140 - צירוף ליניארי: הגדרה

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Summary

In this engaging lecture by Dr. Aliza Malek from the Technion, we explore the concept of linear combinations within vector spaces. Starting with the definition of a linear combination, the discussion progresses to examples demonstrating how vectors can be composed using scalar multiplication and addition within spaces like R^4 and 2x2 matrices. The session delves into the nuances of when a vector can or cannot be represented as a linear combination of given vectors, and touches upon solving systems of nonhomogeneous linear equations to find unique or multiple solutions. The importance of understanding these combinations is set up as a precursor to future lessons.

Highlights

Dr. Aliza Malek introduces the foundational concept of linear combinations in vector spaces in this lesson. 👩‍🏫
The lecture analyses various examples in R^4 and with 2x2 matrices to illustrate the mechanism of forming linear combinations. ➕➖
Solving systems of nonhomogeneous linear equations provides insights into determining vector compositions. 📊
The lecture highlights special cases where vectors cannot be formed from a given set due to their inherent properties, like symmetry in matrices. 🔍
The session promises to build on this foundational knowledge, making it clear that linear combinations are just the beginning of broader vector space explorations. 🚀

Key Takeaways

Linear combinations involve expressing a vector as a sum of scalar multiples of other vectors, using operations native to vector spaces like scalar multiplication and addition. 📐
The ability to form a vector from a set of vectors can be determined using systems of linear equations, which can result in a unique solution, multiple solutions, or no solution at all. 🤔
Not all systems yield a solution, especially if the target vector lies outside the span of the given vectors, as shown with symmetric matrices. 🚫
Understanding linear combinations is crucial as they form the basis for exploring deeper aspects of vector spaces, to be discussed in future lessons. 😎

Overview

In this intriguing lecture by Dr. Aliza Malek, a fundamental topic in linear algebra is tackled: linear combinations within vector spaces. The lecture kicks off with a definition and then gracefully leads into examples that explore how a vector can be expressed through scalar multiples and sums of other vectors. With a focus on spaces such as R^4 and 2x2 matrices, the complexity of identifying linear combinations is unveiled step by step.

Each example in the lecture becomes a puzzle—sometimes easily solved and other times more complex, demanding the setup and resolution of a system of linear equations. These equations can yield no solution, a singular solution, or infinite solutions, highlighting the beauty and challenge of vector combinations. Through engaging explanations, Dr. Malek breaks down the concepts and offers practical insight into each scenario.

The lecture sets the stage for future lessons by establishing why these linear combinations matter. While the current discussion centers on understanding the process and implications of forming vectors through other vectors, it's made clear there's much more to come. We're left with the anticipation of delving deeper into the intricacies of vector spaces, ensuring the journey in this linear algebra course is both comprehensive and exciting.

Chapters

00:00 - 00:30: Chapter 4 - Vector Spaces In Chapter 4 - Vector Spaces, the instructor, Dr. Aliza Malek, builds upon the previous lessons on operations between subspaces to introduce the concept of linear combinations, which is fundamental to understanding vector spaces. The chapter provides a definition of linear combinations in the context of a vector space V over a field F, with S being a set consisting of vectors v₁, v₂, ..., vₖ.
00:30 - 01:00: Introduction to Linear Combinations The chapter introduces the concept of linear combinations in vector spaces. A vector w in a vector space V is considered a linear combination of a set k of vectors if it can be expressed as w = α₁v₁ + α₂v₂ + ... + αₖvₖ, where α₁, α₂, ..., αₖ are scalars in a field F.
01:00 - 01:30: Definition of Linear Combination The chapter explains the concept of a linear combination in a vector space. It describes linear combination as the process of writing vector 'w' using other vectors v₁,...,vₖ by employing operations of scalar multiplication and addition, which are fundamental in vector spaces. The chapter then introduces examples to further illustrate the concept, starting with an example using the vector space R⁴.
01:30 - 03:30: Example 1: Linear Combination in R⁴ The chapter titled 'Example 1: Linear Combination in R⁴' involves the discussion of linear combinations within the context of four-dimensional space, R⁴. The speaker takes two vectors, \( v_1 = (1,1,0,0) \) and \( v_2 = (0,0,1,2) \), and explains how these vectors form a set \( S \) consisting of \( k=2 \) vectors. The task is to express another vector \( w = (3,3,-2,-4) \) as a linear combination of the vectors in set \( S \). The chapter seems to explore the steps and calculations needed to achieve this representation, although the details of the calculations are not explicitly provided in the transcript.
03:30 - 05:30: Example 2: System of Equations in R⁴ In this chapter, the focus is on solving systems of equations in a four-dimensional space, R⁴. It starts with an example demonstrating how a vector (3,3,-2,-4) can be expressed as a linear combination of other vectors. Specifically, it shows how multiplying vector (1,1,0,0) by 3, and vector (0,0,1,2) by -2 will yield the desired vector. The process is broken down step-by-step, explaining that (3,3,-2,-4) equals 3 times (1,1,0,0) added to minus 2 times (0,0,1,2). The chapter closes with a rhetorical question, setting up further exploration into the topic.
05:30 - 08:30: Example 3: No Solution for Linear Combination The chapter discusses the concept of linear combinations in the context of vector spaces, specifically in R^4. It is highlighted that not all vectors can be represented as a linear combination of a given set of vectors. The example provided shows that if you try to construct a vector (1,1,1,1) using two vectors, where taking the first vector gives (1,1), you cannot achieve the desired vector because the fourth component will always be twice the first component regardless of how the second vector is scaled. This exemplifies a scenario where a vector does not lie within the span of the chosen vectors.
08:30 - 11:30: Example 4: Linear Combination with Matrices In this chapter, the concept of linear combinations with matrices is explored. The example begins with a set S consisting of three vectors in four-dimensional space R⁴: (1,2,3,1), (4,5,6,1), and (7,8,9,1). The goal is to express another vector w, given as (1,−1,−3,−2), as a linear combination of the vectors in set S. The chapter explains the process of determining the coefficients needed to achieve this representation, illustrating the principles of linear algebra in the context of vector spaces and matrix operations.
11:30 - 12:30: Conclusion and Summary The conclusion emphasizes the challenge of guessing the answer and proposes an alternative method of finding it. It suggests solving for scalars by comparing the components of a vector and creating a system of equations. The approach involves determining the scalars α₁, α₂, and α₃, which, when multiplied with their respective elements, will yield the desired vector result.

140 - צירוף ליניארי: הגדרה Transcription

00:00 - 00:30 Chapter 4 - Vector Spaces Linear Combination - Definition Linear Algebra Course Dr. Aliza Malek Hello, thank you for coming back. After getting to know operations between subspaces, we are ready to continue learning a few more concepts related to vector spaces. And here is one of the basic concepts we would like to learn: Linear combinations. Let's get going. Here is the definition: I have V a vector space over field F, and I have S, which is v₁,v₂,...,vₖ,
00:30 - 01:00 a set of k vectors in the space V. Vector w in V is called a linear combination, or a linear combination of the vectors in S, meaning that when we say that w is a linear combination of them? If there exist scalars α₁,α₂,...,αₖ in F, such that: w=α₁v₁+α₂v₂+...+αₖvₖ.
01:00 - 01:30 In other words, I can write w using v₁,...,vₖ. What does it mean to write w using v₁,...,vₖ? When I use the only operations I know how to do in a vector space, to multiply by scalars and perform addition. Then, w is said to be a linear combination of v₁,...,vₖ. Let's see examples. Here is the first example: We will take V to be R⁴.
01:30 - 02:00 And we will take S to be v₁=(1,1,0,0), v₂=(0,0,1,2). We took two vectors, k=2. Of all the infinite vectors in R⁴, I only took two in S. Let's write w, which is (3,3,-2,-4), as a linear combination of the vectors in S. How do we do it? I'm sure that you too can guess.
02:00 - 02:30 (3,3,-2,-4) will be equal to 3 times (1,1,0,0), because that will give me (3,3,0,0), plus, minus 2 times the vector (0,0,1,2). Then I have 3+0, which is 3, 3+0, which is 3, 0−2, is −2, and 0−4, is −4. That was really simple. You ask what happens if we want to write (1,1,1,1) using these?
02:30 - 03:00 I guess you too notice that we can't do that. Because once we take the first vector once to get (1,1), it doesn't matter how many times you take the second vector, the fourth component will always be 2 times the first component. Therefore, we cannot get the vector (1,1,1,1). So not everything can be gotten as a linear combination of v₁, v₂, that we have chosen. Let's see another example: We will stick with V being R⁴.
03:00 - 03:30 And now we will take S to be the following set: v₁,v₂,v₃, these are the three vectors written here, (1,2,3,1), (4,5,6,1), (7,8,9,1). Now we have three other vectors in R⁴, and we will try to write vectors using them. We will write w, which is (1,−1,−3,−2), as a linear combination of v₁,v₂,v₃. So we have to write (1,−1,−3,−2), using these three.
03:30 - 04:00 But this time, we can't guess. It will be a bit hard to guess the answer, indeed, it's really hard. So what are we going to do? We will look for scalars where α₁ times the first, plus α₂ times the second, plus α₃ times the third, will give us the w we want. We will have to find these scalars. How do we do it? We will simply compare the components of the vector, and get a system of equations.
04:00 - 04:30 Here is the system we get: α₁+4α₂+7α₃, should be equal to 1. 2α₁+5α₂+8α₃, will be equal to −1. And so on, and we get a system of nonhomogeneous linear equations. Wait, but we have already learned to solve that. We know what awaits us. When solving such systems, we can get one of three:
04:30 - 05:00 Either a unique solution, or an infinite number of solutions, or there will be no solution at all. So if I get a system of nonhomogeneous linear equations, what does the solution mean, in the context of my question? If there is a unique solution, what does it mean? This means that a unique linear combination can be obtained. That means that there will be one and only one way to write w as a linear combination of v₁,v₂,v₃.
05:00 - 05:30 And if we get infinite solutions, what does that mean? That we have infinite ways to write w using v₁,v₂,v₃. And if we get that there is no solution at all, this means that w is not a linear combination of v₁,v₂,v₃. What will happen this time? Let's solve and find out. So I remind you, this is the linear combination, from it we get the system of nonhomogeneous linear equations
05:30 - 06:00 that we see here, and which we would like to solve. How do we solve it? We have already learned that, we construct an expanded matrix of coefficients, and rank it. I will show you the final result of the ranking, of course, and here is what we get: (1,0,−1|−3), (0,1,2|1), and two rows of zeros. Note that I ended up with two equations, with three unknowns, there is no contradicting row, therefore we have infinite solutions. Here are the corresponding equations,
06:00 - 06:30 And here is the general solution of the system: α₁=−3+α₃, α₂=1−2α₃, and a free α₃. Wait a moment... This is the system's solution, but it is not what I was asked for. I was asked to write a linear combination. I still have to answer the question. In the meantime, I just solved the system. So let's answer the question. I need to choose a private solution that I can place here.
06:30 - 07:00 So we choose whichever α₃ that we want. For example, I chose to place α₃=0. You can place whatever you want. After all, we have infinite ways to do it. I chose α₃=0, I will get that α₁=−3, α₂=1, and α₃=0. And so, now that I have specific α₁,α₂,α₃, I can place and get that w=−3v₁+1v₂+0v₃.
07:00 - 07:30 And here is one way to write w as a linear combination of v₁,v₂,v₃. And here is example number three: Again we are with R⁴. And notice that I'm with the same set of vectors, S, v₁,v₂,v₃, just like we had in the previous example. Which one do we want to write this time? (1,1,1,1). Remember we didn't succeed before? Perhaps this time we will succeed.
07:30 - 08:00 I can't guess the answer this time either, and so, I take (1,1,1,1), and write: α₁ times the first, plus α₂ times the second, plus α₃ times the third, and I am looking to find what α₁,α₂,α₃ are, if they exist at all. The system of equations looks exactly the same as before, only this time I compare to (1,1,1,1). Here is the expanded coefficient matrix, I rank it once again, and what do I get? I have a contradicting row over here.
08:00 - 08:30 We have no solution. 0α₁+0α₂+0α₃=1, there is no solution, or, if you recall, (r(A|b)>r(A), and therefore, there is no solution. What does it mean? This means that w is not a linear combination of v₁,v₂,v₃. And here is example four: This time, we have V to be real 2 by 2 matrices. I have S be v₁,v₂,v₃, a set of three matrices,
08:30 - 09:00 from the infinity of 2 by 2 matrices, I chose three matrices, and I put them in set S. Which one do we want to write? (4,−1, −1,5) as their linear combination. What do we do? Just like before. We can't guess the solution, sometimes we can. When we can't guess, we build the appropriate equations. So here is (4,−1, −1,5), equals to α₁ times the first, plus α₂ times the second, plus α₃ times the third.
09:00 - 09:30 This time we compare the components of the matrix. The component in the first spot, which is 4, will be equal to α₁+α₂+α₃, and so on. This is the resulting system of equations. Again, we build an expanded matrix of coefficients. Just note that I happen to have the same equation over here twice, and so I will write it only once, and we will get α₁+α₃=5, α₂=−1, 3α₃=6,
09:30 - 10:00 this time we don't even need matrices in order to solve this, we can directly get a result: α₁=3, α₂=−1, and α₃=2. Therefore w, meaning (4,−1, −1,5) equals: 3v₁−v₂+2v₃. This time we got a unique solution. I have one and only way to write w as a linear combination of v₁,v₂,v₃.
10:00 - 10:30 Now, we would like to write another w, which is (1,2, 3,4), as a linear combination of those same terms in the S set. What do we do? As before, the equations will remain the same equations, the relationship between α₁,α₂,α₃, will remain the same. Only this time, instead of comparing to (4,−1, −1,5), we compare to (1,2, 3,4).
10:30 - 11:00 The same equations, that is, the same matrix of coefficients, but our b changes. It becomes (1,2, 3,4), instead of (4,−1, −1,5). And of course, you directly see that we don't have a solution here. We are being told over here, that it was possible to know in advance that we wouldn't get a solution. Indeed, you know why? Because if you look for a moment at the matrices in the S set, you can see that they are all symmetric matrices.
11:00 - 11:30 We know that 2 by 2 symmetric matrices, are a vector subspace of all 2 by 2 matrices, therefore, they are closed to addition and multiplication by a scalar. That is, when we add and multiply symmetric matrices by a scalar, we will get only symmetric matrices. Symmetric matrices, or subspaces, are closed to linear combinations. Therefore, it is impossible to get a matrix that is not symmetric
11:30 - 12:00 as their linear combination. So once we noticed that there are only symmetric matrices here, and the symmetric matrices are a vector space, there is no way to get a matrix that is not symmetric. Let's summarize what we have seen so far. Given a set S, in a given vector space V, some of the vectors in V will be a linear combination of the vectors in S, and some won't be.
12:00 - 12:30 We will show interest in the collection of all linear combinations, which are possible to be obtained using the vectors in S. This set will be of much interest to us, but I won't tell you why just yet. In the next lesson we will do some more examples, then we will see why these linear combinations are so important to us. That's it for now. Thank you very much, and see you in the next lesson for the practicing of this subject.