WEBVTT mathematics/linear-algebra/hovasapian
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Welcome back to Educator.com and welcome back to linear algebra.
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The last lesson, we talked about a vector space. We gave the definition of a vector space and we gave some examples of a vector space.
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Today, we are going to delve a little bit deeper into the structure of a vector space, and we are going to talk about something called a subspace.
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Now, it is very, very important that we differentiate between the notion of a subset and a subspace.
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We will see what that means, exactly, in a minute. We will see subsets that are not subspaces, and we will see subspaces... well, all subspaces are subsets, but not all subsets are subspaces.
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So, a subspace, again, has a very, very clear sort of definition, and that is what we are going to start with first.
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So, we will define a subspace, and we will look at some examples. Let us see what we have got.
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We will define a subspace here, we will... let us go to a black ink here.
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Okay. We will let v be a vector space... okay? And w a subset, so we have not said anything about a subspace yet... a subset.
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Specify one quality of that subset is very important. It might seem obvious, but we do have to specify it... and w a non-empty... non-empty subset of v.
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Okay. If w is a vector space with respect to the same operations as v, then w is a subspace of v.
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In other words, if I am given a particular vector space, and I am given a particular subset of that vector space... if the w, if that subset itself is a vector space in its own right, with respect to those two operations... then I can say that w is an actual subspace of that vector space.
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Again, subset, that goes without saying, but a subset is not necessarily a subspace. A subspace has very, very specific properties.
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So, we will start off with the first basic example, the trivial example, which is probably not even worth mentioning, but it is okay.
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0... so vector... and v. So, the 0 vector itself, that element alone is a subspace.
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It is non-empty, and it actually satisfies the property, because 0 + 0 is 0, c × 0 is 0, so the closure properties are satisfied, and in fact all of the other properties are satisfied.
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And v itself, a set is a subset of itself. Again, we call these the trivial examples. They do not come up too often, but we do need to mention them to be complete. Okay.
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The second example. Let us do example 2.
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Okay. We will let v = R3, so now we are talking about 3-space, and w be the subset all vectors of the form (a,b,0).
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In other words, I am dealing with all 3 vectors, and I am going to take as a subset of that everything where the z, where the z component is 0.
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In other words, I am just looking at the xy plane, if you will. So, there is no z value. That is clearly a subset obviously this is just the z component is 0, so it is clearly a subset.
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Now, let us check to see if it is actually a subspace.
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Okay. So, with respect to the addition, well, let us see, we will let a1 equal... a1 be (1,0), we will let a2 = (a2,b2,0).
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a1 + a2 = a1 + a2, b1 + b2, 0. Well, yes, that is a number, that is a number, that is a number, this is a three vector. It does belong to w.
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So, the closure is satisfied.
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Now, notice, when we say closure with respect to the subset, that means when I start with something in that subset, I need to end up in that subset.
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I cannot just... if this is the bigger set and if I take a little subset of it, when I am checking closure now, I am checking to see that it stays in here, not just that it stays here.
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If I take two elements, let us say (a,b,0), and another (a,b,0)... if I add them and I end up outside of that set, that cannot be so.
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I might end up in the overall vector space, but the idea is I am talking about a subset. I want closure with just respect to that subset.
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So, be very, very clear about that. Be very careful about that.
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Now, let us check scalar multiplication. c × a = c × (a,b,0).
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Well, it is equal to (ca,cb,0). Sure enough, that is a number, that is a number, third element is still 0, so that is also in 2.
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So, yes, we can certainly verify all of the other properties. It is not a problem.
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As it turns out, this is a subspace. So, w is a subspace of v.
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This gives rise to a really, really nice theorem which allows us to actually not have to check all of the other properties... a,b,c,d,e,f,g,h... we only have to check two... closure.
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Since most of the time we are not talking about the entire space anyway, we are only talking about a part of it, the sub-space, a subset of a given space that we are working with, this theorem that we are about to list comes in very, very handy.
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So, let us see. v is a vector space... v be a vector space with operations of addition and scalar multiplication... let w be a non-empty subset of v, of course.
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Then, w is a subspace if and only if equivalence a, if closure is satisfied with respect to the addition operation and be closure satisfied with respect to the scalar multiplication.
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In other words, if I have a given vector space and if I take a subset of that, if I want to check to see if that subset is a subspace, I only have to check two properties.
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I have to check that if I take two elements in that subset, and if I add them together I stay in that subset, and if i take an element in that subset and multiply it by a scalar, I stay in that subset, I do not jump out of it.
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Checking those two takes care of all of the others. This lets me know that I am dealing with a subspace, and this if and only if means equivalent.
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If I had some subset of a space, and if I check to see that, you know, the closure is satisfied with respect to both operations, I know that I am dealing with a subspace and vice versa.
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Okay. Let us do another example. Example 3.
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We will let v = the set of 2 by 3 matrices, so v = m 2 by 3.
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So, the vector space is the set, the collection of all two by three matrices.
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We will let w be the subset of the two by 3 matrices of the form... the subset of m 2 by 3, of the form (a,0,b), (0,c,d).
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Basically I am taking, of all of the possible 2 by 3 matrices, I am taking just the matrices that have the second column of the first row 0, and the first entry of the second row 0.
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Everything else can be numbers. Okay. We just need to check closure, and we just need to check closure with respect to addition, closure with respect to scalar multiplication.
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So, let us take (a1,0,b19, (0,c1,d1), and let us add to that another element of that subset.
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(a2,0,b2), (0,c2,d2), well when we add these, we get a1 + a2, 0 + 0 is 0, b1 + b2, 0 + 0 is 0, c1 + c2, and d1 + d2.
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Sure enough, I get a 2 by 3 matrix where that -- oops, let me get my red here -- where that entry and that entry are still 0.
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So, yes, this is a member of w, the closure. It stays in the subset.
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If I check closure with respect to scalar multiplication, well c × (a,0,b), (0,c,d) = ca, c × 0 is 0, cb, c × 0 is 0, cc, cd.
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Sure enough, that entry and that entry are 0. There are our numbers. That is also in w.
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So, yes, now I do not have to say all of the other properties are fulfilled. Our theorem allows us to conclude because of closure properties, this is a subspace, not just a subset, a very special type of subset.
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It is a subspace. Okay.
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Let us do example 4. We will let p2 be the vector space of polynomials of degree < or = 2.
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So, t² + 3t + 5... t + 6 - t² + 7.
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All of the polynomials of degree < or = 2, remember a couple of lessons ago, we did show that the set of polynomials is a vector space. I mean -- yes, is a vector space.
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Now, we are going to let w, which is a subset of p2, be the polynomials of exactly degree 2.
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Okay, so not degree 2, degree 1, degree 0, but they have to be exactly of degree 2. Clearly this is a subset of the set of polynomials up to and including degree 2.
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The question is, is this w, the polynomials of degrees 2, is it a subspace of this... well, let us see.
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I mean, we do not know, our intuition might say yes it is a subset, it is a subspace, let us make sure... it has to satisfy the properties.
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Okay. We will take 3t² + 5t + 6, and we will take -3t² + 4t + 2.
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Both of these are of degree 2, so both of these definitely belong to w. Now, we want to check to see if we add these, will they stay in w?
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Okay. Well, if I add them, I get 3t² + 5t + 6 + -3t² + 4t + 2.
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When I add them together, that and that cancel... I end up with 5t + 4t which is 9t, 6 + 8.
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Well, notice, 9t + 8, this is a polynomial of degree 1. It is not a member of w, which was the set of polynomials of exactly degree 2.
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So, closure is not satisfied. Therefore, just by taking a subset, I am not necessarily dealing with a subspace.
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So, you see the subspace is a very special type of subset.
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So, no, w is not a subspace.
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Okay. Now, we are going to deal with a very, very, very important example of a subspace.
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This subspace will show up for the rest of the time that we study linear algebra. It is profoundly important.
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It will actually show up in the theory of differential equations as well. So, it is going to be your friend.
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So, write... example 5.
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Okay. Let v = RN, so we will just... vector space, n space... nice, comfortable, we know how to deal with it... it is the set of all n vectors.
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Okay. We will let w be the subset of RN, which our solutions to the homogeneous system equals 0, where a is m by n.
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The question is... is w a subspace? Okay, let us stop and take a look at what it is we are looking at here.
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We are starting with a vector space RN and we are picking some random matrix a, okay? Just some random m by n matrix a, and we can set up... we know that that matrix a, multiplied by some vector in a... we can set up that linear system.
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We can set it equal to 0 and we can check to see whether that system actually has a solution. Sometimes it will, sometimes it will be unique, sometimes it will have multiple solutions, sometimes it will have no solution at all.
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As it turns out, for those values, for those vectors in RN that are a solution to that particular homogeneous system, that is what we want to check to see.
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So, there is some set, let us say there is 17 vectors that are a solution to a particular homogeneous system that we have set up.
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They form a subset, right? of the vectors in RN. There are 17 vectors in this infinite collection of vectors that happen to satisfy this equation.
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When I multiply a matrix by a vector in that subset, I end up with 0, so there is a collection.
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I want to see if that collection actually satisfies the properties of being a subspace. So, let us go ahead and do that.
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Okay. Now, we need to choose of course 2 from that subset.
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So, let us let... we will let u be a solution, in other words, a × u = 0, and we will let v be a solution.
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So, I have picked 2 random elements from that subset. So av = 0, I know this because that is my subset. It is the subset of solutions to this homogeneous equation. Okay.
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Let us check closure with respect to addition. That means we want to check this is a × u + v.
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In other words, if I add u and v, do I end up back in that set?
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The set is the set of vectors that satisfy this equation.
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Well, that means if I add u and v, and if I multiply by a, do I get 0? Well, let us check it out.
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a × u + v, we know that matrices are linear mapping, right?
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So, I can write this as au + av. Well, I know that u and v are solutions already, so au = 0... av = 0.
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Well, 0 + 0 is 0, so yes, as it turns out, a(u + v) does equal 0. That means it is in... that means u + v is in w... the set of solutions.
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Okay. Now, let us check the other one... c ×... well, actually what we are checking is a c(u).
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We want to see if we take a vector that we know is a solution, if we multiply it by a scalar, is it still a solution... so what we want to check is this.
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Does it equal 0? Well, a c(u), well we know that multiplication by a matrix, again, it is a linear mapping, so I can write this as c × a(u).
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Well, a(u) is just 0. c × 0... and I know from my properties of a vector space that c × 0 = 0.
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So, yes, as it turns out, c(u) is also a solution.
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Therefore, yes, the subset of RN that satisfies a given homogeneous system is not just a subset of RN. It is actually a subspace of RN.
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That is a very, very, very special space. A profoundly important example. Probably one of the top 3 or 4 most important examples in linear algebra and the study of analysis.
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We give it a, in fact, we give it... it is so special that we give it a special name. A name that comes up quite often. It is called the solution space for... this one I am going to write in red... it is called the null space.
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So, once again, if I have some random matrix, some m by n matrix, I can always set up a homogeneous system by taking that matrix and multiplying it by an n-vector.
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Well, the vectors that actually satisfy that solution form a subset, of course, for all of the vectors.
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Well, that subset is not just a subset, it is actually a subspace. It is a very special kind of subset, it is a subspace and we call that space the null space of the homogeneous system. Null just comes from this being 0.
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The null space is going to play a very, very important role later on.
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Now, let us talk about something called linear combinations. We have seen linear combinations before, but let us just define it again and deal with it more formally.
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Definition. We will let v1, v2, and so on all the way to vk be vectors in a vector space.
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Okay. A vector 7 in v, big vector space, is called a linear combination of v1, v2... etc. if z, if the vector itself in the vector space can be represented as a series of constants × the individual vectors.
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These are... these constants are of course just real numbers. Again, we are talking about real vector spaces.
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So, again, a linear combination is if I happen to have 3 vectors, if I take those 3 vectors and if I multiply them by any kind of constant to come up with another vector and I know I can do that... I can just add vectors.
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The vector that I come up with, it can be expressed as a linear combination of the three vectors that I chose.
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Again, because we are talking about a vector space we are talking about closure, so I know that no matter how I multiply and add these, I still end up back in my vector space, I am not going anywhere.
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Let us do an example. Okay. In R3, so we will do a 3-space, we will let v1 = (1,2,1), we will let v2 = (1,0,2), we will let v3 = (1,1,0).
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Now, the example says find c1, c2, c3, such that some random vector z... so for example (2,1,5) is a linear combination of v1, v2, v3.
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So, in other words, I have these three vectors, (1,2,1), (1,0,2), (1,1,0), and I have this random vector (2,1,5).
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Well, they are all 3 vectors. Is it possible to actually express this vector as a linear combination of these 3?
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Let us see. Well, what is it asking?
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It is asking us to find this... c1 × (1,2,1) + c2 × ... this is not going to work.
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Maybe I should slow my writing down a little bit... (1,0,2) + c3 × (1,1,0)... we want it to equal (2,1,5).
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This should look familiar to you. It is just a linear system.
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This is what we want. We want to find these constants that allow this to be possible.
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Okay. So, let us move forward. c... c... c... c2... c2... c2... c3... c3... c3...
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What you have is a linear system. A linear system that looks like this... (1,1,1), (2,0,1), (1,2,0) × (c1,c2,c3) = (2,1,5).
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That is what we want to find out. We want to find c1, c2, c3... solving a linear system. Okay.
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Well, let us just go ahead and form the augmented matrix... (1,1,1,2), (2,0,1,1), (1,2,0,5), and let me actually show the augmentation.
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I am just taking the coefficient matrix and I am taking the solutions... and this one I am going to subject to Gauss Jordan elimination and turn it into reduced row echelon form.
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I end up with the following... (1,0,0,1), (0,1,0,2), (0,0,1,-1), there we go.
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c1 = 1, c2 = 2, c3 = -1. There you have it, these are my constants.
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So, I end up with 1 × v1 + 2 × v2 - ... oh, erasing everything... - v3 = (2,1,5).
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In this particular case, yes, I was able to find constants such that any random vector could be expressed as a linear combination of other vectors in that space.
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Okay. The definition here.
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If s is a set of vectors in v, then the set of all linear combinations of the elements s is called the span of s.
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In other words, if I have 6 vectors and I just chose them randomly from the space, if I arrange those in any kind of combination I want, okay?
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5 × the first one + 10 × the second one - 3 × the first one... in any combination.
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All of the vectors that I can generate from the infinite number of combinations, I call that the span of that set of s, which sort of makes sense.
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You are sort of taking all of the vectors, seeing what you can come up with, and if it spans the entire set of vectors. That is why it is called the span of s.
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So, let us see. Our example, we will let v = R3 again and we will choose 2 vectors in R3.
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We will choose (1,2,3), and we will choose (1,1,2).
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The span of s = the set of all linear combinations of these 2 vectors... c1 × (1,2,3) + c2 × (1,1,2).
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If I choose 1 and 1, it means I just add these... 1 + 1 is 2, 2 + 1 is 3, 3 + 2 is 5, that is the vector.
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The set of all possibilities, that is what the span is. Okay.
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Let us see. Let us now list a very, very important theorem.
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Okay. Let s be a set of vectors from a vector space... and we said k... let s be a set of vectors in a vector space v.
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Then, the span of s is a subspace. In other words, let us just take R3.
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If I take 2 vectors in R3, like I did here, the span of that is all of the linear combinations that I can come up with of those 2 vectors.
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Well, this set right here, this span of s is actually a subspace of R3.
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That is extraordinary. No reason for believing that should be the case, and yet it is.
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Okay. Let us finish up with a final detailed example.
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Once again, we will deal with p2. Do you remember what p2 was? That was the vector space of polynomials of degrees < or = 2.
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Okay. So, we are going to choose... I am going to go back to black, I hope you do not mind... we will let v1 = 2t² + t + 2.
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We will let v2 = t² - 2t... we will let v3 = 5t² - 5t + 2.
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We will let v4 = -t² - 3t - 2.
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So, again, this is a vector space and I am just pulling vectors from that vector space.
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Well, the vectors in the space of polynomials happen to be polynomials of degree < or = 2. Here I have a 2, 2, 2, 2, they all happen to have a degree of 2, but I could have chosen any other ones.
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Now, I am going to choose a vector u, I am going to pull it randomly, and it is t², this one I am actually going to write in red, excuse me.
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u... it is t² + t + 2, well, this is also in p2.
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Here is the question. Can we find scalars (a,b,c,d), such that u is actually a linear combination of these other 4.
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In other words, can I take a random vector and express it as a linear combination of these other 4.
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Well, you are probably thinking to yourself... well, you should be able to, you have 4 vectors and you have another one... let us see if it is possible, it may be... it might not be.
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a1v1 -- I am sorry it is av1 -- + bv2 + cv3 + dv4.
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In other words, is this possible? Can we express u as a linear combination of these vectors.
00:36:34.000 --> 00:36:45.000
Well, let us find out. Let us write this out and see if we can turn it into a system of equations and solve that system of equations and see if there is a solution for (a,b,c,d).
00:36:45.000 --> 00:37:36.000
Okay. Let me go back to my blue ink here. So, av1 is... one second, I will do blue here... a × 2t² + t + 2 + b × v2, which is t² - 2t + c × v3, which is 5t² -5t + 2, and then + d × -t² - 3t - 2.
00:37:36.000 --> 00:37:48.000
So, again, I have this thing that I am checking, and I just expand the right side to see what I can get. Okay.
00:37:48.000 --> 00:38:06.000
Another question is... is all of this is equal to u? Which is t² + t + 2.
00:38:06.000 --> 00:38:18.000
Okay. I multiply all of these out. When I multiply this, when I distribute and simplify and multiply, I am going to end up with the following.
00:38:18.000 --> 00:39:06.000
When I combine terms, 2a + b + 5c - d... t² + a - 2b - 5c - 3bt + 2a + 0b + 2c - 2d, again, multiplied all of these out, expanded it, collected terms, put all of my t²'s together, all of my t's together, my nothings together. Okay.
00:39:06.000 --> 00:39:12.000
I need that to equal t² + t + 2.
00:39:12.000 --> 00:39:23.000
Well, this is an equality here. I need to know if this is possible. So, here I have a t², let me go to red... here I have t², and here I have t².
00:39:23.000 --> 00:39:27.000
The coefficient is 1. That means all of these numbers have to add to 1.
00:39:27.000 --> 00:39:32.000
Here I have a t, here I have a t, the coefficient is 1, that means all of these number have to add to 1.
00:39:32.000 --> 00:39:40.000
This nothing at all, it is just a constant, that means all of these numbers have to add to 2. That is what this equality means.
00:39:40.000 --> 00:39:48.000
I have myself a system of equations. Let me write that system of equations.
00:39:48.000 --> 00:40:00.000
Okay. I have got... 2a + b + 5c - d, I need that to equal 1.
00:40:00.000 --> 00:40:10.000
I have a - 2b - 5c - 3d, I need that to equal 1.
00:40:10.000 --> 00:40:22.000
Then I have 2a + 0b, it is up to you whether you want to put the 0 there or not... it does not really matter, + 2c - 2d = 2.
00:40:22.000 --> 00:40:53.000
Okay. I have a linear system. I am going to set up the coefficient matrix... (2,1,5,-1,1), (2,1,5,-1,1), (1,-2,-5,-3,1), (1,-2,-5,-3,1), (2,0,2,2,0), (1,-2,2,-2,2,)
00:40:53.000 --> 00:41:03.000
Of course my solution is there, so I just remind myself... and again I subject this to reduced row echelon form.
00:41:03.000 --> 00:41:14.000
Let me, well, let me do it this way. So, Gauss Jordan elimination to reduced row echelon for which I use Maple, and I get the following.
00:41:14.000 --> 00:41:30.000
(1,0,1,-1,0), (0,1,3,1,0), (0,0,0,0,1). Okay.
00:41:30.000 --> 00:41:43.000
There is a problem with this. These are all 0's, this is a 1. This tells me that 0 = 1, this is not true.
00:41:43.000 --> 00:41:54.000
This is inconsistent. That is what it means, when we... remember when we reduced to reduced row echelon, if we get something like this, we are dealing with a system that is inconsistent. There is no reduced row echelon form.
00:41:54.000 --> 00:42:18.000
So, because this is inconsistent, that means there is no solution. In other words, u, that vector u that we had... it does not belong to the span of v1, v2, v3, v4.
00:42:18.000 --> 00:42:28.000
Just because I can pick 4 vectors at random does not mean a vector in that space can necessarily be represented as a linear combination.
00:42:28.000 --> 00:42:41.000
In other words, it is not in the span. So, that is what we did. We took those 4 vectors, we set up the equality that should be the case, we solved the linear system, and we realized that the linear system is inconsistent.
00:42:41.000 --> 00:42:51.000
Therefore, that polynomial, t² + t + 2 is not in the span of those vectors. Very, very important.
00:42:51.000 --> 00:43:02.000
Okay. So, today we discussed subspaces and again a subspace is a subset of a vector space, but it is a very special space of a subset that satisfies certain properties.
00:43:02.000 --> 00:43:07.000
In order to check to see that something is a subspace, all you have to do is check two properties of that subset.
00:43:07.000 --> 00:43:14.000
You have to make sure that closure with respect to addition is satisfied, and closure with respect to scalar multiplication is satisfied.
00:43:14.000 --> 00:43:25.000
Again, what that means is that this might be our bigger vector space... if we take a subset of that, closure means if I take two elements in that space, I need to end up back in that space.
00:43:25.000 --> 00:43:34.000
So, I do not end up outside in the bigger space, I need to stay in that subset. That is what turns it into a subspace. Very, very important.
00:43:34.000 --> 00:43:37.000
Okay. Thank you for joining us here at Educator.com, we will see you next time. Bye-bye.