128 - תת מרחב וקטורי: דוגמאות כלליות

Summary

This lesson dives into the world of vector subspaces, explaining the foundational properties that determine whether a subset is indeed a subspace. Dr. Aliza Malek outlines the criteria necessary for identifying subspaces within vector spaces, employing illustrative examples to reinforce the concepts. She emphasizes the importance of verifying properties like non-emptiness, closure to addition, and scalar multiplication. Through detailed calculations, the lesson demonstrates how these principles apply in practice, particularly with simple and more complex linear conditions. The session concludes with a discussion on homogeneous versus inhomogeneous systems of equations, juxtaposing their suitability as vector subspaces.

Highlights

• W1, a subset of R2, is used to illustrate subspaces by satisfying specific conditions like 'a + 2b = 0'. 📏
• Dr. Malek emphasizes checking for the zero vector, as it simplifies confirming whether a set isn't empty and serves as a vector's existence proof. 🔍
• Through rigorous demonstration, vector addition and scalar multiplication are confirmed for W1, showing it as a subspace of R2. ➕
• The lesson highlights that homogeneous systems have solution sets naturally forming vector spaces, unlike their inhomogeneous counterparts. 🔑
• Engagement with audience Q&A enriches the comprehension of conditions affecting subspace validity. 🗣️

Key Takeaways

• Understanding vector subspaces is crucial: They must not be empty, and need closure under addition and scalar multiplication to qualify as subspaces. 📘
• The zero vector plays a vital role: It's essential to start with the zero vector as it's the only vector guaranteed to exist in any subspace. 0️⃣
• Homogeneous linear equations naturally lead to vector subspaces, while inhomogeneous ones do not due to lack of closure properties. 🔍
• Learning through examples: Subspaces can form from sets like R2 with specific conditions, but they must meet fundamental vector space properties to be verified. 🧩
• Curiosity and questions drive deeper understanding, as evident by the intriguing inquiries into additional conditions and complex equation systems. 💡

Overview

In this enlightening session, Dr. Aliza Malek breaks down the necessary characteristics that define vector subspaces using practical examples and rigorous proofs. We begin with a fundamental review of subspaces, emphasizing the condition requirements like non-emptiness, closure under addition, and scalar multiplication. The lesson is crafted with a showcase of how to identify such subspaces, pivotal for understanding more complex vector spaces.

Focusing on the subset W1 within R2, Dr. Malek demonstrates the analysis required to confirm if it serves as a legitimate vector subspace using detailed calculations. This exploration includes calculating vector additions and verifying the zero vector's role within these spaces. The importance of focusing on zero solutions is highlighted, revealing deeper insights into why it's a recommended step during such analysis.

The class wraps up with an engaging examination of homogeneous versus inhomogeneous systems of linear equations. Here, Dr. Malek outlines why homogeneous systems naturally align as vector subspaces due to their closure properties, whereas inhomogeneous systems fall short. This dichotomy in understanding fosters a richer grasp of linear algebra concepts, ultimately cementing a foundational understanding essential for further study.

Chapters

• 00:00 - 00:30: Chapter 4 - Vector Spaces Chapter 4 discusses Vector Spaces, presented by Dr. Aliza Malek. The chapter explores the concept of vector spaces and vector subspaces. After defining these terms and outlining how to verify them, the chapter turns to examples. It initially presents general examples of vector subspaces and illustrates these within familiar vector spaces. This chapter serves as a foundational exploration into vector subspaces, paving the way for further detailed examples in subsequent lessons.
• 00:30 - 02:00: Properties of Vector Subspaces A vector subspace is a subset of a vector space that meets certain criteria.
• 02:00 - 05:30: Checking if a Set is a Subspace The chapter discusses how to determine if a given set is a subspace of a vector space. It uses the example of the vector space R2 over the field R to illustrate the concept. The set W1 is defined as the set of all vectors (a, b) in R2 such that the linear equation a + 2b = 0 holds.
• 05:30 - 11:00: Example: Subspace in R2 This chapter discusses the concept of subspaces in the context of R2, illustrating the conditions under which a subset of vectors, such as W1, is not the entire space R2. It uses the example of the vector (1,1) to show that if vectors don't satisfy a certain condition (e.g., 1 + 2*1 ≠ 0), they are not part of W1. The chapter further investigates whether W1 can be considered a subspace of V by examining the three properties required for a vector subspace, beginning with the necessity that W1 is not an empty set.
• 11:00 - 15:30: Example: Solutions of Homogeneous Systems The chapter "Solutions of Homogeneous Systems" delves into the properties of the vector space W1, which is not equivalent to all of R2. The discussion includes an example of a vector V, given as (-2,1), that satisfies the criterion for being a part of W1. The condition is expressed as a linear combination: a*(-2) + 2b + 2*1 = 0, which simplifies to -2 + 2*1 = 0, affirming that V belongs to W1. Additionally, the chapter notes that the zero vector (0,0) also meets the necessary condition to belong to W1, underscoring the inclusion of the zero vector in this subset.
• 15:30 - 16:00: Questions on Homogeneous and Inhomogeneous Systems The chapter explores the concept of homogeneous and inhomogeneous systems, highlighting the role of the 0 vector in these systems. It explains that the 0 vector, defined as 0 plus 2 times 0 equals 0, fulfills a specific condition that makes it belong to the set W1. This indicates that W1 is not an empty set.

128 - תת מרחב וקטורי: דוגמאות כלליות Transcription

• 00:00 - 00:30 Chapter 4 - Vector Spaces Vector Subspace - General Examples Linear Algebra Course Dr. Aliza Malek Hello, thank you for coming back. After we have seen what a vector space is, and what a vector subspace is, and how to check it, it is time to see examples. In the upcoming lessons we will see many examples. Let's start with something general, and after that, in any standard vector space we have seen, we will see some examples of subspaces, so let's begin. First let's recall, V is a vector space, W is a subset,
• 00:30 - 01:00 W is a vector subspace of V if it meets the following three properties: It is not a set, it's closed to addition, and closed to multiplication by a scalar. And as we said, it can also be written in a slightly shorter way, properties B and C together become B-tag. Alpha u plus beta v belongs to W. We wrote properties B and C in a slightly shorter way. For some of the examples we will examine properties A, B, C, and for some, we will check A and B-tag.
• 01:00 - 01:30 Property A must never be given up on, remember? Let's see some examples. We will start from V equal to R2, a vector space over the field R, a well-known, familiar, standard vector space. We will take W1 to be the following set: All a,b in V, all a,b in R2, so that a plus 2b is equal to 0.
• 01:30 - 02:00 And this is the condition, so that it will not be the entire space. It is clear to you that W1 is not all of R2. For example, 1,1, 1,1 is not found in W1. Because 1 plus two times 1 is not equal to 0. And so it is not the entire space R2. We want to know whether W1 is a subspace of V. We will see the three properties of a vector subspace. The first property, we must make sure that W1 is not an empty set.
• 02:00 - 02:30 We said that W1 is not all of R2. Does anything even live in there? Sure, here it is: V, which is equal to minus 2,1, is in W1, how do I know that? Because the condition is met. a minus 2, plus two times b, plus 2 times 1, is equal to 0. Minus 2, plus 2 times 1, equals 0. Therefore, this is a vector that is in the set W1. Wait, the 0 vector, which is equal to 0,0, also fulfills the condition.
• 02:30 - 03:00 Because 0 plus 2 times 0, equals 0. Pay attention, there are a lot of zeros here, don't get confused. 0 equals 0,0. This is the 0 vector, and these are the number 0 in each element, and it is what fulfills the condition. Therefore, the 0 vector belongs to W1. It is not an empty set.
• 03:00 - 03:30 I am being asked me here whether it matters how we check? To show that the set is not empty, does it matter if we show a vector like minus 2,1, or show the 0 vector? Does it matter? And the answer is absolutely not. But, we recommend checking that the set is not empty by having the 0 vector. Why is it recommended to check the 0 vector? We have several reasons why we check the 0 vector, and not something else.
• 03:30 - 04:00 It's not mandatory, it's just a recommendation. What are the reasons? Let's see several of them: First of all, every subspace must have the 0 vector. And so it's something we already know should be there. So let's look for it. This is the only vector whose existence in the space can be guaranteed. Last lesson we saw the 0 space, remember? One of the trivial subspaces. In the 0 space, we only have a 0.
• 04:00 - 04:30 Sometimes, the condition is so complicated that only the 0 vector will satisfy that condition. That is, there is a situation in which only the 0 vector lives in there, and even if we search really hard, we will not find anything else that is in this space. Therefore, the 0 vector must definitely be in there. So look for it in particular. Here's another reason: If the 0 vector does not exist, then the set is definitely not a vector space, and there is no point in continuing to search.
• 04:30 - 05:00 Because property 4 of a vector space will not hold. Therefore, it is worth looking for it, because maybe, in some case, it is not there, and therefore we can stop. This is the easiest way to check. You saw it that 0 plus two times 0, being equal to 0, is a little easier to check than minus 2, plus 2 times 1, no need to start looking, just check for the 0. It is the easiest, the safest, and the recommended way. Though, not mandatory.
• 05:00 - 05:30 So let's continue our example, for now we just saw that W1 is not an empty set. Now we move to property B. We want to show a closure to addition. What does it mean? Let's take two vectors u and v, that are in W1. We want to show that the sum is in W1. So first of all, let's try to understand. u,v belongs to W1, what does that mean? u is of the form a and b, because it is a subset of V, that was R2.
• 05:30 - 06:00 That's why u looks as a,b this is its general form. It belongs to W1, what does that mean? That a plus 2b must equal 0. It fulfills the condition of W1. v is also such a vector. We will call it x,y. Note that we must take different letters, because there is no relationship between the vector u to the vector v, except that both are in W1.
• 06:00 - 06:30 Therefore, x plus 2y is equal to 0, pay attention. a plus 2b means, the first element plus two times the second element. In v, the first element is called x, the second element is called y, Therefore the condition becomes x plus 2y, is equal to 0. What do we want to prove? That amount is in W1. What is the meaning of this, let's see. First of all let's add the vectors and see which is the new vector that needs to be checked.
• 06:30 - 07:00 u plus v is a,b plus x,y, we know how to add them, because we use the addition of R2. After all, W1 has the same operations, that's why we use the operations of v. We add, and we get a plus x, b plus y. What does it mean to belong to W1? Which is what we want to show. What is the meaning of this? The meaning is that the first element plus two times the second element, is equal to 0.
• 07:00 - 07:30 What does it look like on here? Here: The first element, which is called a plus x, plus two times the second element, called b plus y, should be equal to 0. Then we have satisfied the condition of W1, and we are sure that u plus v will really be in W1. So let's see if it actually happens. a plus x, plus two times b plus y, is equal to... Let's do the math, we want to get 0 in the end. If we get a 0, we succeeded.
• 07:30 - 08:00 If we don't get a 0, we stop and declare that W1 is not a vector space. So here is the calculation, first of all we will open brackets: a plus x, plus 2b, plus 2y. Now remember, a, x, b, y, these are all real numbers. We are in W1, a subset of R2, ordered pairs of real numbers. Therefore, here I can use the properties of the field. For example, I will use the commutative law, and I will get
• 08:00 - 08:30 a plus 2b, plus x, plus 2y. Here, we used the commutative law, and in the set law, I will first add, a plus 2b, and, x plus 2y, why? Because pay attention, what do I know about, a plus 2b? That it is equal to 0. And the same goes for x plus 2y, it is also equal to 0. That is why it is written here, according to the given data,
• 08:30 - 09:00 0 plus 0, which is 0. That's it, we were able to show that the first element plus two times the second element, in the new vector, u plus v, is equal to 0, and it fulfills the condition, therefore, the sum lives in W1. Next, now we need property C. Do you remember what C is? Closure to multiplication by a scalar. I will take a vector in W1, I will take some scalar in the field, my field is R, and I would like to show that alpha u is in W1.
• 09:00 - 09:30 So let's write these things. u belongs to W1 means, a plus 2b is equal to 0. Now let's take alpha u. Let's write for ourselves what alpha u is. Alpha u equals alpha times ab, which is, alpha a, alpha b, belongs to W1. What does it mean, what do I have to prove?
• 09:30 - 10:00 I need to prove the first element plus two times the second element, equals 0. First element, alpha a, here, plus two times the second element, alpha b, two times alpha b, equals 0. How do we prove it? We will step out on one side, we will do the math, until we get to the other side. So, alpha a plus two alpha b, equals: Here is alpha, and here is alpha, Arithmetic within a field, let's take alpha out of the brackets.
• 10:00 - 10:30 It's alpha a plus 2b. But wait, a plus 2b equals 0. According to the given data, I can place, alpha times 0. And this is of course equal to 0. That's it, we showed that alpha a plus two times alpha b, is equal to 0, and therefore, alpha u also belongs to W1.
• 10:30 - 11:00 That's it, we showed that properties A to C hold. So, W1 is a subspace of V, and what was V? R2. This means that if I look at W1 as a world in itself, it also is a vector space. In the future this will tell us, that all the theory we know about the large vector space R2, this theory will also be true for the vector space W1,
• 11:00 - 11:30 which is a slightly smaller vector space, within the larger vector space R2. I am being asked here, what would happen if there was another homogeneous linear equation system. Pay attention, a plus 2b, equals 0. It is some kind of a homogeneous linear equation. Maybe they would have given me another condition here, 2a minus 5b, equals 0. What would happen then?
• 11:30 - 12:00 Would I still have a vector subspace? An interesting question. Let's answer it by a generalization. Let's look at the general case. And this is example number two: I take a matrix A of order m by n, no matter the order, the elements are in the field F, no matter the field. I want to look at the system A times x equals 0, a homogeneous linear system. Let W denote its collection of solutions.
• 12:00 - 12:30 And I ask myself whether W is a vector space? For knowing a vector space, or a vector subspace, first, I need to identify where W is, whose subset is it? Let's check it out. We want to check the subspace. We know how many unknowns we have, n, right? m equations, n unknowns.
• 12:30 - 13:00 Therefore, the solution is some kind of a column with n elements. So, columns with n elements, in which vector space are they known to be located? Indeed, they are in Fn, and so, with W is the group of solutions of the system Ax is equal to 0, when A is an m by n matrix, W is in Fn.
• 13:00 - 13:30 Now I know that W is a subset of Fn. So if I want to know if it is a vector space, it is enough to check a subspace of Fn. So let's see that indeed W is a subspace of Fn. How will I show this? Properties A to C. Property A, W is not an empty set, why? Which is surely a solution of a homogeneous system of equations? That's right, the 0 vector, it always solves it.
• 13:30 - 14:00 A times 0, equals 0. There is always a solution. The group of solutions, the group of solutions of a homogeneous system is not an empty set. Now we also need to show closure to addition and multiplication by a scalar. But wait, we already saw that before. When we talked about a homogeneous system of equations, we already saw that if we have a solution, scalar multiplication is also a solution. And if I have two solutions, the sum is also a solution. That's it, so we already know
• 14:00 - 14:30 that there is a closure to scalar multiplication and closure to addition. Therefore, A to C are true, what follows from this? That W, that is, the group of solutions of a homogeneous system of equations, and it doesn't matter what this system is, as long as it is homogeneous linear, the group of solutions will always, always, always, be a vector space. We have proved it in a general way. I am being asked, what happens if the system has a single solution.
• 14:30 - 15:00 Okay, a single, in a homogeneous system solution, means the 0 solution, the trivial solution. So what do we get? What W would be then? In W we only have 0, only 0 means the 0 space, one of the trivial subspaces we saw in the previous lesson. A good question. What about the solutions of a system of inhomogeneous equations? Is it also a subspace?
• 15:00 - 15:30 Well, regarding this we have already seen that the group of solutions is not closed to addition. A solution plus solution of an inhomogeneous system will not give us a solution. It is also not closed to multiplication by a scalar. Sometimes it is even an empty set, and it doesn't even contain the 0 vector. And so the answer is simply that it is not a subspace, therefore, it is not a vector space either. That's it, here we finish.
• 15:30 - 16:00 In the following lessons we will see examples which are subsets of standard vector spaces we already know. Thank you very much.