134 - תרגילים מעל שדה סופי

Summary

Join Dr. Aliza Malek in this insightful session as we delve into the intricacies of vector subspaces over finite fields. The lesson focuses on the concept that in a finite field, the closure to scalar multiplication condition can often be omitted if other conditions are satisfied. Throughout the video, Dr. Malek illustrates this point with detailed examples and exercises, ultimately showing that understanding finite fields can simplify problems in linear algebra. By exploring these unique properties, you'll gain a deeper comprehension of how vector spaces operate in finite fields, with real-world implications for mathematical and scientific applications.

Highlights

• In finite fields, two conditions - non-emptiness and closure to addition - can replace the usual three for subspaces. 😲
• Scalar multiplication in finite fields is like repeated addition, except when multiplying by zero. 🔄
• The tutorial explores why some conditions might be omitted in finite fields but not in infinite fields. 📏
• Interesting exercises demonstrate the differences between finite and infinite fields. 💪
• Provides comprehensive examples showing why certain conditions don't apply to finite fields. 📈

Key Takeaways

• Learn the unique properties of vector subspaces in finite fields. ✨
• Explore why closure to scalar multiplication isn't always necessary in finite fields. 🚀
• Understand the implications of finite fields in simplifying linear algebra problems. 🔍
• Discover the difference in subspace conditions between finite and infinite fields. 🚧
• See how the tutorial composes a list of all subspaces for a specific finite field example. 📊

Overview

The video kicks off by introducing vector spaces over finite fields. Dr. Aliza Malek outlines the definitions and conditions that usually apply to vector subspaces, setting up the scenario for finite fields where some conditions can be omitted. She emphasizes that, in finite fields, establishing a subspace requires only that the space is non-empty and closed under addition.

As the lesson progresses, Dr. Malek delves into scalar multiplication in finite fields, showing how it can be expressed as repeated addition. This simplification doesn't hold in infinite fields due to the complexity introduced by negative numbers and fractions. Through examples, she illustrates that a non-empty set, when closed under scalar multiplication, does not guarantee closure under addition.

The tutorial concludes with a fascinating exercise where participants explore the number of subspaces within a finite vector space. This hands-on example reinforces the learned principles, demonstrating the finite nature of options in these mathematical structures. Overall, the video effectively demystifies the relationships and rules applicable to vector spaces over finite fields.

Chapters

• 00:00 - 00:30: Chapter 4 - Vector Spaces Chapter 4 provides a detailed exploration of vector spaces, specifically focusing on exercises over a finite field in the context of linear algebra. Dr. Aliza Malek guides the lesson, emphasizing a break from traditional lectures to engage with exercises. The chapter starts with a reminder of the fundamental definition of a vector space V over a field F and the criteria for identifying W as a subspace. The distinction between the ten properties of a vector space and the three properties necessary for a subspace is highlighted as a foundation for the exercises.

134 - תרגילים מעל שדה סופי Transcription

• 00:00 - 00:30 Chapter 4 - Vector Spaces Exercises Over a Finite Field Linear Algebra Course Dr. Aliza Malek Hello, thank you for coming back. In this lesson we will take a little break, and we see exercises of a vector subspace over a finite field. So let's begin. I remind you of the definition, I have V, a vector space, over the field F, W is a subset of V. W is a subspace, if and only if, W satisfies the following three properties. Remember, we have a vector space, then it's ten properties, A subspace, then only three properties, let's recall what they are.
• 00:30 - 01:00 W is not an empty set, W is closed to addition, and W is closed to scalar multiplication. Now we will focus on the case of a finite field, that is, F is equal to Zp, where p is a prime number. We will only work with finite fields. So here is the first exercise: V is a vector space over Zp, W is a subset of V, then W is a subspace of V, if and only if...
• 01:00 - 01:30 Let's see what we want to assume here, W is not an empty set, and W is closed to addition. Indeed, we lack closure to scalar multiplication. This exercise means that if the field is finite, just two conditions are enough. A non-empty set, and closure to addition. Closure to scalar multiplication can be omitted, let's see this.:
• 01:30 - 02:00 If W is a subspace, then we know that it satisfies properties A, B, and C, is not empty, closed to addition, closed to scalar multiplication. Wait, but, non-empty, and closed to addition, are exactly our two conditions. So if it meets all three, it surely meets the two that are given. Therefore, the tricky part is the other direction. We only know A and B, they are the 1 and 2 given to us.
• 02:00 - 02:30 That is, out of the three conditions that need to be proven, A, B, and C, A and B I already have, that's a given, I only have to prove C. So let's show that W is really closed to scalar multiplication. What does it mean? We take some v in W, we take some scalar in the field F, which is Zp. And we want to show that α times v belongs to W.
• 02:30 - 03:00 If α happens to be not 0, What does that mean? This means that α is equal to 1, or 2, up to p-1, these are the terms found in Zp. Without 0, it's from 1 to p-1. Then, what does αv mean? αv means v+v+v, and so on, α number of times. Either 1 time v, or 2 times we take v+v, or v+v+v, which is α=3, and so on.
• 03:00 - 03:30 We add v, α number of times, to itself. And how do I know all this is in W? Because I am given that W is closed to addition. Now we only have to check the case where α equals 0. Because I can't say, v+v+v, zero times. Therefore, α=0 should be checked separately. So if α is equal to 0, we know that v belongs to V, which is a vector space.
• 03:30 - 04:00 Therefore, α times v, which is actually 0v, equals 0. We saw that in a vector space, no matter over which field, 0 multiplied by a vector gives the 0 vector. So, if my scalar is 0, showing closure to multiplication by scalar 0, means showing that 0 is in W. That's what we need to show, because 0 times v gives the zero vector. So we only have to show,
• 04:00 - 04:30 that if there is closure to addition, and the set is not empty, we are guaranteed that 0 is in W. So we will use the fact that W is not an empty set. If I know that it is not an empty set, surely there is some vector in W. Only, I don't know which one it is. It could be that this vector is directly the zero vector, then I'm done. Because I only have to show that 0 is inside. So if you happen to tell me that 0 is inside, then I'm done.
• 04:30 - 05:00 But if this vector I took outside is not necessarily the zero vector, If my vector v is different from 0, what do we do? We saw a moment ago that p-1 times v, does belong to W, because there is closure to addition, v+v+v, p-1 times, is definitely inside. So I certainly have the vector (p-1)v in W. So what do I have now in W? v, and p-1 times v, are both in W.
• 05:00 - 05:30 And W is closed to addition, so let's add them. So I also know that (p-1)v plus v, also belongs to W. But what does (p-1)v plus v equal to? It is equal to pv. But above Zp, p is 0, that's why it says 0v over here. 0v is of course is the zero vector, and it is in W, Because the sum (p-1)v plus v, is in W.
• 05:30 - 06:00 And so I was able to show that the zero is also in there. In fact, we got that in a finite field, scalar multiplication is the same as repeated addition, except for multiplying by zero, but the zero we tested separately. Here is an interesting question: Out of A, B, and C, we only took A and B, a non-empty set, closed to addition, and here we got a subspace.
• 06:00 - 06:30 A very nice question is asked here, and if we take A and C? That is, a non-empty set that is closed to scalar multiplication. Will this also give us a subspace? So the answer is no. A non-empty set that is closed to scalar multiplication, will not necessarily be closed to addition. Because repeated addition means we add the same vector to itself over and over again, and it will stay inside all the time.
• 06:30 - 07:00 But I am not guaranteed that if I add two different vectors, that I will stay inside. Let's see an example of this: I take the field F to be Z2, 0 and 1, the smallest one there is. And I take my space to be Z2, F squared. It's Z2 squared, that is, all terms of the form a,b, the ordered pairs a,b,
• 07:00 - 07:30 while a and b belong to F, F means Z2. Let's take W, some subset, 0,0, 1,0, and 0,1. W is not an empty set, here are three terms, and it is also closed to scalar multiplication. How do I know that it is closed to scalar multiplication? Here is a vector, 1,0,
• 07:30 - 08:00 if I multiply it by a scalar, I have only two options, either to multiply it by 1, itself, and here it is, or to multiply by 0, and get 0, and the 0 is already here. The same goes for the second vector. If I take 0,1, I can multiply it by 1, itself, multiply it by 0, get 0, and so I have closure to scalar multiplication. In fact, over Z2, every set containing the zero vector, is closed to scalar multiplication. Why?
• 08:00 - 08:30 Because the vector is already inside, we multiply it by 1, then by itself, we multiply it by 0, then it is the 0 vector. So if the 0 vector is already inside, that's it, we have closure to scalar multiplication. But, W is not a subspace, because W is not closed to addition. If I now take the two vectors, 1,0 and 0,1, and add them, I get 1,1, and it is no longer in W. So I have a non-empty set, it is closed to scalar multiplication,
• 08:30 - 09:00 but it is not closed to addition, so it is not a subspace. How would we build an example over Z3? Over Z2 is easy, we only multiply by 0,1, and we are done. So here is how we do it: Take two different vectors, which are not a multiple of each other, like these two, and add all their multiples to the set, let's see: I first take the zero vector, which is one of the multiples, no matter whose,
• 09:00 - 09:30 here is the first vector, 1,0. So 1,0 I multiply by 1, get itself, multiply by 0, get itself, but in Z3 I can also multiply by 2, so let's add the result, 2,0, as well. That's it, I'm out of multiples. Now we will take a new vector that is not here in the list, here, 0,1. We will add its multiples. What multiples can I have? Multiplication by 0 is already inside, multiplication by 1 is itself, and multiplication by 2, that we are missing, we add it, and we are done.
• 09:30 - 10:00 Now we get that the set is not empty, there are many vectors here, and it is closed to scalar multiplication, that's how we built it, we took all the scalar multiplications. But again, we don't have closure to addition, because when I take the two basic vectors that I started with, 0,1 and 1,0, and I add them, I get something that is already outside, 1,1 is no longer in the set. That's it, I showed you how to build an example in general.
• 10:00 - 10:30 And if we take only A and B in an infinite field, will it be a subspace? Definitely not. In an infinite field, scalar multiplication is really not equivalent to repeated addition, only natural numbers multiplication. To take seven times is to add it to itself seven times, but half number of times? It is not possible to add a term to itself half number of times, so it won't work.
• 10:30 - 11:00 There are negative numbers, there are fractions, it is impossible to convert scalar multiplication to repeated addition, and so closure to addition won't be enough. And think about it, if that was enough, we wouldn't have made it a condition. Here is another exercise, again I take F to be Z2, which is the terms 0,1, and I take V to be the same thing, F squared, which is all the ordered pairs with terms in the field F, in Z2, that is,
• 11:00 - 11:30 and we want to find all the subspaces that exist in V. Pay attention, we are over a finite field, everything here is finite, we can literally count how many subspaces we have here. Let's do just that. So we start from the trivial spaces, the 0 space, and the entire space. Those are surely there. Now we will start taking subspaces that are not the trivial space.
• 11:30 - 12:00 How do we do that? Suppose W is a non-trivial subspace, what does that mean? That there must be some vector v in W, that is not the 0 vector. Because if only the 0 vector is there, it is trivial. So we take a vector v in W, which is not the zero vector. Now we will run through all the possibilities for v to be different from 0, and we get all the following subspaces. W1 will be 0,0 with 1,0.
• 12:00 - 12:30 This is a set that is closed to scalar multiplication, and is not empty. And what? Well, being closed to scalar multiplication doesn't give us a subspace, remember? We want closure to addition, and that the 0 is inside. So here is the 0 inside: 1,0, we will add 1,0 to it, because there is nothing else to add, we will get 0,0.
• 12:30 - 13:00 Now let's take W2, 0,0 with 0,1. And what are we left with? There is one more vector we didn't take, it's 0,0 with 1,1. These are all subsets, not empty, the 0 is inside, are closed to addition, and are also closed to scalar multiplication, although we don't have to check that as well. Do we have more? Pay attention.
• 13:00 - 13:30 If W has another vector u, which is not v, and is not 0, that means we already have at least three. But don't forget, that if I took u which is neither v nor 0, for example, neither 0 nor this, I also have to take u plus v, which means, how many vectors will I have there? I will have 0, I will have v, I will have u, and I will have u+v, It's four vectors. Wait, but four vectors is the entire space, right?
• 13:30 - 14:00 a has two options, and b also has two options, This means that in F2, when F is Z2, there are only four vectors. So once I added another vector to the pair, 0,1 for that matter, I also have to add v+u, so that there is closure to addition. It doesn't happen automatically, remember? So we already have four vectors, but four vectors is the entire space.
• 14:00 - 14:30 That is, it is the trivial space V, and so there are no more options to get subspaces, except for what we have already listed. What does it mean? 0, V, and W1, W2, W3, that we listed over here. So in total, F2, when F is Z2, has five subspaces. That's it friends, we are done for now. Thank you very much.