a a nn n n deta 0, a polynomial of degree n in By the Fundamental Theorem of Algebra, we know that the equation deta I 0hasn, not necessarily distinct, roots in C Therefor, A has n eigenvalues of A Clearly, the property in Theorem does not even hold in For example, eigenvalues 0 0 are i which are not in For this project, we consider in Z p 0 Naturally we do not expect this property hold For example, the matrix A not have an eigenvalue in Z 0, Observe that the characteristic equation of A is 0 Clearly, Neither 0 nor is a solution of this equation Now let us go back to equation (*) above This equation allows us to establish two relationships between the eigenvalues and the trace and the determinant of the matrix does 4

5 a b Theorem Let A Then () the sum of the eigenvalues are the trace of matrix c d A (tracea a d); and () the product of the eigenvalues are the determinant of A(detA ad bc) Proof Let n, a a, a b, a c and a d in (*) We have: (**) a d ad bc 0 Let and be eigenvalues of A Then the characteristic polynomial can also be expressed as (***) deta I Comparing coefficients of and the constants in (**) and (***), we have: ad bc deta and a d tracea Note that properties () and () in Theorem hold for eigenvalues in all fields, and therefore in Z p Example 3: LetA Consider eigenvalues of A in Z 3 Then deta I 4 3 0or Solving the equation, we have and 0 Note that, 3, 0 Hence, A has eigenvalues tracea 4 3, deta A common property of eigenvector and eigenvalue for a n n matrix in is that if then their eigenvectors x and x are linearly independent This property holds for any n n matrix in the infinite field, but does it also holds in a finite field? We first show that this property holds for all n n matrices in Theorem 3 If,,, k are distinct eigenvalues of an n n matrix A with corresponding eigenvectors x,x,,x k,thenx,x,,x k are linearly independent Proof Let r be the dimension of the subspace of R n spanned by x,x,,x k Ifr k, then x,x,,x k are linearly independent Suppose that r k Without loss generality, we assume that x,x,,x r are linearly independent Since x,x,,x r,x r are linearly dependent, there exist scalars c,c,,c r,c r not all zero such that () c x c x c r x r c r x r 0 Note that c r must be nonzero, otherwise, x,x,,x r would be dependent Since 5

6 c r x r 0, c,c,, and c r cannot all be zero Assume that c j 0forsome j k Multiplying () bya, wehave () c Ax c Ax c r Ax r c x c x c r r x r 0 Subtracting r times () from(), we get c r x c r x c r r r x r 0 Since x,,x r are linearly independent, c i i r 0fori,,,,r Because c j 0, r j that contradict to the assumption that,,, r are distinct eigenvalues of A In a finite field, if eigenvalues of a matrix exist and distinct then their corresponding eigenvectors are linearly independent Yet what if the eigenvalues are not distinct? The following examples illustrate that it is not certain as to the linear independence of the eigenvectors in such a case vectors Example 4: The identity matrix 0 and has two eigenvalues and Clearly, are eigenvectors of A corresponding to eigenvalue and they are linearly independent On the other hand, the matrix A also has non-distinct 0 eigenvalues of and To find eigenvectors corresponding to eigenvalue, we solve the system of linear equations A Ix x x x 0 for a nontrivial vector x Solution vectors are of the form t Hence, A does not have two linearly x 0 independent eigenvectors for eigenvalues and There is a very important class of matrices that have quite nice properties concerning eigenvalues and eigenvectors A symmetric matrix A is a square matrix with the property that a ij a ji for all i j or equivalently A T A Matrices are symmetric Observe that for a symmetric matrix the entries above the diagonal and entries below the diagonal of the matrix are mirror images of each other and Theorem 4 Let A be an n n symmetric matrix Then the eigenvalues of A are real and the eigenvectors corresponding to distinct eigenvalues are orthogonal 6

7 Proof Let be an eigenvalue of A and x be an eigenvector corresponding to Denote by x the conjugate transpose of x and by the conjugate of SinceAx x, x Ax x x Now we take the conjugate transposes of both sides of the above equation Because A T A, the left side of the equation x Ax x A x x A T x x Ax Observe the right side of the equation: x x x xsincex x x Ax x x, Hence, is real Let and be two distinct eigenvalues of A and x and x be eigenvectors corresponding to and, respectively Because A, and are real, x and x are real We want to show that x T x 0 By definition of eigenvalue and eigenvector, we have (#) Ax x and (##) Ax x Multiplying (#) by x T,wehavex T Ax x T x SinceA T A, x T Ax T x T A T x x T Ax x T x Because of x T x x T x and, 0 x T Ax x T Ax x T x x T x x T x implies x T x 0 It can be shown through similar steps that properties in Theorem 4 for symmetric matrices still hold for n n matrices in Z p Example 5: Consider eigenvalues and eigenvectors of A in Z 3 Eigenpairs for A are,x 4, and,x 0, in InZ 3, the eigenpairs for A become,x 3, and,x 3 0, InZ 3, x T x Eigenvalues and Eigenvectors of A Over Z p In this section, we will study eigenvalues and eigenvectors of A over the extended field Z p ;, Z p For some matrices with entries in Z p, their eigenvalues do not exist in Z p Example 6: Consider eigenvalues and eigenvectors of A in Z 3 7

8 deta I det The equation 0 does not have a solution over Z p What if we extend the finite field Z 3 to Z 3 ;, Z 3 0,,,,,,,, where is a solution of the equation 0? Because 4 3 0, and are eigenvalues of A over Z 3 Notice the eigenvalues and of A form a conjugate pair This will always be the case regardless of the field, so long as the conjugate pairs exist Theorem 5 If is an eigenvalue of A, then is also an eigenvalue of a b A c d Proof Let be an eigenvalue of A over Z p From Theorem 4, we know is an solution of the equation deta I a d ad bc 0 Since the solutions are of the form: a d and exists, a d 4ad bc The eigenvalues for A above exist in the newly extended finite field Z 3 Dothe corresponding eigenvectors also exist in Z 3? The answer is yes Let us revisit Example 6 for eigenvectors of A A I X x x 0 Row Row Row RowRow So, x and x satisfy the equation: x x 0 Let x Then 8

9 X 3 Through similar steps, X is an eigenvector corresponding to Now we know the eigenvectors and eigenvalues of A exist in Z 3, do eigenvalues and eigenvectors of A satisfy the properties given in Theorems, 3 and 4? The following verifications show that they do Properties in Theorem : tracea and deta tracea, and 4 3 deta Property in Theorem 3: Since, X X for any constant Therefore, X and X are linearly independent Observe that X and X form also a conjugate pair Property in Theorem 4: Since X T X X and X are orthogonal T , In the following, we will take a closer look at our Z p cases We will examine matrices that have eigenvalues and eigenvectors in Z p and any properties that these matrices have in common We start our research with matrices in Z 3 The brute force calculations have been done and counted and the results stand as such From the 8 possible matrices there are 8 matrices that have eigenvalues and eigenvectors that will be in Z 3 Why is this? We shall put forth some theories So, Let A a c b d a d a The characteristic equation of A is: a d ad bc 0 d 4ad bc a d a d 4bc Since belongs to Z 3, a d 4bc is not a perfect square Thus we are looking for the matrices in which a d 4bc or a d 4bc for the eigenvalues to fall in Z 3 Theorem 6 For a matrix A with entries in Z 3, the eigenvalues of A are in Z 3 if the column sums or row sums of A are equal a r d Proof Let A Since r a d a d 4bc a d 4r dr a r a d a d r a d a perfect square, r in Z 3 Similarly, we can show r a r a when A Since a d and is in Z 3, is also in Z 3 r d d 9

10 Thus is only a sufficient condition The opposite is not exactly true If the column sums are not equal then the eigenvalues are not always not in Z 3 One example of such instance is an upper or lower triangular matrix whose eigenvalues are its diagonal entries Such a matrix with unequal column sums will still have eigenvalues in Z 3 since the entries are in Z 3 Yet,if we look at the theorem in this manner and then remove any upper and lower triangular matrices we will find all possible matrices from Z 3 with eigenvalues not in Z 3 The following table is a breakdown of possible matrices that have eigenvalues not in Z 3 based on the results given in Theorem 6 a c a,c b,d b,d b,d b d 0 0 0,0,0,, 0, 0,,,,0,0 0,0, 0,, 0,,,,0, 0,0,0 0,, 0,,0,, From the table above we have narrowed the possible matrices from 8 to 54 that could have eigenvalues that do not belong to Z 3 Now we will go through these and remove all upper and lower triangular matrices Such matrices have a 0 in either c and/or b Therefore: 0

11 a c a,c b,d b d 0 0,0,,,,,0, 0,,,,,0,,0 0,,,, and the number of matrices has now been reduced from 54 to 4 possible matrices in which the eigenvalues do not belong to Z 3 Now we have reduced are possibilities to 4 The last six matrices we need to eliminate are a little difficult to figure out Through brute force the six can easily be found amongst the 4, a much easier task than starting a search with 8 possibilities Yet, what is special about these six? What makes them different from the 4 that remain and yet similar to the 57 that have already been cast aside? An intriguing possibility is the row sum Taking the remaining possible matrices we are able to remove the following matrices:, 0, 0 0, 0 These six have equal rows sums in Z 3 Thus leaving us with the 8 matrices with entries in Z 3 in which their eigenvalues do not belong to Z 3 This research just scratches the surface of finding the Z p s required of Z p and the matrices that belong to them for eigenvalues to exist The filters used to find the 8 matrices of Z 3 need to be further researched on larger finite fields Will these criterion always work or is there a necessary and sufficient condition that will produce them all in one pass? We now understand the properties that eigenvalues and eigenvectors posses in finite fields It is time to find where all of these eigenvalues exist

12 References: ) Linear Algebra with Applications, Steven J Leon, Prentice Hall, Uppersaddle,NJ, 006, ) Advanced Engineering Mathematics,Zill and Cullen, Jones and Barlett Publishers, Sudbury, MA, 997 3) Physics for Scientists and Engineers, Serway and Jewett, Thompson Brooks/Cole, Belmont, CA, 004, 6th edition 4) Webster s Dictionary, Hauffman, Nickel Press, USA, 99