WEBVTT mathematics/linear-algebra/hovasapian
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Welcome back to Educator.com and welcome back to linear algebra.
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Today we are going to talk about something called the span of a set of vectors.
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It means exactly what you think that it means. If I have a collection of vectors, 2, 5, 10, the number does not actually matter that much.
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We want to talk about all of the possible linear combinations of those vectors, that are possible... all of the vectors that can be built from that particular set.
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So, for example, if I take R2... the normal plane, I know that I have the vector in the x direction, I know that I have the vector in the y direction, and if I take any collection of those, multiplied by constants, let us say 5i + 6j, I can represent every single vector in R2.
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So, those two vectors, we say it actually spans R2.
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So, that is just sort of the general description that a span is.
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Unfortunately, in this case, the name actually gives you an idea of what it is that you are talking about, so it is not strange. So, let us start with a couple of definitions and let us see what we can do.
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Okay. Now, in a vector space, there is an infinite number of elements and then the reason for that is if I have at least one element in that space, and I know there is at least one, I know that I can multiply that element by any number that I want. Any constant.
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Therefore, since that constant is just a real number, there are an infinite number of elements in that vector space.
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However, what we want to do is we want to see if we can find a finite number of elements in that vector space that when I take certain combinations of them, linear combinations of them, that we can describe the entire space.
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That means all infinite vectors based on just that finite set of vectors.
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So, let us actually write out definitions down.
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Okay... vectors v1, v2 and so forth onto vk are said to span v, which is our vector space.
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If every vector in v can be written as a linear combination of the v1, v2, v3.
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SO, now we have actually written it down. If I have vectors v1 through vk, let us say 6 of them.
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And... if any linear -- excuse me -- if any combination of those vectors, they do not have to all be included, you know some of the constants can be 0, but if some combination of those vectors can represent very single vector in that vector space, then we say that that set of vectors actually spans the vector space.
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Okay. Now, let us list the procedure to check to see if a set of vectors actually spans a vector space.
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Procedure to check if the set of vectors spans a vector space, so vs, for vector space.
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So, let us see... you know what... let us leave it as blue for right now.
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Choose an arbitrary vector v in the vector space, so when you are given a vector space, just choose some arbitrary vector.
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So, if you are given 4-space, R4, then just choose the random vector (a,b,c,d), or you can call it (x,y,z,t), just some random vector and label it... to determine if v is a linear combination of the given vectors.
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So, this is basically just an application of the definition, which is what definitions are all about.
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Let me just take a quick second to talk about definitions real quickly.
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Often in mathematics, we begin with definitions. They are a basic element.
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We use those definitions to start to create theorems and we sort of build from there, build our way up from the bottom if you will.
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If you find that you have lost your way in mathematics, more often than not, you want to go back to your definitions, and 90% of the time, the problem is something is either missing from a definition, or there is a definition that the student has not quite wrapped his mind fully around.
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Again, mathematics is very, very precise. It says exactly what it wants to say, no more and no less.
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Okay. Determine if v is a linear combination of the given vectors... definition.
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If so, if it is a linear combination, then yes, the vectors actually span the vector space.
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If not, if there is no way to form a linear combination, then no. Okay.
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So, again, when you are forming a linear combination, you are taking a constant, multiplying it by a bunch of vectors and setting it equal to some, in this case an arbitrary constant.
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So, once again, we are going to investigate the linear system, the linear systems are ubiquitous in linear algebra. Okay.
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So, let us start with an example here.
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Let us go to the next page. So, first example. Let us consider R3.
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So, regular 3 space... (x,y,z), the space we live in.
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We are going to let v1 = (1,2,1), we will let v2 = (1,0... -- oops, excuse me -- (1,0,2), and we will let our third vector be (1,1,0).
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I wrote these out in the form of a list, in terms of their coordinates... I could write the vertically, I could write them horizontally without spaces, however you want.
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Now, the question is these three vectors that I have chosen randomly... do they... so do v1, v2, v3 span R3?
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Are these vectors enough to represent all of R3.
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Is a linear combination, any linear combination, any of these three vectors... can I find any vector in R3 and use these three to represent it?
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Well, let us see. Okay.
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Well, first thing we do from our procedure, let us choose an arbitrary vector in R3... arbitrary v in R3.
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So, let us choose, let us say that v is just equal to (a,b,c,d), and again variables do not matter. It is just some random vector.
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Okay. Now, we want to see the second thing that we are going to check is are there constants c1, c2, c3 such that, well, c1v1 + c2v2 + c3v3, a linear combination equals v.
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That is what we are doing. That is all we are doing. We are taking an arbitrary vector, we are setting it equal to the constant times the vectors that we have in our set, and we are going to solve this linear system.
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So, this is the vector representation, the simplest representation... now we are going to break it down a bit.
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This is, of course, equal to c1 × the first vector, which is (1,2,1)... I am going to go ahead and write them vertically.
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That is just a personal choice of mine. You are welcome to write them any way you please. I like to do them vertically because it keeps the coefficient of the ultimate matrix that we are going to do... systematic.
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These becomes the columns of the particular matrix.
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Plus c2 × (1,0,2)... I hope that 2 is clear... + c3 × third vector, our third vector is (1,1,0), and we are setting it equal to (a,b,c).
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We want to find out if this linear system actually has a... so, let us write this system... when we actually multiply these c1's out, we get this... I would like you to at least see one of them.
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We get c1 + c2 + c3 = a.
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We get 2c1, c20, I will just leave that over here, + c3 = b, and I get c1 + 2c2, and there is nothing over here = c.
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This is, of course, equivalent to the following augmented matrix... (1,1,1,a)... (1,1,1,a).
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Again, I am taking the coefficients of c1, c2, c3... c1, c2, c3 is what I am looking for.
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Can I find a solution to this? If I can, then yes... (2,0,1,b), (1,2,0)c, so this is the system that we are going to solve.
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We are going to subject it to reduced row echelon form. I will not go ahead and show you the reduced row echelon form... I of course did this with my computer, with my Maple software.
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Fast and beautiful. As it turns out, this does have a solution. In other words, it has a non-trivial solution, and here is what it looks like.
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You end up with c3 = (4a - b - 2c)/3 -- oops, these strange lines that show up on here.
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Let us see... c2 = (a - b + c)/3... let me go ahead and put parentheses around so that you know the numerator is that way, because I am writing my fractions not two dimensional, but in a line.
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c1 = (-2a + 2b + c)... and again, all divided by 3.
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So, for any choice of a, b, or c, I can just put them in here and the constants that I get end up being a solution to this system.
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So, because there is a solution to this system... this system... that means there is a solution to this because these are all equivalent.
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That means that any vector that I choose, and again, I just chose it (a,b,c)... is... I can find constants for them and these are the explicit values of those constants no matter what vector I choose.
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So, yes... let us try this again... so here the answer is yes... v1, v2, v3 do span R3.
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Turn this and make this red... that vector, that vector and that vector are a perfectly good span for R3.
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Now, you know, of course, that R3, or if you do not, I am telling you right now... that R3, the standard 3 vectors that we use as the set which spans the space is of course the i vector, the j vector in the y direction, and the k vector in the z direction.
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Those are mutually perpendicular. Well, as it turns out, you can have any collection of vectors that may span a vector space.
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There is no particular reason for choosing one over the other, so there are an infinite number of them, but in certain circumstances... it makes sense to choose one over the other.
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In the case of the i,j,k, we choose it because they have the property that they are mutually orthogonal -- excuse me.
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Which actually -- excuse me -- makes it easier to deal with certain things.
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Okay. Let us consider another example. Let us consider... we will do this example in red... p2.
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If you recall p2 is the vector space of all polynomials... all polynomials of degree 2 or less.
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So, we will let our set s, this time we will actually do it in set notation... p1t, p2t, oh this p2 right here has nothing to do with this p2.
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This is a general symbol for p2, the space of polynomials of degree 2. This just happens to be number 2 in the list.
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p1... let us define that one as t² + 2t + 1, and we will say that p2 of t is t² + 2.
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Okay. We want to know does s... do these 2 polynomials, are they enough to span all of p2.
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In other words, can I take two constants c1 and c2, and multiply them by this... by these 2... c1 × this one, c2 × this one.
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Can I always find constants such that every single polynomial, every second degree polynomial, or first degree polynomial, remember it is degree 2 or less... can be represented by just these 2 vectors... well, let us find out.
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So, again, the first thing that we do is we choose an arbitrary vector in p2... and an arbitrary vector in the space of polynomials looks as follows... at² + bt + c, right? because it is t², right?
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And now... we want to show the following... we want c... c1 × p1t + c2 × p2t = our arbitrary vector that we picked.
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This one up here equals at² + bt + c, so again, we are just setting up a basic equation... a linear combination of the vectors that we are given... does it equal our arbitrary vector?
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Well, let us go ahead and expand this out based on what they are... so, we have c1 × p1t which is t² + 2t + 1 + c2 × t² + 2... okay.
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We want that to equal at² + bt + c... let us multiply this out.
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Now, when you multiply this out, this is going to be c1t², 2c1t + c1, and then 2c2² + 2c2.
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I am going to skip that step... and just... imagine that I just multiplied it, basic distribution, you know, something from algebra 1.
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Then I am going to combine terms in t², in t, and in t to the 0 power.
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So, it is going to look like this when I actually expand it. It is just one line that I am skipping.
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It is going to be c1 + c2 × + 2c1... sorry... t²... 2c1 × t + c1 + 2c2 = at² + bt + c.
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Well, we have an equality sign here. We have some -- change this to blue -- this is the coefficient of t² here.
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This is the coefficient of t² in the equality, so that equals that.
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That is our first equation... c1 + c2 = a.
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Well, 2c1 is the coefficient of t, here b is the coefficient of t. They are equal on both sides, so this one becomes 2c1 = b.
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Then, we do this one over here. We have c1 + 2c2 = c... c1 + 2c2 = c.
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This is going to be equivalent to the following... (1,1,a)... I am just taking coefficients... (2,0,b), and (1,2,c).
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Now, when I subject that to reduced row echelon form, this one I do want you to see...
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So, we subject that to Gauss Jordan elimination for reduced row echelon... we get (1,0,2a-c).
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(0,1,c-a), and we get (0,0,b-4a+2c). We get this as the reduced row echelon of the system that we just took care of.
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Now, this is possible if and only if this thing right there... so you see this 0 right here, this... that means this thing right here... the b - 4a + 2c has to equal 0 for this thing to be consistent.
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The only way that that b - 4a + 2c is equal to 0 is if b = 0, a = 0, c = 0. That is just the trivial solution.
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So, there is no solution to this system. So, the answer to this one is no.
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p1 and p2, t, those two polynomials do not span p2, which is the vector space of polynomials of degree 2 or less.
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So, it is not enough. I need some other vector, I do not know, maybe 2 or 3 of them.
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Let us try... let us go back to red here for example number 3.
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These will just list, you know these already... i and j span R2.
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We know that i,j, and k, the three unit vectors in the (x,y,z) direction span R3, and so on, onto R4... R5.
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So, e1, we do not give the letters anymore, after 4... we just call them e.
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e1, e2, and so on all the way to eN... They span n-space, which is RN.
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Now, you have probably already noticed this, but notice, I have 2 vectors that span R2, three vectors that span R3, N vectors needed to span RN, that is a general truth.
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We will talk more about that in a minute... well, actually the next lesson.
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Okay. Now we are going to get to a profoundly, profoundly important example.
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This one, we want to do very, very carefully... therefore, so, let us consider the following homogeneous equation... ax = 0, such that a, and we are actually going to explicitly list this vector... this matrix, excuse me, (1,1,0,2), (-2,-2,1,-5), (1,1,-1,3), (4,4,-1,9). Okay.
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So, we have this homogeneous system, matrix × some vector x is equal to the 0 vector. Here is our matrix a.
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In other words, what are my x values that will actually make this true such that when I multiply this by some vector x, which I will put in blue...
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Let us just say this is going to be x1, x2, x3, x4... what are the values of x that will actually make it true so that when I multiply these, I end up with 0.
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It might be one vector, it might be an infinite number of vectors, it might be no vectors.
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So, if you remember what we called the set of vectors that actually make this true, we called it the null space.
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The solution space of the homogeneous system... in other words, the vectors that make this space be true. We called it the null space.
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So, let me just write that down again... the null space is a very, very important space. Okay.
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Now, one thing that we also know about the null space is the null space is a subspace... of... in this case, R4, remember?
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So, if we happen to be dealing with a null space in R4, you have these 1, 2, 3, 4... 4 by 4, we are looking for a vector which has 4 entries in it, so it is from 4-space.
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And... remember that we proved that it is actually a null space, it is a subspace of R4.
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Not just a subset... it is a very special kind of subset... it is an actual subspace.
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In other words, it is a vector space in its own right. It is as if I can ignore the rest of R4, just look at that, and I can treat it the same way I would the rest of that vector space, it has special properties... all of the properties of a vector space.
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Okay. Now, can we find a set of vectors... here is our problem here... can we find a set of vectors that stands the null space?
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Let me write that down... can we find a set of vectors... set which spans... singular because it is set... this null space.
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So, again, we are not looking to expand the entire R4. We are just looking for something that will span this particular null space. The null space based on this particular matrix.
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Well, let us solve this system... Let us see what x's we can come up with.
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So, when we solve the system, we create the augmented matrix. When I take this and I put 0's over on the final column in matrix a, and I end up subjecting that to reduced row echelon form, which I will not do... well, actually you know what, let me go ahead and write it all out. It is not a problem.
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So, I have 1...1... yes, (1,1,0,2,0), (-2,-2,1,-5,0), (1,1,-1,3,0)... and (4,4,-1,9,0).
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So, this is our augmented matrix. This is just this thing in matrix form, and then I am going to subject this to Gauss Jordan elimination to convert it into reduced row echelon, so let us write out the reduced row echelon form of this.
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It is going to be (1,1,0,2,0), (0,0,1,-1,0), (0,0,0,0,0)... 0's everywhere.
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Okay. So, this is our reduced row echelon, and remember... this represents x1, x2, x3, x4. This is the -- we are looking for a vector... a 4 vector.
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So, let us take a look here. This one is fine. This one has a leading entry -- let me go to red -- so, this one is good, and this one has a leading entry.
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This one does not have a leading entry. The x2. So... x2... x3, x4, yes.
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So, that second does not have a leading entry and the fourth does not have a leading entry.
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Remember, when we do something, when we convert it to reduced row echelon, the columns that do not have a leading entry actually are free parameters... I can call them s, t, x, y, they can be any number I want.
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Then I solve for the other two. So, let us go ahead and set x2 = r, and we will set x4 = s, they can be any numbers, they are free parameters.
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Then, what I get is x1 + x2 + 2x4, that is what this line here says... x1 + x2 +2x4 = 0.
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Therefore, x... x2 is R, x4 is S, so x1 = -R - 2s.
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So, I have x2, I have x4, I have x1 that I just calculated, and now I will do x3... x3 here... x3... this right here...x3 - x4 = 0.
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So, x3 = x4, which is equal to x... s... I am sorry.
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So, we have x2 is r, x4 is s, x3 is s, excuse me, and x1 is -r - 2s. Let me rewrite that a little bit differently.
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I am going to write that as the following -- let me go back to blue here.
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x1 = -r - 2s, x2 = r, and there is a reason why I am writing it this way, you will see in a minute... x3 = s -- I will put the s over there in that column.
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x4 = s, well take a look at this x1, x2, x3, x4... I have r's here, 0, I have 2, 0, s, s.
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I can rewrite this as the following in vector form. This is equivalent to (x1, x2, x3, x4) equal to, let me pull out the r from here, and I can just take these coefficients (-1,1,0,0).
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And... plus, now I can pull out an s from here, and I can take these coefficients, (-2,0,1,1), if that is not clear, just stop and take a look at what it is that I have done.
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I will just treat this as a column... treat this as a column... Okay.
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So, notice what I have done. I have taken the solution space which is x1, x2, x3, x4, this is because i have solved the system... these are all the possible solutions.
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There is an infinite number of them because r and s can be anything. I have written it in vector form this way, as a linear combination of this vector and that vector.
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These are just arbitrary numbers, right? The r and the s are just arbitrary numbers, therefore I have expressed the solution set, which is this thing of the homogeneous system based on that matrix.
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I have expressed it as the linear combination of this vector and this vector.
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Therefore, those 2 vectors (-1,1,0,0) and (-2,0,1,1), they actually span the null space.
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The null space has an infinite number of solutions, that is what our system tells us here.
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Well, I know that I can describe all of those solutions by reducing it to two vectors, any linear combination of which will keep me in that null space.
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It will give me all of the vectors, all of the vectors are represented by a linear combination of this vector and that vector.
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They span the null space of the homogeneous system.
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Again, profoundly important example. Go through this example again carefully to understand what it is that I did.
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I had a homogeneous system, I solved that homogeneous system for the solution space. I represented that solution space... once I have that... I represent it as vectors, and these vectors that I was able to get -- because these are arbitrary constants -- well, that is the whole idea behind a span.
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I was able to represent the higher solution space but only 2 vectors. That is extraordinary.
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Okay. Thank you for joining us here at Educator.com for the discussion of span, we will see you next time.