135 - תרגילים מסכמים

Summary

In this video, Dr. Aliza Malek takes us through summarizing exercises on vector subspaces as part of a linear algebra course. The focus is on understanding the operations between vector subspaces, calculating the sum and intersection of subspaces, and determining if these sums are direct. The exercises involve working with specific vectors in R4 and matrices, where viewers learn step-by-step methods to solve these problems, including finding the intersection of space and ensuring the completeness of sums.

Highlights

• Exploring operations between vector subspaces in R4 🇷🇺
• Detailed explanation of direct vs. non-direct sums 🌟
• Visual walkthroughs for solving intersection and sum problems 🖥️
• Insight into matrix operations within vector subspaces 📊
• Answers and proofs for conditions of direct sums ✅

Key Takeaways

• Deep dive into vector subspaces and their operations 🧠
• Understand the concept of direct and non-direct sums in vector spaces ➕
• Practical breakdown of finding intersections and sums between subspaces 🔍
• Explore how conditions and parameters affect vector spaces and matrices ↔️
• Learn through multiple techniques, enhancing problem-solving skills 🎓

Overview

In the latest lecture by Dr. Aliza Malek, students dive into the world of vector spaces, focusing specifically on operations between subspaces—a central topic in linear algebra. The session is packed with exercises that guide learners through calculating intersections and sums of subspaces, with an emphasis on determining whether sums are direct.

Dr. Malek uses the example of vector subspaces in R4 to explain the concepts. She shows how to calculate the intersection of subspaces by forcing the conditions of one space onto another. Additionally, through example problems, she illustrates what makes a sum direct or not, using both computation and theory as tools.

The session further explores the use of matrices within vector spaces. Dr. Malek clarifies how to decompose matrices into sums of alternating multiplicative and symmetric matrices. Viewers gain insights into how direct sums provide unique decomposition methods and how different subspaces can be manipulated to test and prove these properties.

Overall, the lecture demonstrates Dr. Malek's method of combining theoretical concepts with practical examples, fostering a deeper understanding of linear algebra and vector spaces.

Chapters

• 00:00 - 10:30: Vector Space Operations Between Vector Subspaces - Summarizing Exercises In Chapter 4, titled 'Vector Space Operations Between Vector Subspaces - Summarizing Exercises', Dr. Aliza Malek introduces exercises related to operations between vector subspaces. The exercises focus on operations such as intersecting and summing subspaces. In the first exercise, two subspaces of V=R4 are considered, with the first subspace U consisting of vectors in the form of a, b, a+b, and 2b, where a and b are real numbers.
• 00:00 - 11:30: Exercise 1: Vector Subspaces U and W The chapter titled 'Exercise 1: Vector Subspaces U and W' explores the concept of vector subspaces with a specific focus on two subspaces, U and W. Subspace W is defined as consisting of elements in the form (a, b, 2b, 2a), where 'a' and 'b' are real numbers. The exercise requires determining the sum of these two spaces, denoted as U+W, and checking if this sum is a direct sum. The chapter addresses the fundamental steps needed to solve these problems by calculating both the sum of the subspaces and their intersection.
• 01:00 - 07:30: Calculating Intersection The chapter 'Calculating Intersection' outlines the process of determining whether a sum is a direct sum by calculating the intersection. It explains the methodology of applying the conditions of one vector space (U) onto another (W) to align a general term in W with a general term in U.
• 07:30 - 11:30: Finding the Sum U+W In this chapter titled 'Finding the Sum U+W', the concept of understanding a general term in a sequence denoted as U is explored. The discussion highlights the significance of the components of U, particularly noting that while the first two components can be any letters, the third component is specifically defined by the condition that it must be the sum of the first and second components. This establishes a rule for determining elements within the sequence or set U.
• 11:30 - 18:00: Exercise 2: Matrix Subspaces U and W In "Exercise 2: Matrix Subspaces U and W," the focus is on manipulating terms within subspaces. A condition is introduced, requiring the third term to equal the sum of the first two terms, expressed as 2b = a + b. This forms a constraint within the exercise, emphasizing the relationship between components in a matrix subspace. Additional conditions are hinted at but are not explicitly detailed here, leaving some aspects for further exploration or calculation.
• 18:00 - 24:30: Exercise 3: Polynomial Subspaces U and W The chapter explores polynomial subspaces U and W, focusing on the conditions for a vector to belong to subspace U. A critical condition is that the fourth component must equal twice the second component. The discussion elaborates on transforming elements from subspace W (with components a, b, 2b, 2a) to align with the structure of subspace U (with components a, b, a+b, 2b). The narrative emphasizes understanding the merging of these conditions to comprehend the relationship between the two subspaces.
• 24:30 - 24:30: Conclusion and Further Exploration In this closing chapter, the text explores the necessity of equating certain variables within specific sets, namely W and U. The discussion leads to the conclusion that for an element to exist in both sets, it must satisfy certain conditions where 'a' is equal to 'b'. This equivalence results in a repetitive pattern or sequence of 'a, a, 2a, 2a', symbolizing a general term in the intersection of these sets. The chapter encourages further exploration and understanding of these mathematical principles.

135 - תרגילים מסכמים Transcription

• 00:00 - 00:30 Chapter 4 - Vector Space Operations Between Vector Subspaces - Summarizing Exercises Linear algebra course Dr. Aliza Malek Hello, thanks for coming back. It's time to practice. In the next two lessons, we will solve exercises related to vector subspaces, and the operations between subspaces. Intersecting, and summing up, and so on, so let's start. Here is the first exercise. I have two subspaces of V=R4. The first space U is all terms of the form a, b, a+b, 2b, where a and b are real.
• 00:30 - 01:00 The second space will be W, all terms of the form a, b, 2b, 2a, where a and b belong to R. What do we want to do with it? Find U+W, we want to find the space of the sum, and determine whether the sum is a direct sum. So let's start solving. First, we need to understand what we should do. We have to calculate both U+W, because it is being asked, and also the intersection.
• 01:00 - 01:30 Why? Because this will help us determine whether the sum is a direct sum. First, we will start calculating the intersection. So the first way to do this is to force the conditions of the U space on the W space. We are going to take a general term in W, and force it to look like the general term of U.
• 01:30 - 02:00 For this we need to understand what a general term in U looks like. So what do we have? Here we are looking at a term of U, and we see a, which just some letter, it doesn't matter, b, another letter, also doesn't matter which. But here it does not say c, it is no longer any just any letter. Here it says that the third component, should be equal to the first component plus the second component.
• 02:00 - 02:30 So we will now take the term of W, and force the third term to be equal to the first plus the second. 2b will be equal to a+b. The third component, 2b is equal to the first plus the second, to a+b. There is another condition here. It does not say what I want over here. Here it says: two times the second component.
• 02:30 - 03:00 That is, the second condition, to be in U, is that the fourth component is equal to twice the second component. The fourth component, 2a, should be equal to twice the second component, which is b, twice b. 2a = 2b. This is how we have taken a term of W: a, b, 2b, 2a, and we made it be like the term of U: a, b, a+b, 2b. What comes out of combining these two conditions?
• 03:00 - 03:30 We accept that a must equal b. Here we will place a equal to b, because it is forbidden to choose a and b within W as we wish. If we want the term of W to also exist in U, It must hold that a is equal to b. And so here we will place a=b, and get a, a, 2a, 2a. Therefore a general term in the intersection is of the form a, a, 2a, 2a.
• 03:30 - 04:00 What is a? Any term within R. So what did we get? That the sum is not direct. Why? Because the intersection did not come out as 0. Here we have an infinity of vectors in the intersection, it is not the zero space, so the sum is not a direct sum. That is, a term in the sum space can be written in infinite ways, as a term from U, and a term from W. Now we will see another way to find a general term in the intersection.
• 04:00 - 04:30 Here is the second way. What will we do this time? We will take a general term of U, we will take a general term of W, and force them to be equal. We will compare components of the general terms. So I remind you, a term of U and a term of W, and yes, here we are offered to try to solve it ourselves. Stop the video, and try to find a general term in the intersection this way by yourself.
• 04:30 - 05:00 Okay, and here's how we do it. I need to compare a term of U, here it is, and I want to compare it to a term of W. And pay attention, when I take a term of W, we will change the letters to other letters. And you have to be ready for that. Do not give the same letter names within U and W.
• 05:00 - 05:30 If you take the same letter names, you may make very critical mistakes. It won't happen in this example, but it will in other examples. It is very important, if in U we use the letters a and b, It is very important to use other letters in W. Therefore, the letters a and b that were given here in the definition, I changed to c and d. Instead of a I wrote c, instead of b I wrote d,
• 05:30 - 06:00 so over here I have 2d, and over here I have 2c. Use different letters. So, when we get some term of the form we saw before, a=b, we will know exactly where to place. If I get a condition of the a=b style, I can place it where it says a and b. And if I get a condition of the c=7d style, I can place where it says c and d. Therefore, always make sure to use different letters.
• 06:00 - 06:30 Now we will compare. The first term to the first, the second to the second, and so on. And what do we get? These are all the equations, we compared the four components. If you look at all the equations, in the end what stems out is that a equals b, equals c, and equals d. Please note, we received conditions for four letters, but I don't need all four. Because if I know the connection between a and b, I will place it here, and I will derive from this a general term for the intersection.
• 06:30 - 07:00 And if I know the relationship between c and d, I will place it here, and from here I will derive the general term for the intersection. So we don't have to the same thing twice. It is enough to place a,b in U, or c,d in W. If we do both, we just get the same result twice. It can be nice as a check, to make sure we didn't make a mistake, that in the end we got a term of the same form.
• 07:00 - 07:30 Okay, so what does the intersection result in? So if over here I place a=b, how do I know that I chose to place a=b? Because I introduced the intersection using a and b. b does not exist of course, because it is equal to a. So here you will place a=b, and you will get: a, a, a+a which is two a, and 2a, which is what is written here. Of course, this is the same result we got in the first way as well. And of course as we said before, the sum is not direct because the intersection is not 0.
• 07:30 - 08:00 Now we are done with the intersection, we did it in two ways, now we will move on to find the sum as was asked in the exercise. So what will I do? To calculate the sum we want to write a general term, x, y, z, w, because I don't know which is in there, I just know it's part of R4. So I take some vector in R4, and I want to write it as a sum of a term from U and a term from W, so we will write it as a sum of a general term from U plus a general term from W.
• 08:00 - 08:30 Here is what we get: That x, y ,z, w is equal to a, b, a+b, 2b plus c, d, 2d, 2c. Lucky that we used different letters, because if we had taken the letters a and b here also, we would get a totally wrong answer. Make sure to use different letters, I will remind you of this more than once. And here is what we get: We add, we get a+c, b+d, a+b+2d, 2b+2c.
• 08:30 - 09:00 And this should be equal to x, y, z, w. In fact, we get some kind of system of equations with lots of letters. We ask ourselves, what do we want to do, what are we looking for? We are looking for a, b, c, d using x, y, z, w. Because for every x, y, z, w you give me,
• 09:00 - 09:30 I want to find the a, b from here, and the c, d from here. And so I am looking for the a, b, c, d that match x, y, z, w. Let's write the system of equations. x equals a plus c, y equals b plus d, and so on. We will build a matrix of coefficients, these are the coefficients of a, b, c, d. x, y, z, w is the free variables vector, x, y, z ,w,
• 09:30 - 10:00 and we want to make sure we have a solution for that. Why? So that the vector is within the sum. Go ahead and rank it, I suggest you try it yourself, and compare with my solution. For the system to have a solution, that is, for x, y, z, w to actually be in the sum, we must make sure that w-2x-4y+2z is equal to 0, otherwise there is no solution.
• 10:00 - 10:30 What does it mean there is no solution? That x, y, z, w is not in the sum. That's it, we found the condition to be in the sum space. The conditions is that w-2x-4y+2z will be equal to 0. In other words, we will isolate one of the variables, for example the w, so we want w to be equal to 2x+4y-2z.
• 10:30 - 11:00 Hence a generalized vector will look like this: x, y, z, and instead of w we placed what we got above. This is how a general term in the sum space looks like. Therefore, in general, U+W is equal to the collection, to the space, of all the terms of the form x, y, z, 2x+4y-2z, so that x, y, z belong to R.
• 11:00 - 11:30 This is the sum space. The sum is a subspace of V, and not all of V, how do I know that? Because not all vectors in V will satisfy the condition that their w, the fourth component, will equal to twice the first, plus four times the second, minus twice the third. And therefore U+W is not a direct sum because the intersection turns out not to be 0, and it is not the entire space, it is an actual subspace of V.
• 11:30 - 12:00 Our mascot has noticed a very important thing. In U+W, when we look at the general term here, we see three parameters, x, y, and z, three parameters. In U and W, if you remember, I only had two, a and b in U, and c and d in W. As you recall, in the intersection we got a, a, 2a, 2a, one parameter. Try to find a connection between all these numbers.
• 12:00 - 12:30 I promise you that this connection is not accidental, we will see it later. Here is exercise number two, I have two subspaces, U is all the matrices in V, meaning, all the 2 x 2 matrices that hold: A times 0, 1, -1, 0, equals to 0, 1, -1, 0 times A. In other words it says over here that U is all the matrices in V,
• 12:30 - 13:00 which alternate in multiplication with the given matrix, 0,1, -1,0. Let's see what W is. W is all matrices A in V, so that A t is equal to A. What is W in other words? All symmetric matrices. What do we want to do? To find U+W, and determine whether the sum is a direct sum. An exercise in the same style as we have already seen, but now we are dealing with matrices.
• 13:00 - 13:30 So here, we will start as before, by finding U+W and the intersection. This, because it is what is asked, and this, to know whether the sum is a direct sum. Let's start with the intersection, we will immediately know if the sum is direct or not. We are searching, we take a term in W, a symmetric matrix. It's just a,b,b,c, and we force her to meet the condition of U.
• 13:30 - 14:00 Meaning, we will take the symmetric matrix a,b, b,c, multiply one side by 0,1, -1, 0, multiply by the other side, and demand an equality, here. We force it to fulfill the condition of U. a,b b,c times 0,1, -1,0 equals, as we said, 0,1, -1,0 times a,b, b,c. Perform the multiplication on the left side, perform the multiplication on the right side, and here is what we get. This is what we get on the left side, this is what we get on the right side.
• 14:00 - 14:30 Now of course we compare, b is equal to minus b, and so on. Here are the conditions: b is equal to minus b, and a is equal to c. Each such condition appears twice. And pay attention, b is equal to minus b, when we are in real numbers above R, meaning b=0. A number in R that is equal to its opposite, is just zero. Now I take these terms, these conditions, and I place backwards.
• 14:30 - 15:00 Once I place backwards, I will be able to tell what a general term in the intersection looks like. So I will take a symmetric matrix, a,b,b,c, and I will place b equals 0, and a equals c, and here is a general term in the intersection. Note that we also have a name for this group here, right? All scalar matrices. And the sum of course is not direct. That is, when we want to decompose a matrix with a sum
• 15:00 - 15:30 as the sum of one from here and one from there, we will have a great many ways to do this. And now let's see that U+W is actually the entire space. It is clear to us that U+W is contained within V, that is, when we take matrices from U and W and add them up, we get some kind of 2 by 2 matrix, this means the sum is a subspace of V, we always know that. What needs a bit more effort here is showing the other direction,
• 15:30 - 16:00 that every 2 by 2 matrix in V, belongs to U+W, which means that every matrix here is a sum of one from U, meaning which alternates in multiplication with 0,1, -1,0, and one from W, and one symmetrical. So we take a general term in V, x,y, z,w, and we write it...
• 16:00 - 16:30 We will show that it belong to the sum by writing it as one from U plus one from W. A general term in U, check how it looks. We will come to the conclusion that it is of the form a,b, -b,a. This is how all the matrices that alternate in multiplication with 0,1, -1,0, look like. Just check it out, take a general matrix, a,b, c,d, multiply on the left, multiply on the right, compare, and that is what you will get.
• 16:30 - 17:00 We already know about W, it's a symmetric matrix, it's a known space. And now we will take the matrix x,y, z,w, and decompose into a sum of one like this one, the same term on the main diagonal, and opposite terms on the secondary diagonal, and another symmetrical one. I am about to give you the final answer, you will check it by yourself. Here, x,y, z,w is equal to... Here you go.
• 17:00 - 17:30 Check out my solution by yourself. Zero, y-z divided by 2, z-y divided by 2, pay attention, I have the same term on the main diagonal, and opposite terms on the secondary diagonal, and the additional matrix is a symmetric matrix. The diagonal is free in a symmetric matrix, the term that is here, w+z divided 2, is equal to the term that is over here, y+z divided 2. Is this the only way to write it?
• 17:30 - 18:00 Obviously not, because the sum is no direct. I will give you another way to look at it right away. Take x from here, x,w, write a 0 over here. Place x here, but if I have x here, I also have x here, and therefore over there, will be w minus x instead of w. We will take x out, and bring w back. But here we need the same term, so if I place x here, there will also be an x over here.
• 18:00 - 18:30 And so here I want w, then what I will do is I will take the w I want, and I will take out the x I don't want. Try it by yourself. And so there are a great many ways to do it, infinite ways to do it, because the sum is not direct. Given the following subspace of R2[x], this is our next exercise with polynomials.
• 18:30 - 19:00 All polynomials in V such that their derivative at the point -1, is 0. First we will derive, place -1, compare to 0, this is our space. What are we being asked? Is there a subspace W in V, so that V is equal to the direct sum of U plus W? That is, I want to find a subspace whose intersection with U is 0, to get a direct sum,
• 19:00 - 19:30 and also that together they will give us all the entire space. The sum will give the entire space. Let's have a look. First of all we will take note of how terms in the space U look like. We will write it explicitly. Here, we take some polynomial in V, since it is R2[x], it is a term of the form ax^2+bx+c. Because it belongs to U, I know that when I derive it,
• 19:30 - 20:00 2ax+b, and I will place -1, I will get 0. Therefore, 2ax, that is, 2a times -1 plus b, is equal to 0. If so, what is the condition to be within U? b should be equal to 2a. So if b equals 2a, what does it say here? It does not say ax^2+bx+c, because b is equal to 2a,
• 20:00 - 20:30 we will place this inside and get: That a general term in U is of the form ax^2+2ax+c. So which do I take as W? Pay attention, ax^2, it does not matter what the coefficient of x^2 is, c, no matter what the free coefficient will be. But in U, we have a little problem with the coefficient of x, it must be equal to twice the coefficient of x^2.
• 20:30 - 21:00 Therefore, I need to play around with the coefficient of x, so that I can get any polynomial in the space, and not only those whose x coefficient is twice the coefficient of x^2. And so I choose dx as W, when d is free. That is, I want to take the space of all, so called, monomials, all polynomials that are multiples of just x.
• 21:00 - 21:30 That way I can play with the coefficient of x. Indeed, of course you cannot just say that this W fits, I really need to show you that it works. So first of all, let's see that the intersection comes out as 0, so that the sum is direct. So why is the intersection really 0? Let's take a term in U, and force the W conditions upon it. What is the condition of W?
• 21:30 - 22:00 If I see dx here, it means that the x^2 coefficient is equal to 0, why? Because there is no x^2 here. And the free coefficient is also 0, because there is no free coefficient here. Wait, so if the coefficient of x^2 is 0, a is equal to 0, and the free coefficient is 0, c is equal to 0. What does this tell me? Indeed, that the intersection is 0, so the sum really is a direct sum. Now I have to show that the sum is the entire space, that it is really possible to get any polynomial of the form ax^2+bx+c.
• 22:00 - 22:30 So let's show this. I take some polynomial, we chose the letters αx^2+βx+γ in R2[x]. Now, first we will write it as a polynomial of the form of U, ax^2+2ax+c means, if here I want to get αx^2, and the coefficient of x is conditioned to be twice the coefficient of x^2,
• 22:30 - 23:00 there is no choice, the part of U must be αx^2+2αx+γ. Now we need to get to βx. Lucky that I can play as I wish with the coefficient of x. As we said before, I throw away the 2α, and I bring the β I want. And therefore I choose to multiply x by β-2α. The β I need, minus the 2α which I don't want,
• 23:00 - 23:30 which I get from the term of U. Can I do it in another way? Is there another way for me to write a term from U plus a term from W? No. How do you know, you didn't check? I don't need to. The fact that the intersection is 0 tells me that the sum is direct. When a sum is direct, it means there is only one way to do the decomposition. If a decomposition is found, that's the only decomposition available.
• 23:30 - 24:00 And this is exactly the strength of a direct sum, that it tells me that this decomposition is unique. Now I am asked if there are other options for W? For this W I have chosen, the decomposition is unique, but perhaps another W can be chosen, what do you think? So I will tell you what the answer is, and you check it out. There are many more options to choose W. In fact, there are infinitely many ways to choose W,
• 24:00 - 24:30 so that the intersection is 0, and the sum gives the entire space. Later we will learn methods to do it very, very quickly. That's it guys, we are done for now, so thank you very much. And in the next lesson we will continue to solve additional exercises in the subspace, and operations with subspaces. Thank you very much.