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Theory: The trace of a field element |

You probably already heard of the trace of a matrix. It is defined as the sum of the diagonal elements of a square matrix. What is so special about that? Well, as explained by that link, the trace is basis invariant. Let L : K be a finite field extension, like is a field extension of 2. Given some linear transformation A: L L you can write that linear transformation in the form of a matrix equation (because it is linear) after chosing some basis for L that associates L with a vector space. This equation is linear in x because y is fixed. We already saw discussed one such basis for our field , elsewhere: (1, t, t2, t3, ..., tm-1).

Lets work out a simple example using this polynomial basis. Let m = 4, using reduction polynomial t4 + t + 1. Let y = t, some element of our field 24. The linear transformation that sends x yx is then given by, Atx = tx. The trace of the matrix should be independent of the chosen basis, so that we might as well talk about "the trace of the linear transformation", or even about "the trace of y", for y uniquely determines this transformation. Multiplying with t means that this transformation sends 1 to t, t to t2, ... and tm-2 to tm-1. Only the transformation of the last member of our polynomial basis is a little different: tm-1 ttm-1 = tk + 1. Writing out the matrix for the given example gives therefore,

where xi is the coefficient corresponding to the basis element ti.

We find that in this case the trace is Tr(At) = Tr(t) = 0 (the sum of the diagonal elements of the matrix). Please note that Tr(y) is an element of K (being 2 in our case): adding the diagonal elements (which are all element of K) is still done modulo 2!

The more general matrix (for arbitrary m for which there exists an irreducible trinomial tm + tk + 1 that is used as reduction polynomial) leads to the matrix

where the lower 1 in the right-most column appears in the k-th row, and hence we can immediately see that

Since libecc forbids values of k > m/2, we can safely assume that in our case the trace of t is always 0.

Lets investigate that this trace is indeed independent of the chosen basis by applying a basis transformation. The only possible basis transformation happens to be applying Frobenius a number of times. Lets investigate applying Frobenius once.

The Frobenius map sends 1 1, t t2, t2 t4 = t + 1 and t3 t6 = t3 + t2. From which follows that the corresponding matrix is

the inverse of which is

Applying Frobenius to the equation Atx = tx gives Frob(Atx) = Frob(tx) = t Frob(x), where the last equality holds because that t is a constant. Furthermore, Frob(Atx) = Frob(At Frob-1(Frob(x))) = Frob(At Frob-1(z)), which brings us to the matrix equation

Working out the matrix multiplication we get

where the zi are the coefficients relative to the basis (1, t2, t + 1, t3 + t2). And indeed, still we have Tr(t) = 1 + 1 + 1 + 1 = 0.

More in general, any change of basis can be represented with some non-singular matrix (which means that its determinant is non-zero and thus that it has an inverse). Let the matrix B represent a change of basis, then the matrix At will change into BAt/B, preserving the value of the determinant (see formula 18 on this page) and the value of the trace (see formula 7 on this page).

The following is not a proof in any way, but it gives things a bit a place, and therefore making it easier to understand more mathematical texts that actually derive and prove things, I hope.

Consider the linear equation (using some basis)

Then i are the eigen values of A and xi are the corresponding eigen vectors. Note that for a fixed i the above represents the linear transformation (Aixi = ixi).

For the story about eigen values and corresponding eigen vectors, i has to be an element of the base field K (2), so we really only have one non-trivial eigen value: 1. Plugging that value into the equation turns A into the identity matrix and trivally all of the vector space into the corresponding eigen space (that is, every value of x is an eigen vector of the identity matrix with scaling factor 1; what else is new). But we can also consider values for y = i that are element of L, in that case we simply get the equation Ayx = yx which holds per definition for any value x L. Nevertheless, we can still call y eigen value and related theorems still hold, like that the trace of the matrix is the sum of all its eigen values (now elements of L), and the norm of the matrix (its determinant) is the product of all the eigen values. Below we assume that the eigen values y L and you can forget about the concept of eigen vectors: the equations are trivially true for any x. This paragraph was just added to take away possible confusion about that.

Obviously (Ay - yI) can't have an inverse when (Ay - yI)x = 0 for every x . The matrix will be singular therefore, its determinant will be zero and we have Ay - yI = 0. The determinant is a polynomial in y of degree m and the equation is called the characteristic equation (or polynomial) of Ay; the roots are the eigen values of the matrix.

If we choose y = t and consider

or

then the equation At - tI = 0 must be the minimal polynomial of t (that is, a monomial of degree m): it is the reduction polynomial! The corresponding At will have eigen values that are precisely the roots of the reduction polynomial. What are those roots?

We already have one root (per definition): t itself. It is near impossible to write out t in its complex form, that would result in huge formulas if at all possible to find, but we don't have to do that. We can express the other roots easily in t by repeatedly applying Frobenius.

For example, let the reduction polynomial be t4 + t + 1 = 0. Then by replacing each t with its square, the equation still holds: (t2)4 + t2 + 1 = t4t4 + t2 + 1 = (t + 1)2 + t2 + 1 = t2 + 1 + t2 + 1 = 0. And doing that again, it still holds: (t4)4 + t4 + 1 = (t + 1)4 + t4 + 1 = 0, and so on. After all, the Frobenius map is an automorphism. If t is a generator of the field and its order is n = 2m - 1, hence n is the smallest positive integer such that tn = 1, then applying Frobenius m - 1 times will lead necessarily to m different values. The roots of the reduction polynomial are therefore given by the set (t, t2, t22, ..., t2m-1) and they represent the eigen values of the matrix At in the linear transformation Atx = tx, independent of the basis. If t is not a generator of the field, which is possible when m is non-prime, it might happen that the same eigen value is shared by more than one eigen vector, but in that case we have to add the same eigen value multiple times. Therefore, the trace of t is given by the sum of this set and the detAt is given by the product of this set.

The same story holds for an arbitrary value of y (not zero), and we find

Also note that which makes us jump of joy because it means that every non-zero y has an inverse, as should be the case for field elements!

The roots of the reduction polynomial are not always linear independent and we can't use them as a basis, but it can be proven that there will always exist some element such that (, 2, 22, ..., 2m-1) is linear independent. Such a basis is called a normal basis.

As an example, let the reduction polynomial again be t4 + t + 1. The set (t, t2, t4, t8) = (t, t2, t + 1, t2 + 1) is clearly not linear independent. However, we can chose = t3 and find a normal basis (t3, t3 + t2, t3 + t2 + t + 1, t3 + t). If next we want to find the matrix that corresponds with Atx = tx (multiplication with t) then we have to first figure out how the basis elements are converted. t(t3) = t4 = t + 1 = (t3 + t2) + (t3 + t2 + t + 1). t(t3 + t2) = t3 + t + 1 = (t3) + (t3 + t2) + (t3 + t2 + t + 1). t(t3 + t2 + t + 1) = t3 + t2 + 1 = (t3) + (t3 + t2 + t + 1) + (t3 + t). t(t3 + t) = t2 + t + 1 = (t3) + (t3 + t2 + t + 1).

And thus we have

Note that again Tr(t) = 0, as it should be. Lets have a look at an element whose trace will not be 0 for change, like t3. Using the above formula for the trace we can immediately calculate it.

Note that this is exactly the sum of the normal basis that we used above.

Lets do one more check with a matrix, using this same basis. The basis elements are converted as follows: t3(t3) = (t3 + t2). t3(t3 + t2) = (t3 + t). t3(t3 + t2 + t + 1) = 1 = (t3) + (t3 + t2) + (t3 + t2 + t + 1) + (t3 + t). t3(t3 + t) = (t3 + t2 + t + 1).

And thus we have

and as expected, the trace of this matrix is 1.

We can make an interesting observation here. A vector relative to a normal basis (, 2, 22, ..., 2m-1) allows us to denote every single element of the field of course, otherwise it wasn't a basis. The zero is always given by the vector (0, 0, 0, ..., 0), and thus it is impossible that the vector (1, 1, 1, ... 1), which corresponds to adding all elements of the normal basis, would be zero as well. Moreover, adding all elements of a normal basis means adding all the roots of , it is equal to the trace of , and a trace is element of the base field and can therefore only be equal to 0 or 1 in our case! Hence, in order for (, 2, 22, ..., 2m-1) to be a normal basis we must have 0 Tr() = 1 !

Now consider the following magic. When , in combination with Frobenius, can be used to form a normal basis, then so can 2 because if you replace with 2 in the basis you simply get (2, 22, ..., 2m-1, ). Therefore, it must be that Tr(2) = 1 too. The inverse is also true, if 2 can be used to generate a normal basis then so can . Hence, we can conclude that for any arbitrary element y

This result is not too weird, considering that Frobenius is an automorphism that basically just changes the basis - and as we saw before, the trace is not dependent on the basis.

In other words, Tr(y - Frob(y)) = Tr(y) - Tr(Frob(y)) = 0. And therefore we can conclude that the equation x = y + y2 can only have solutions when Tr(x) = 0 ! Note that Frob(y) = y2 and that, since we are working with characteristic 2, y + y2 = y - y2.

And this is the result I needed to prove Hypothesis 1.

THEOREM 5

Let x . Then Tr(x) = 0 iff there exists a y such that x = y + y2.

PROOF

There is another, easy way to prove theorem 5. Suppose there is a solution to the equation x = y + y2, so that for a given x and y the equation is true. Then applying Frobenius to both sides (squaring both sides) results again in in an equation that is true of course. We can repeat squaring both sides precisely m - 1 times at which point the term y2 will turn into y because y2m = y. If we subsequently add up all those equations we get immedeate proof that the trace of x is 0.

where the sum of all left-hand-sides is precisely the trace of x and the sum of all right-hand-sides is zero!

Weew, you have no idea how glad I am that I finally got to this point without using sophisticated mathematics! I worked five days on the above. Hopefully I reached my goal and you are now reasonably comfortable with Frobenius and traces. Just for the kicks (and so you appreciate my efforts a bit more), here is the "sophisticated" derivation:

Let G be the Galois group of the Galois extension L/K, then (Hilbert's theorem 90). Therefore, if G is cyclic with generator g, this is the same as Norm(x) = 1 x = y/g(y). And Tr(x) = 0 x = y - g(y).

Also trace (of field elements) is usually explained with Galois theory. Maybe I'll add a chapter about Galois Theory later to this project.

A week after I wrote the above, I found emperically an extremely fast way to find the trace of an element of . I showed this method on an IRC channel on 10 December 2004 and posted it to the sci.math news group on 11 December 2004. I managed to find a prove for it on 18 December 2004 (see below). Unfortunately, then I found a (yet unpublished) article on the net that proved the same thing (theorem 7 of On the Number of Trace-One Elements in Polynomial Bases for F2n).

Although we found this independently, and although the other article will probably not be published before 2005, I suppose I can't publish it myself anymore; too bad.

THEOREM 6

Let tj with reduction polynomial tm + tk0 + tk1 + + tkn + 1, where each ki < m / 2. Then the trace of tj is 1 iff j is odd and there exists an i such that j = m - ki, or when j = 0 and m is odd.

Please give me credit if you use it anywhere.

PROOF

We start this proof with a repetion of the theory of eigenvalues and the trace, but slightly more rigorous because we need that to understand how the eigenvalues of a power of t relates to the eigenvalues of t.

Let be the degree m extension field of 2. Then = 2[t]/(r(t)) where r(t) is an irreducible polynomial of degree m over 2. The element = t is a root of r(t) in , and { 1, , 2, ..., m-1 } is a basis for over 2, called the polynomial basis. The trace of an element x was shown before to be .

Simultaneously, is a vector space V of dimension m over 2. And the element x can be considered to be a linear transformation x: V V, representing multiplication with x, carrying any element y to xy. Using the polynomial basis, let T be the matrix that carries any element y to ty. Thus, Ty = ty for any vector y V and where t is the special element as defined by the reduction map. Then T has the general form

where the ri 2 are the coefficients of the reduction polynomial r(t) = tm + rm-1tm-1 + ... + r1 t + r0. Note that always r0 = 1 or the reduction polynomial would not be irreducible.

Also note that the trace of this matrix (the sum of the diagonal elements), and thus of t, equals rm-1 (a well-known fact, as is proven for example by Bhargav (lemma 3.2)), Tr(t) = rm-1.

The characteristic polynomial of a linear transformation T is defined as f(e) = det(T - eI), where I is the identity matrix. Obviously it has coeficients in 2, the ground field of V. The equation f(e) = 0 has a solution if and only if (T - eI)(v) = 0 for some nonzero vector v; in general, for any linear transformation A, det(A) = 0 if and only if Av = 0 is satisfied by a nonzero vector v. We are taking the special case of A = (T - eI). Now note that (e, v) is an eigenvalue - eigenvector pair of T if and only if Tv = ev, which is true if and only if (T - eI)(v) = 0. e is an eigenvalue of T if and only if f(e) = 0, where f is the above mentioned characteristic polynomial.

Now let T again be the matrix that carries an element y to ty. Then we will show that the characteristic polynomial of T, f(e), is precisely the reduction polynomial r and thus e will be an eigenvalue of T if and only if r(e) = 0, and the eigenvalues of T turn out to be the roots of the reduction polynomial.

The characteristic polynomial of T is det(T - eI) is

It is easy to see that this determinant is the reduction polynomial because it is equal to

and the second underdeterminant can be worked out to

and thus

then use this formula recursively to find that this is equal to

Ok, almost equal then. However, in our case with characteristic 2, the -1 is congruent 1 and in any case the equation r(e) = 0 remains the equation that needs to be satisfied in order for e to be an eigenvalue of T.

We already have one root of the characteristic polynomial of T therefore, because r(t) = 0 by definition. As was shown before, the other roots of the reduction polynomial can be obtained by applying Frobenius m - 1 times and we find that the eigenvalues of T are e { t, t2, t4, ..., t2m-1 }. Note that the eigenvector v that belongs to a given eigenvalue e is (1 e e2 e3 ... em-1)T (and any vector that can be obtained by multiplying this vector with a non-zero scalar of course; there will be 2m - 1 different vectors that satisfy Tv = ev).

The trace of a linear transformation T is known to be equal to the sum of the eigenvalues of T. So, we find again that . Note that the norm of T (its determinant) is known to be equal to the product of the eigenvalues of T. And thus we have T = = t(2m-1) = 1.

Now we are ready to look new things.

Consider the matrix Tn, then it is obvious that this matrix will have the exact same eigenvectors as T; for the eigenvalue-eigenvector pair (e, v) of T we find that Tn v = en v and thus v is an eigenvector of Tn with eigenvalue en.

From this we can conclude that the trace of tn equals the sum . Note that when t is a generator of the field so that any x can be written as a power of t, then we arrive again at the general formula for the trace of a field element . If the field is not primitive (t is not a generator), then we just have to find some generator g and use (g, g2, g3, ..., gm-1) as basis and will get the same result. And while we're at it, also note that Tn = = Tn = 1.

We found above that the eigenvalues of T are precisely the roots of the reduction polynomial. That means that we can write the reduction polynomial as r(t) = (t - e1)(t - e2) (t - em)

Where the ei, the m eigenvalues of T, are the zeros. Expanding and using induction, we see that r(t) = tm - S1tm-1 + S2tm-2 + + (-1)mSm

Or, since we have characteristic 2, r(t) = tm + S1tm-1 + S2tm-2 + + Sm where the Si are the Elementary Symmetric Polynomials of the m variables e1, e2, ..., em.

Moreover, using the fact that , we can write

Tr(tn) = Nn where the Nn are the Newton Symmetric Polynomials of the m variables e1, e2, ..., em.

If we restrict ourselfs to reduction polynomials of the form tm + tk0 + tk1 + + tkn + 1, where each ki m / 2, then the first floor((m - 1) / 2) ESP's are zero.

Using the fact that the characteristic of the field is 2, the relationship between NSP's and ESP's becomes

From this we see immediately that, since Si = 0 for all i floor((m - 1) / 2), also Ni = 0 for all i floor((m - 1) / 2). Whence the sum is zero for any value n < m.

For example, when m = 6 (even m is most critical) then only the first two ESP's and NSP's are garanteed zero: S1 = S2 = N1 = N2 = 0. And the sum exists of a maximum number of terms when n is maximum, n = m - 1 = 5, resulting in the sum = S1N4 + S2N3 + S3N2 + S4N1 = 0.

In other words, again remembering that the characteristic is 2, we find that for any 0 < n < m

This doesn't cover n = 0, but we already know that Tr(1) = 1 iff m is odd.

QED

Copyright © 2002-2008 Carlo Wood. All rights reserved.…...

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...As for my first assignment for entrepreneur’s subject, I would like to discuss about the criteria and how the entrepreneur should think. First and foremost the entrepreneur character can be build, can be groomed from experienced and some are born with the DNA in their blood!. Who is the entrepreneur? Entrepreneur is someone can see opportunities and situation in which new goods, services, raw materials and organizing methods can be introduced and sold at greater than their cost of production. Someone with the character of entrepreneur will always see problems, challenges and uncertainties as an opportunity for them. The only job that don’t have any qualification to apply is an entrepreneur. Anybody can be an entrepreneur, once they sustain in this field over failure because failure is the only thing guaranteed. How they respond to failure determines their success. Successful entrepreneurs are paid for their high tolerance for stress and pain. Setbacks, obstacles and challenges are painfully common elements of entrepreneurship. Most people react to these hurdles with stress and pessimism, with an attitude that obstacles are negative experiences that only hinder progress. As an entrepreneur, they encounter so many challenges they simply can’t afford to react this way. Instead, successful entrepreneurs view challenges as opportunities. Each challenge or setback reveals a key opportunity to grow -either to improve upon an existing weakness or take measures to avoid experiencing a...

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...perceived as a living and breathing human being, who act as a social animal and is growth with the sensibility to opportunities and threats around.This article is mainly talking about the importance of sensing and manipulating as a entrepreneur. Firstly, two pictures of entrepreneurship will be illustrated with understanding about the principal of entrepreneurship. Afterward, two main differentiate characteristics of entrepreneurship as a life will be summarized together with evidences. Besides, how future entrepreneurship create values will be speculated associating with the business trends. Entrepreneurship——as a living body or as a robot Entrepreneurship, just like a navigation over red oceans, is a strategic operation in a market filled with fierce competition. It is not enough for an entrepreneur to have merely the determination of success and knowledge, and that is what a manager do to act as a interest-directed professional talents. According to Gibb’s theory, Entrepreneurship is no longer understood as a business-thinking components but a dynamic human being. In conventional understanding, referred to textbook (p210, Figure9.3) Entrepreneurship is a sub-set of business in a context of business. According to Gibb’s summary, entrepreneurs are always behave like a manager of a large business, they will have a forehead preparation such as a comprehensive plan and a clearly predicted interest goal, they behave more resource-oriented in their plan practices.For......

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...• An entrepreneur is someone who can take any idea, whether it be a product and/or service, and have the skill set, will and courage to take extreme risk to do whatever it takes to turn that concept into reality and not only bring it to market, but make it a viable product and/or service that people want or need," • Most people think being an entrepreneur is all about coming up with an idea, but that's just one part • entrepreneurship entails recognizing the right opportunity, finding resources — such as funding and tools — to pursue the opportunity and creating the right team to do so . • Although there are no specific traits all entrepreneurs share, there are certain characteristics that most successful entrepreneurs possess • Ability to plan: Entrepreneurs must be able to develop business plans to meet goals in a variety of areas, including finance, marketing, production, sales and personnel. Communication skills: Entrepreneurs should be able to explain, discuss, sell and market their goods or services. Marketing skills: Good marketing skills, which result in people wanting to buy goods or services, are critical to entrepreneurial success. Interpersonal skills: The ability to establish and maintain positive relationships with customers and clients, employees,......

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...them together. In 1869, Edison comes up with a Boston steamship, landed in New York City. This project left him poor and penniless, and in debt. Today with Thomas Edison electrical invention we now have the ability to have electricity to turn appliances on and off like dishwasher, washers and dryers. According to Anita Ruddick, founder of the hugely successful fair trade outlet The Body Shop’ said, “Nobody talks of entrepreneurship as survival, but that’s exactly what it is and what nurtures creative thinking. Innovation plays a large role in the study of entrepreneurship and the two ideas are closely linked. Entrepreneurship involves the creation of new products with new ideas at the forecourt of new products and it is important that entrepreneurs have the imagination and vision in order to make their ideas a success. Anita Roddick develops and markets natural cosmetic products during the research on Anita Roddick I learned she looks for employees who show their interest for what they are doing rather than their qualification for the job. She employs people with bid ideas. Employ people that will contribute to or detract from their association’s presence in the marketplace. Anita Roddick makes it easy for people to under and how their contribution to the work program makes a positive difference in the world and productivity will increase dramatically. Anita Roddick said “Nothing is more motivating than giving staff, employees, and associates the opportunity to express their......

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...then says that this was also one of the best decisions he ever made. I believe that any decision he makes will end up working out because he believes in himself. He says you can’t connect the dots looking forward, only looking backwards afterwards, so you have to trust that they will connect. You have to trust in something; your gut, karma, destiny, something. He believed in something whatever that something was. Steve Jobs said he used to walk downtown to receive one good meal a week. He slept in his friends dorms every night. He made the sacrifice, he took the risk, never settled, and he never gave up. He did what he loved and did “great work”. Every characteristic of an entrepreneur I can think of, Steve Jobs has it. As a future entrepreneur, this video inspires me to make sacrifices. Every good entrepreneur must make some sort of sacrifices in their life to do what they love and truly believe in. Whether that’s living off ramen noodles for a couple years, sleeping on the floors of your friends dorm room, or giving up social life sacrifices must be made in order to succeed. “If you live each day as if it were your last, someday you’ll most certainly be right.” This is a funny quote, but at the same time it is powerful. Far too often I find myself being lazy and not how I would want to live my life. He said he’s looking in the mirror every morning and said, “if today were the last day of my life, would I want to do what I’m about to do today.” This is something that I......

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