WEBVTT mathematics/linear-algebra/hovasapian
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Welcome back to Educator.com and welcome back to linear algebra.
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In the last lesson we talked about something called the span of a given set of vectors.
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In other words, any linear combination of those particular vectors represents, all the vectors that can be represented by what actually spans.
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Today we are going to be talking about a related concept, again, a very, very profoundly important concept called linear independence, and its dual, linear dependence.
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So, let us write out some definitions and get started right away and jump into the examples, because again, examples are the things that make it very, very clear... I think.
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So, let us define linear dependence. Even though we speak about linear independence, the actual definition is written in terms of linear dependence.
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Well, let us actually -- okay -- vectors v1, v2... vk are linearly dependent.
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We will often just say dependent or independent without saying linear. We are talking about linear algebra so it goes without saying.
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If there exists constants, c1, c2... all the way to ck... not all 0 -- that is important, because that is easy -- such that c1v1 + c2v2 + ckvk = 0, or the 0 vector.
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So, let us look at this again. So, if vectors v1 to vk, if I have a set of vectors and if I can somehow, if there exist constants -- no matter what they are, but not all of them 0.
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If I can arrange them in such a way, some linear combination of them, if they add up to 0, we call that linearly dependent, okay?
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A couple of vectors are linearly dependent, and let us talk about what that actually means. Okay.
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Oh, by the way, if it is not the case, that is when they are linearly independent.
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If you cannot find this, or if the only way to make this true is if all of the individual c's are 0, or there is no way to find it otherwise, that means that they are linearly independent.
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Here is what the meaning is, just so you get an idea of what is going on.
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If we solve this equation... actually, I am going to number this equation. I have not done this before but this is definitely 1 equation that we are going to want to number.
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We will call it equation 1, because we are going to refer to it again and again. It is a very important equation.
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If you solve this equation for any one of these vectors, so let us just choose one arbitrarily, let us choose that.
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I am going to write that as c2v2 = well, we are going to move all of these over to the right... -c1v1 - c3v3 - so on, all the way to the end... -ckvk.
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Then I am going to go ahead and divide by c2, so I get v2 = this whole thing.
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Let me just... I hope you do not mind, I am going to call this whole thing the capital Z.
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Z/c2, well, as you can see, if this is true, then you can always solve for one of these vectors and this vector is always going to be some linear combination of the other vectors.
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That is what dependence means. Each of those vectors is dependent on the other vectors.
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In other words, it can be represented as some combination of the others. That is why it is called dependent.
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So, that is all that means. It is nothing strange, it makes perfect sense, and if this relationship does not exist, then it is independent.
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In other words, 1 vector cannot be represented as a combination of the others. It is independent. That is what independence means.
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Okay. Let us jump into examples, because that is what is important.
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Actually, let me talk about it... let me list at least the procedure. It is analogous to what we did before.
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The procedure for determining if a given list of vectors is linearly independent or linearly dependent... LD, LI... abbreviations.
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The first thing we do, well, we form equation 1. Remember when we were dealing with span over on the right hand side?
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We did not have 0, we had some arbitrary vector. Now, for linear independence or linear dependence, we set it equal to 0.
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So, we form equation 1, which is a homogeneous system.
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Then, 2, then we solve that system, and here are the results.
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If you find out it only has the trivial solution, that means all of the c's the constants are 0... that implies that it is independent.
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The other thing is, if there exists a non-trivial solution... just one, could be many, but if there is just one, so again, you remember this reverse e means there exists.
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So, if there exists a non-trivial solution, that implies that it is dependent.
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Again, we are just solving linear systems. That is all we are doing, and the solutions to these linear systems give us all kinds of information about the underlying structure of this vector space.
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Whether something spans it, whether something is linearly independent or dependent, and of course, all of these will make more sense as we delve deeper into the structure of the vector space. Okay.
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So, let us start with our example... this is going to be not a continuation of what we did for the span, but I guess kind of a further discussion of it.
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You remember in the last example of the last lesson, we found we had this homogeneous system and we found solutions for x, and we found 2 vectors that actually span the entire solution space, the null space.
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Those vectors were as follows... (-1,1,0,0), (-2,0,1,1).
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So, we know that these two vectors span the solution space to this particular equation based on what a was.
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I will not write down what a is. It is not necessary.
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Now, the question is... we know that they span the null space... the question is are they linearly independent or dependent.
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So, our question here is... are these two vectors LI or LD.
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Our procedure says form equation 1. So, form equation 1.
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That is just c1 × this vector, (-1,1,0,0) + c2 × (-2,0,1,1)... and we set it equal to the 0 vector which is just all 0's.
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So, we do not need the vector mark anymore... (0,0,0,0). Okay.
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Then, what we end up having is the following... this is equivalent to the following... (-1,-2,0), we are just taking the coefficients, that is all we are doing.
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(-1,0,0), (0,1,0), (0,1,0), and when we subject this to Gauss Jordan elimination, reduced row echelon, we end up with the following... c2 = 0, c1 = 0.
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That means that all of the constants, we only have two constants in this case, so this is only the trivial solution.
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Therefore, they are independent. There you go, that is it. It is that simple.
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You set up the homogeneous system, you solve the homogeneous system, and you decide whether it is dependent or independent. Fantastic technique.
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Okay. Let us consider the vector space of polynomials again. Let us consider p2, again.
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p2, the set of all polynomials of degree < or = 2.
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Let us look at 3 vectors in there... we have t² + t + 2.
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We have p2t, which is equal to 2t² + t, we have p3t, which is equal to 3t² + 2t + 2.
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So, they are just, you know, random vectors in this particular space, in other words random polynomials.
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Well, we want to know if these three as vectors are the linearly dependent or independent.
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Well, do what we do. We set up equation 1, which is the following. We take arbitrary constants... c1 × p1t + c2 × p2t, we will write everything out here... we want things to be as explicit as possible.
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Plus c3 × p3t, and we set it equal to 0, that is our homogeneous system.
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Now, we actually expand this by putting in what these p1, p2, p3 are. Okay.
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We get p1 × t² + t + 2 + c2 × 2t² + t + c3 × 3t² + 2t + 2 = 0.
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Now, let us actually... this one I am going to do explicitly... there is no particular reason why, I just decided that it would be nice to do this one explicitly.
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So, I have c1t² + c1t + 2c1 + 2c2t² +c2t + 3c3t² + 2c3t + 2c3 = 0.
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Algebra makes me crazy, just like it makes you crazy, because there are a whole bunch of things floating around to keep track of it all.
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Just go slowly and very carefully and be systematic. That is... do not ever do anything in your head.
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That is the real secret to math, do not do anything in your head. You will not be impressing anyone.
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I collect the terms... the t² terms, so I have that one, that one, and that one, and I end up with... so let me write these out as t² × c1 + 2c2 + 3c3.
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Then, I will take the t terms... there is a t, there is a t, there is a t, and I will write that as a second line here, just to be clear what it is that we are doing.
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c1 + c2 + 2c3... then I have plus the... well, the rest of the terms.
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That one... and that one... and is there one that I am missing? No. It looks like it is okay.
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So, it is going to be + 2c1 + 2c3 and all of this... sum... is equal to 0.
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Again, that means that this is 0, this is 0, this is 0. That is what this system is.
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So, we will write that, because everything is 0 on the right, so all of these have to be 0 in order to make this left side 0.
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So, I get c1 + 2c2 + 2c3 = 0.
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Note, we do not want these lines floating around. We want to be able to see everything here.
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c1 + c2 + 2c3, is equal to 0.
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2c1 + 2c3 = 0, this is of course equivalent to... I will just take the coefficients... (1,2,3,0), (1,1,2,0), (2,0,2,0), okay.
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So, this is the system that we want to solve, and we are going to subject that to reduced row echelon.
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So, I put a little arrow to let you know what is happening here and what you end up with is (1,0,1,0), (0,1,1,0), (0,0,0,0).
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So, let us take a look at our reduced row echelon. We have this is fine, yes. That is a leading entry... that is fine, that is a leading entry.
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There is no leading entry here. Remember when we solved reduced row echelon for a homogeneous system, this means we have infinite number of solutions, because this one can be any parameter.
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If this is any parameter, well, I can choose any number for this one and then that this means these two will be based on this.
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Therefore, we have infinite solutions. In other words, there does exist a non-trivial solution.
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So, there exists a non-trivial solution, which implies dependence... that means that those three polynomials that I had, one of them can be expressed as a linear combination of the other two.
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So, they are not completely independent. At least one of them depends on the others.
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So, we have dependence. Again, today we talked about linear independence and dependence.
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The previous lesson we talked about the span, so, make sure you recall... we are still studying a linear system when we do that, but with a span we choose an arbitrary vector... that is our solution on the right hand side of the equation, that linear combination that we write.
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For linear dependence and independence we are solving a homogeneous system. We just set everything equal to 0. Make sure to keep those straight.
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Thank you for joining us here for a discussion of Linear Algebra at Educator.com. We will see you next time.