131 - פעולות בין תמ: הפרש

Summary

In this video, Dr. Aliza Malek explains the concept of the difference between vector spaces within linear algebra. The video discusses how the difference between two subspaces is defined and their properties. It provides examples and proves that the difference, except for trivial cases, cannot form a vector space. By illustrating with the example of matrices, and using diagrams in R2 space, it clarifies how subspaces interact and the inherent limitations in forming vector spaces through their difference.

Highlights

• Difference between vector spaces is denoted as U - W or U\W. 🎓
• If a space is discarded and 0 is returned, it's not a vector space. This is highlighted with the example of upper and lower triangular matrices. 🧐
• The theorem establishes that non-trivial subspaces can't form a vector space via their difference. 📘

Key Takeaways

• The difference between vector spaces focuses on what is in one subspace but not in the other, denoted as U\W. 🤔
• The result of a difference between subspaces is generally not a vector space, except in trivial cases like when W is the whole space or just the zero vector. 🎯
• Using matrices to visualize, the difference often discards crucial components, thus failing to meet the criteria of a vector space. 🔄

Overview

Vector spaces are fundamental structures in linear algebra, and understanding their interactions is crucial. Dr. Aliza Malek engages students in a discussion about the difference between subspaces, U and W, answering the question of what vectors are excluded from one subspace compared to another. The process is not just theoretical; it involves practical implications and logical reasoning.

Through examples involving matrix operations, the video illustrates why taking the difference between vector spaces often leads to losing essential properties that define a vector space. The removal of dimensions—represented as lines or points in visual models—plays an integral role in understanding why these differences don't form vector spaces in most cases.

This concept extends into proof, utilizing common algebraic tools to reaffirm the theorem that only trivial subspaces, like the entire space or just the zero subspace, allow the difference to be a vector space. For learners, this video is a deep dive into how we examine, demonstrate, and prove the characteristics and limitations of vector spaces in higher-dimensional contexts.

Chapters

• 00:00 - 00:30: Vector Spaces Chapter 4, titled 'Vector Spaces', discusses vector subspace operations, specifically focusing on the concept of difference in vector spaces. Dr. Aliza Malek explains the operations on vector spaces using a mathematical framework that is similar to set operations. The chapter emphasizes understanding the difference between sets, specifically vector spaces, by illustrating that vector spaces are not merely sets. The vector space 'V' over a field 'F' has two subspaces 'U' and 'W', which are used to explore these operations.
• 00:30 - 01:00: Difference between Two Subspaces The chapter explains the 'difference' operation between two subspaces, U and W, which results in a new set. This set consists of all vectors that belong to subspace U but do not belong to subspace W. The notation used to denote this difference is U\W. To be included in the set resulting from U\W, a vector must satisfy the condition of being in U and simultaneously not being in W.
• 01:00 - 01:30: Discarding the Zero The chapter discusses the concept of vector spaces, highlighting the importance of the zero vector. Initially, the zero vector was discarded, leading to the realization that the absence of a zero vector means the space cannot be considered a vector space. The chapter concludes with the reintegration of the zero vector into the discussion, emphasizing its necessity for the structure of a subspace.
• 01:30 - 02:00: Union of the Zero Chapter Title: Union of the Zero This chapter explores the concept of forming a union of sets, focusing on the addition of vectors. The text emphasizes that U minus W is not simply all vectors in U that are not in W, due to the introduction of a new vector into the union set. Consequently, the new set comprises all elements from both U and W, along with the additional vector, rather than just the vectors from U not in W. Therefore, the union extends to include this zero vector, creating an expanded set.
• 02:00 - 02:30: Checking Vector Subspace with Matrices Example The chapter discusses the concept of a vector subspace through the lens of an example involving matrices. It begins by posing the question of whether a given set is a vector subspace. The example involves n by n matrices, which provide a convenient way to explore the concept. Specifically, the chapter examines upper and lower triangular matrices of order n, designated as U and W respectively, and contemplates the nature of another set, Z, in this context. The approach highlights the utility of matrices in demonstrating the properties and structure of vector subspaces.
• 02:30 - 03:00: Characteristics of Difference Not Being a Vector Space The chapter discusses the concept of difference not being a vector space, focusing specifically on the set U minus W. It questions whether this set qualifies as a vector space after certain operations, like reintroducing the zero element. The set in question consists of upper triangular matrices, excluding the lower triangulars, resulting in the loss of diagonal matrices which represent the intersection between the two discarded sections. The chapter emphasizes the implications of these exclusions and the structural characteristics of the resulting set.
• 03:00 - 03:30: Understanding Non-Trivial Subspaces This chapter focuses on understanding non-trivial subspaces, specifically dealing with upper triangular matrices that exclude diagonal matrices. It emphasizes the process of identifying intersections within these matrices and the concept of differences by discarding intersecting elements. An example is provided to illustrate matrices that belong to this subset, showcasing upper triangular matrices that do not include diagonals.
• 03:30 - 04:00: Proof and Containment in Subspaces In this chapter, the discussion revolves around the results of adding non-diagonal triangular matrices, resulting in a diagonal matrix. This observation leads to the hypothesis that the sum is not part of the original set (Z), except in trivial cases. The chapter explores how the difference fails to form a vector space, illustrating the concept through examples.
• 04:00 - 04:30: Example with R2 and Subspace W This chapter discusses the concept of difference between vector spaces, particularly focusing on a vector space V over a field F and its subspace W. It highlights that the difference will not form another vector space unless in trivial cases, which is explained briefly. A theorem about the difference between the vector spaces is introduced, emphasizing that if W is a non-trivial subspace of V, one does not need to consider an additional subspace U for the difference calculation.
• 04:30 - 05:00: Addition and Illustration with Vectors The chapter discusses the concept of a non-trivial subspace in vector spaces. It defines a non-trivial subspace W of a vector space V as a subspace that is neither the zero space nor the entire vector space V. Furthermore, the transcript explains that the set Z, defined as V minus W union {0}, is not a vector space and can never be a vector space. This concept emphasizes the restrictions and properties that define vector spaces.
• 05:00 - 05:30: Conclusion and Preview of Next Lesson The chapter discusses the implications of W being a non-trivial space. It explores scenarios where W is zero or takes up all of V, emphasizing the importance of avoiding these cases. The proof of why W must not be trivial is outlined, leading to understanding the implications in the vector space context.

131 - פעולות בין תמ: הפרש Transcription

• 00:00 - 00:30 Chapter 4 - Vector Spaces Vector Subspace Operations - Difference Linear Algebra Course Dr. Aliza Malek Hello, thank you for coming back. After we have seen intersection and union between spaces, it is time for the difference, let's begin. What would be the difference between vector spaces? Again, I remind you, this is a difference between sets, only that the sets are spaces, and not just sets. V is a vector space over F, U and W are two subspaces of V.
• 00:30 - 01:00 The difference between two subspaces is the collection of all vectors in V, So that... Remember what a difference is? Being in U, but not in W. How do we write it? In a normal way. U minus W, or as we write, U\W, is equal to all v in V, so that V in both U, and also, V is not in W, because we discarded everything in W.
• 01:00 - 01:30 Indeed, it is not a vector space, why? Which terms did we discard? We discarded the 0. If I take the the U, and I throw W, W is a subspace, 0 is sitting in there. That's why we threw 0 out, it has no chance of being a vector space. So you know what? Let's return the 0.
• 01:30 - 02:00 The union of the 0. Note that no U minus W is contained in the new set, because I added a vector that is not there. So in fact, U minus W is not equal to all v's in U, neither is v in W, the union of 0, but rather, it sits in there, it is all of them, plus the 0 that we added, and so we got a new set.
• 02:00 - 02:30 Are you asking if it is now a vector subspace? Let's check an example before that. I take V to be n by n matrices, of course, we are already used to looking at it. The convenience of matrices is that we can talk about this as sets with names, and so it is much easier to demonstrate it. I take U to be upper triangular matrices of order n, I take W to be lower triangulars of order n, as usual, and I ask myself, what Z will be,
• 02:30 - 03:00 that is, what will be U minus W after we returned the 0 as well, because we threw it away, and we immediately return it. After we returned it, did it come out as a vector space? What is this set that we got? It is in fact a collection of all the matrices that are upper triangular, U, that is, but, if we discard the lower triangulars, remember what it means, what we really throwing away? The intersection between the two. But the intersection between the two is the diagonal matrices.
• 03:00 - 03:30 Therefore, these are all upper triangulars, but without diagonal matrices. We discarded the diagonal ones. The intersection, and we have already seen that the difference means taking and throwing away the intersection. Let's see an example. Here A, a matrix that is in Z, because it is upper triangular, and is not diagonal, and here is another one as such, an upper triangular, and not diagonal, because we discarded the diagonals.
• 03:30 - 04:00 Let's add them, what did we get? We got a diagonal matrix. We took two non-diagonal triangulars, added them together, and got a diagonal. This means that the sum is part of those we discarded, and therefore is not found in Z. If so, it seems that the difference is never a vector space, but, except for the trivial cases. And that we will see.
• 04:00 - 04:30 The difference is never going to come out as a vector space, unless it's the trivial cases. What does trivial cases mean? We will immediately formulate it. So here is the theorem regarding the difference between vector spaces: I have V, a vector space over F, and now let's take W, a non-trivial subspace, of V. There is no need to take both U and W, because V is already a space, and we will simply get the difference between V and W. We don't need to grab U as well, it's the same thing,
• 04:30 - 05:00 the same idea. Now, what does it mean that W is a non-trivial subspace of V, remember? W is not the 0 space, and it is not the entire space V. In this case, Z as defined before, which is V minus W, union of 0, This set, Z, is not a vector space, it will never be a vector space,
• 05:00 - 05:30 as long as W is non-trivial. Let's see the proof... But before that, let's understand why W must not be a trivial space, what will happen if it is 0, and what will happen if it is V? What's the problem with that? Let's check. Suppose that W is all of V, what will Z be in this case? I take V, I discard W, that is V, and I return the 0. What comes out of all this? From V I discard W, since W is sitting there,
• 05:30 - 06:00 and from V I threw everything, I was left with nothing, an empty set. We returned the 0, what did we get? 0, the 0 space. Wait, but the 0 space is a known vector space, so in this case, we will get that Z is indeed a vector space. Okay, let's put that case aside. Let's see the other extreme case. Let's assume this time that W is the 0 space. What will happen then, what will Z be equal to then? Z, which is V minus W, union of 0, means:
• 06:00 - 06:30 I take V, I discard W, which means I discarded 0, and I returned the 0. If we take V and discard 0, and then return the 0... I returned what I discarded, I was left with the entirety of V. Wait, but V is a vector space, that's a given, so with this fact it is impossible to argue either. That's why these two extreme cases work. If by chance, W it is the 0 space, or the entire space,
• 06:30 - 07:00 Z comes out a vector space. Any other case, meaning any case where W is a non-trivial subspace of V, it won't work, we won't get a vector space. Let's see the proof first. So now I will assume that W is a non-trivial subspace of V, meaning it's not V, and it's not 0, and we will check out Z. I will assume by negation that Z comes out of a vector space, because I want to show that it does not. So let's assume it is, get a contradiction, and we are done.
• 07:00 - 07:30 So let's say it's a vector space, what does that mean? So let's look at Z union W. Z union W, remember? Remember what the theorem said, when Z union W is a subspace? When there is containment, Z in W, W in Z, this was the theorem we proved in a previous lesson. So let's see what Z union W is in our case. Here's Z, I took V, discarded W, and returned 0.
• 07:30 - 08:00 After that, I look at Z union W, I discarded W, and returned W, what did I get? The entirety of V. We take Z, discard W, return 0, then return all of W. Well, then we didn't discard anything. We got all of V. But V is a vector space. A union of spaces is a space, if and only if, there is containment. So if Z will be a vector space,
• 08:00 - 08:30 Z union W, is V, and which is also a vector space, therefore there must be containment. According to the theorem it is true that Z is contained in W, or W is contained in Z. Wait, that doesn't make sense. The way we defined Z, we discarded W, so it can't be that Z sits in W, or that W sits in Z, because that is precisely the difference. And so we got a contradiction to the theorem we proved in a previous lesson. If Z were a space,
• 08:30 - 09:00 Z union W was a union of spaces, a union of spaces came out space, then there must be a containment. Pay attention, Z union W is V, Z union W truly came out a space. But, because there is no containment, it means that what we did is not a union between subspaces. Because if there was a union between subspaces,
• 09:00 - 09:30 there must be a containment. Therefore, our union is not between subspaces. But wait, W is a subspace, then we have no choice, it remains that Z must be the one which is not a subspace, so that we cannot say that we have a union and no subspaces. Therefore, the difference does not come out a subspace. That's it, we finished the proof. Let's now see an example.
• 09:30 - 10:00 We will choose V to be R2. And let's take W to be the following subspace: W is equal to all a0, so that a belongs to R. Now, let's look at what Z is. Before we see the answer I will first remind you: All the V's, all of R2, without W, and we returned the 0. What are we left with, what is Z? As our mascot says, we are left with 0b.
• 10:00 - 10:30 We had ab, we threw away a0, we are left with 0b. That's not quite right... Let's examine it. Here is V, V is R2, the entire plain. a0 means this axis, which is called the X-axis, for us, it is called the a-axis.
• 10:30 - 11:00 If I take the plain, and discard the a-axis, do I stay with 0b? Do I stay with the b-axis? Definitely not. Here, that's what is left. This whole area: We discarded the a0 line, we returned the axis origin, so the origin stayed, everything in blue, including the b-axis, that's what is left, that is Z.
• 11:00 - 11:30 These terms can be described like this... How can we describe the entire blue set, with the b-axis, and without the white line? It's in fact V minus W, union 0, we discarded the line, the axis, and we returned the axis origin, then we can say that we have all the terms of the form ab in the plain, all except those where b equals 0,
• 11:30 - 12:00 and so it's all terms where b is different from 0. Because all those where b equals 0, that are on this line, we discarded, except for 0, and therefore, b must be different from 0. That's the way to describe what is left. Okay, of course that this set won't be closed to addition, why? Let's see why. I take two vectors, 1,1, 1,-1, this is the point 1,1 that is here, and 1,-1 that is over here.
• 12:00 - 12:30 Here they are: We will take the arrows that lead to these points, and now let's add them, and see what happens. When I add them together, I get 2,0. 2,0 does not belong to Z, because recall that in Z it is ab, so that b is different from 0, and here b is equal to 0. and so this point is not in Z. Let's see it on the drawing. How do we add two vectors, two arrows? I need to move the red arrow...
• 12:30 - 13:00 When I move it, it has no effect, just don't change direction, and don't change the length. We will move the arrow connected to the end of 1,1. Then the diagonal is their sum, let's do it. Here. Now we connected them, that's the sum. The green line, the green arrow, is 2,0. The point that's here represents the vector 2,0, but wait, it came out on the white line,
• 13:00 - 13:30 and therefore the sum suddenly got discarded, and it is not in Z. So you see that we have no closure to addition. When we find the difference, when these are non-trivial spaces, what we are left with is not a vector space. Here we have this whole region with a section on the a-axis, it is not a vector space. This is friends, with that, we finish. In the next lesson we have to discuss the sum. Thank you very much.