128.1 - דוגמאות לתמ של F^n

Summary

בהרצאה זו, הדוקטור עליזה מלאך מציגה דוגמאות לתת-מרחבים בקורס אלגברה ליניארית. הדגש בהרצאה הוא על המרכבים בתמ של Fn. המטרה היא להראות מתי קבוצה יכולה להיחשב כתת-מרחב ולהזכיר את התנאים הנדרשים לכך - שהיא לא תהיה ריקה, תהיה סגורה לחיבור ולכפל בסקלר. במהלך ההרצאה, מוצגות דוגמאות מתמ אחדות, שם נבדקות הקבוצות W1, W2, W3, W4 ו-W5 ואותם מגדרים אם הם מתאימים להיות תתי-מרחבים או לא.

Highlights

• הדוקטור עליזה מלאך מסבירה על תתי-מרחבים בחללים וקטוריים 🌟
• הוצגו חמש דוגמאות לבחינת קבוצות כמתאימות להיות תתי-מרחבים 🚀
• ב-W1 הוכח שהיא תת-מרחב, לעומת W2 שלא עונה על הקריטריונים 📚

Key Takeaways

• לסיווג קבוצה כתת-מרחב נדרשת לא להיות ריקה, להיות סגורה לחיבור ולכפל בסקלר.
• חללים הומוגניים תמיד מהווים תתי-מרחבים בעוד שהחללים הלא הומוגניים לא.
• תנאים חשובים להוכחת תת-מרחב: סגירות לחיבור, סגירות לכפל בסקלר ואי ריקנות.

Overview

בהרצאה זו, ד״ר עליזה מלאך מהטכניון, מתמקדת במחלקות של תתי-מרחבים בחללים וקטוריים במיוחד בתמ של Fn. מוסבר על התנאים שעל קבוצה לעמוד בהם כדי להיחשב כתת-מרחב: לא ריקה, סגירות לחיבור וסגירות לכפל בסקלר.

הדוגמאות מתמקדמות מתמיהן ספציפיות כמו W1 ו-W2, ומראו את השימושיות של ההבנה הזו. כל דוגמה מנותחת וממוחשת על מנת להדגיש בצורה מעשית את החשיבות של הבנת מתמטיקה ואלגברה ליניארית בחיי היום-יום.

בסיום ההרצאה הדגש הוא על המשמעות של מושגים אלו וכיצד הם מתקשרים לחללים וקטוריים. הדוגמאות כוללות מציאת מטריצות למערכות הומוגניות וגישות לבדוק אם הקבוצה היא תת-מרחב, מעשירה את התלמידים בידע מתמטי עשיר ומועיל להמשך הלימודים.

Chapters

• 00:00 - 06:00: Vector Spaces This chapter titled 'Vector Spaces' is a part of a Linear Algebra course by Dr. Aliza Malek. It focuses on providing examples of subspaces, specifically subspaces of Fn. The chapter begins with a review of the theory, reminding us that a subspace W of a vector space V must satisfy three properties: it must be non-empty, it must be closed under addition, and it must be closed under scalar multiplication.
• 06:00 - 09:00: Subspaces and Subgroups The chapter discusses subspaces and subgroups, highlighting properties that define them. A key focus is on the expression 'B plus C' which simplifies to 'alpha u plus beta v', aiding in verification processes. Various examples are examined to differentiate between subgroups that qualify as subspaces and those that do not. Specifically, subgroups failing to meet properties A to C, or A and B, are not considered subspaces.
• 09:00 - 13:00: Vector Space Criteria This chapter discusses the criteria for a set to be considered a vector space, specifically in the context of R3. The instructor begins by acknowledging that the size of the vector space (R3, R4, R100, etc.) is arbitrary for the purposes of discussion but chooses R3 for simplicity. The chapter then outlines a process for examining subgroups of V (the vector space) such as W1 and W2. For each subgroup, the chapter explores whether or not it qualifies as a subspace by applying the necessary and sufficient conditions for subspace verification. These conditions typically include checking if the subgroup is closed under addition and scalar multiplication, among other criteria.
• 13:00 - 17:30: Examples of Vector Spaces The chapter discusses vector spaces, focusing on a particular example where V is the vector space R3. The condition for a vector to belong to a subspace W1 is that the y-coordinate must be zero, leading to a representation of vectors in W1 as (x, 0, z) in R3.
• 17:30 - 21:30: Special Cases and Non-Vector Spaces The given transcript discusses the concept of subspaces within the context of R3, a three-dimensional real vector space. It explains that certain conditions must be met for a subset of a vector space to qualify as a subspace. Specifically, in this section, the focus is on the condition y=0, which suggests that the subspace described contains vectors where the y-component is zero. The idea of writing the subspace differently but maintaining the same meaning is also highlighted, emphasizing the different representations and conditions that can define a subspace. The overarching theme is understanding and identifying subspaces under specific constraints in vector spaces.
• 21:30 - 25:30: Homogeneous Systems The chapter titled 'Homogeneous Systems' discusses the examination of vector groups and their general terms. Specifically, it explains how placing conditions inside the vector form allows for easier verification of condition fulfillment. The text references zero as a significant term within the vectors, highlighting its role in defining the group.
• 25:30 - 30:00: Scalar Multiplication The chapter titled 'Scalar Multiplication' begins with an assertion that W1 is a vector space, which is also referred to as a vector subspace. The discourse highlights the importance of confirming that W1 is not empty by ensuring that the zero vector is part of W1. This initial step is crucial and emphasized throughout section A of this chapter.
• 30:00 - 35:00: Important Vector Spaces The chapter "Important Vector Spaces" explores the concept of vector spaces, focusing on the importance of the zero vector within these spaces. The text emphasizes that it's not enough to just state the presence of the zero vector; it must be demonstrated to be present. The argument is made that if all the components of a vector are zero, then each individual component, including any specified component, is zero—hence verifying its presence. Additionally, the text discusses using vectors (x1,0,z1, x2,0,z2) from a vector space W1 and scalars in R, suggesting a demonstration or proof involving these elements to illustrate properties of vector spaces.

128.1 - דוגמאות לתמ של F^n Transcription

• 00:00 - 00:30 Chapter 4 - Vector Spaces Examples Of Fn Subspace Linear Algebra Course Dr. Aliza Malek Hello, thank you for coming back. We continue with examples of subspaces, and this time subspaces of Fn, let's begin. First, as usual, we will recall the theory. I have a vector space V, a subgroup W. W is a subspace of V only if the following three properties hold: It is not empty, it closed for addition, closed for multiplication by scalar,
• 00:30 - 01:00 and properties B plus C can be shortened to alpha u plus beta v. It simply shortens the process of checking for us in some cases. Now we will see several examples, all of which will be subgroups of Fn. Some will be subspaces, and some will just be subgroups, that don't deserve to be named a subspace. Why wouldn't they deserve to be named this way? Because they will not hold properties A to C, or A and B.
• 01:00 - 01:30 So we take V to be R3. R3, R4, R100, it doesn't matter. In R3 we write a little less, we only have three components, and we will start solving all kinds of examples of V. We will take W1 and W2, all sorts of subgroups and check each one on its own, whether it is a subspace, or not a subspace. Here is the first group: W1 will be all terms of the form x,y,z in V,
• 01:30 - 02:00 again we shall recall that V is R3, and what is the condition for it to be in W1? That y=0, this is our condition. We can write this group a little differently. If I know that y=0, then it doesn't actually say x,y,z here, after all y is zero, what does it say? x,0,z. Therefore I can describe the group W1 as x,0,z, in V.
• 02:00 - 02:30 V is R3, so I know that x and z belong to R. I write it a little differently. What is written here, and what is written here, is the same. If I want to show that this is a subspace, the condition is y=0. If I want to show that this is a subspace, the condition is for it to look like this: What does that mean to look like this?
• 02:30 - 03:00 A digit, a zero, a digit, or in other words, the second term is zero. It's just another description of the same group. We took the condition and placed it inside, what did we get? A so-called general term of W1. This is the form of all the vectors in W1, now it will be a little easier for us to check the fulfillment of the conditions.
• 03:00 - 03:30 So let's show that W1 is indeed a vector space. A vector space, a vector subspace, it is the same thing. Sometimes I will say it like this, and sometimes I will say it like that. So first thing that must not be overlooked of course: Section A says W1 is not an empty group, and it is advisable to check this using the zero vector. And so, in section A we will always write: W1 not an empty group because zero belongs to W1.
• 03:30 - 04:00 Wait, but we need to explain and show why zero is indeed there. But it's not enough to say it's there, we have to show that it's really there. So let's check it out, it's really simple. Zero means the zero vector, all the components are zero. So if all the components are zero, surely the y component, surely the second component is zero, and thus it is there. What will I do now? I take two vectors in W1, x1,0,z1, x2,0,z2, and I take two alpha beta scalars in R.
• 04:00 - 04:30 We show condition B, and we want to show that alpha u plus beta v, are in W1. In other words, when I finish the math here, what will I want to see? That the second component, its y-component, is zero. Let's see what we get. We will place, instead of u, I place x1,0,z1, instead of v, x2,0,z2,
• 04:30 - 05:00 then I multiply by alpha, by beta, add them together, according to the definition of scalar addition and multiplication, what do we get? That it is alpha x1 plus beta x2, alpha times 0, plus beta times 0, which is zero plus zero, and alpha z1 plus beta z2. This is according to the definition of the operations that we already know. Wait, what does it say here? I have alpha x1 plus beta x2, zero, alpha z1 plus beta z2.
• 05:00 - 05:30 This means that I indeed have a vector that is in W1, term, zero, term. True, this is a special case of a system of homogeneous equations. Do you remember? We once had a+2b=0, All kinds of conditions of a system of homogeneous equations. y=0 is a homogeneous linear system of equations. That's why it is truly a special case, and we have already seen that these things work.
• 05:30 - 06:00 Here is the next question: This time in W2, I have x,y,z in V, in R3, so that y=1. This time we will see that W2 is not a vector space. Wait, it's really simple. Zero is not found in W2, it does not fulfill the condition that the second component is 1. True, this is a special case of a system of inhomogeneous equations. And we saw it in the previous lesson,
• 06:00 - 06:30 a homogeneous system is always a vector space. An inhomogeneous system of equations is always not a vector space. Since this is a special case of an inhomogeneous system, we very easily saw that it does not work, that it is not a vector space. Here is section C: All of x,y,z in V, in R3, such that x=0, or y=0. This time we have a case that is not a vector space either.
• 06:30 - 07:00 Odd. After all, it's also a homogeneous system, but pay attention, in a system of homogeneous linear equations, we have equations that should exist together simultaneously. Here I have that either this will exist, or this will exist. In a linear system we have the word "and". If it were written x=0 AND y=0,
• 07:00 - 07:30 then we would have a homogeneous linear system of equations. But here the word "or" is used, it means the first or the second, I doesn't matter. And this is where the trouble begins. In this case, although zero is really inside, W3 will not be closed to addition. This or that, when we add, neither this nor that. Let's see an example:
• 07:30 - 08:00 Here, v is equal to 0,1,1, because x is zero. u is equal to 1,0,1, y is zero, this or that. What happens when we add? In u+v we have 1,1,2, neither x is zero, nor y is zero, so the sum no longer belongs to W3, and we have no closure to addition. Here is the next section, D: This time I have x,y,z, so that x+y=0.
• 08:00 - 08:30 Familiar? Right... W4 will turn out to be a vector space. It is familiar, very similar to examples that have worked for us before. We will see, there is one way to show this, simply sections A to C, or A and B, as we did before. I leave that to you to do on your own. I will show it to you in a slightly different way. We keep saying that x+y is a special case of a homogeneous system of equations.
• 08:30 - 09:00 But wait, when we saw a homogeneous system of equations, what we had was a matrix. We had a matrix of order m by n, and we had the group of solutions as a subgroup of Fn. Fn we have, R3. But where is the matrix here? Therefore, for this collection to be a subspace, it should be a subspace of Fn, and we want to find a matrix.
• 09:00 - 09:30 Here n is equal to 3, F is equal to R, and we want to know whether it is possible to find a matrix... Equations, no matter how many, the main thing is that we have three unknowns. Why does it not matter how many equations there are? Because we want to be in Fn, in R3. How many equations do I have, It really doesn't matter, the main thing is to have three unknowns, so that the solutions will be in R3.
• 09:30 - 10:00 So we are looking for the matrix which W4 will describe the solution space of its homogeneous system. Can we find such a system where when we do A times x, meaning, x,y,z, A*(x,y,z)=0, and visualize the system of equations written here, x+y=0. Here is an example matrix:
• 10:00 - 10:30 I will take a 2 by 3 matrix, two equations. I could continue with twenty more equations, it doesn't matter, because when I multiply by x,y,z, I just need it to be three over here, no matter how many equations I have. I chose 1,1,0, 2,2,0, Why? Think, what will happen when you take A and multiply it by x,y,z? You will get, x+y+0z=0, 2x+2y+0z=0.
• 10:30 - 11:00 So what really are our equations in this system with this being the matrix of coefficients? Indeed, in the end we only wrote the equation x+y=0, which is the equation written right over here. Therefore, when we see equations over here, we are actually asking ourselves what is the matrix that will suit us, so that when we write the appropriate equations, we will get what we have over here. We can always find many such matrices, we just need to know that we have a matrix in the background.
• 11:00 - 11:30 Ax=0, the breaking down of that matrix means, the equations that are written for us in the condition here. Therefore, when we see it, we immediately associate it with a system of homogeneous linear equations. That's it, let's move to section five. All of x,y,z, such that x>=0. What will we see this time? That W5 is not a subspace and please note and be careful:
• 11:30 - 12:00 W5 is not an empty group, let's check the zero, because if zero is not there, we can stop. But it says x>=0. In the zero vector, this zero is indeed greater than,equal to zero. Luckily they didn't write much greater, because then already on section A we would have gotten it wrong when we would look for zero. This time, the condition has worked. Closure to addition?
• 12:00 - 12:30 Of course it holds. Because if x1 and x2 are greater than,equal to zero, when I add them, x1+x2 will be present in the first component, and it will be greater than,equal to zero as well. Both A and B hold. And what happens with section C? It does not hold anymore. Because when I multiply by a scalar, I am allowed to choose any scalar I like, and so I want a negative scalar. What happens when I multiply by a negative scalar?
• 12:30 - 13:00 For instance, I will take 1,0,0. 1 is greater than equal to zero, that's fine. Let's choose alpha to be -1, and here is what we get? Alpha u is -1 times 1,0,0, which is -1,0,0, and it's no longer in W5, so indeed, each property counts. No property should be missed. If we had tried to check section B, it might have gotten a bit complicated, we might have missed
• 13:00 - 13:30 the closure to multiplication by a scalar. Another difference between checking B, C and B: When it works, it shortens. When it doesn't work, we have to be careful, every property is important, everything should be checked. Here is question number two: V is Fn, we are doing examples of Fn. A is an n by n matrix, some sort of a square matrix. We want to show that the next group...
• 13:30 - 14:00 What is the next group? All the v's in Fn, such that A times v equals 2v. Please note that this is not a homogeneous system. A homogeneous system is Av=0. Here Av=2v, this is some very, very special case. We are not looking at an inhomogeneous system Ax=b, absolutely not.
• 14:00 - 14:30 We take a special case where we take A, multiply it by a vector, and the result is twice that same vector. This is the group, a bit of a strange group, isn't it? Are there any such vectors at all? Luckily, the first check we do for the subspace is that the group is not an empty group, that there are vectors that satisfy this property. So we want to show that this somewhat strange group is indeed a subspace of v, of Fn, that is. And it's not just a subspace, it's a very important subspace.
• 14:30 - 15:00 First we will check. We will show properties of a subspace. This time we will check two properties, because when it works and you test two properties, it is a bit of a shortcut. So first of all, u is not an empty group, there is at least one term for sure that fulfills this condition, and that is the zero vector. Because A times zero is zero, and it is also equal to twice zero. That's it, the zero vector is inside. Now we will take two vectors in u, meaning:
• 15:00 - 15:30 Au=2u, Av=2v. Let's take two scalars alpha and beta, and let's see what happens for the vector alpha u plus beta v? So we multiply by A, A times alpha u, plus beta v equals... Let's do some calculations. We know that A is a matrix, u and v are column matrices, Fn is a special case of the matrices, remember? Therefore, according properties of matrix multiplication, I simply open the brackets.
• 15:30 - 16:00 It's alpha A times u, plus beta u times v. Now I will use the given data, remember? Au will be 2u. Av will be 2v, let's place it. We will get alpha times 2u, plus beta times 2v. All this is scalar multiplication, and addition of matrices, because u and v is a special case of matrices,
• 16:00 - 16:30 I am still continuing with matrix properties. With scalars I can do whatever I want, let's take the 2 out of the brackets, and what did I get here? Twice alpha u plus beta v. Pay attention to the beginning and the end: A times the vector alpha u, plus beta v, equals to 2 times that same vector alpha u, plus beta v. And so this exactly fulfills the condition that A multiplied by a vector is equal to twice the vector, and so, we got that alpha u plus beta v, belongs to u,
• 16:30 - 17:00 and therefore, both A and B hold, and we got a vector space, or a vector subspace, it doesn't matter. You are asking what would happen if the number wasn't 2, if it was a different number? An excellent question. Note that throughout the proof, the number 2 had no meaning. Any number you write here will appear both in this term, and in this term, and we will be able to take it out of the brackets, and so it will be true for any number you give me,
• 17:00 - 17:30 for 2, for 3, for 0.5, for minus pi, for any number you give me, it will work. As we said, these are very, very important spaces. These spaces even have names, but we will come to that much, much later. I will remind you of this example yet. That is it. In the meantime, thank you very much. In the next lesson we will see more examples of other standard spaces. That's it for now, thanks.