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 do the first part of a two-part lesson and we are going to discuss something called the rank of a matrix.
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This is a profoundly, profoundly important concept.
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Not that anything else that we have discussed or that we are going to be discussing is not important, it is, but this is where linear algebra starts to take on a very -- I do not know what the word for it is.
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I know for me personally, this point when I studied rank that linear algebra became mysterious and at the same time extremely beautiful.
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This is where the real beauty of mathematics was for me. This was the turning point. This particular topic.
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I would definitely, definitely urge you to think very carefully about what is going on in these two lessons, and of course, using your textbook, maybe going through some of the proofs, which are not all together that difficult.
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To make sure that you really, really wrap your mind around what is happening with this notion of a rank, and with the notion of the column space and the row space, which we are going to define and discuss in a minute.
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So, do not worry, you do not have to know what they are yet.
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It speaks to the strangeness that I can just take some random rectangular array of numbers, 3 by 6, 5 by 7, 10 by 14, and just throw any kind of numbers in a rectangular array.
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Just by virtue of arranging them that way, there is a relationship that exists between the columns and the rows of that matrix that no one would ever believe should actually exist... and yet it does.
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That is what makes this beautiful.
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So, anyway, let us dive in, and hopefully you will feel the same way. Okay.
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So, we are going to start off with a definition, and let us go ahead and define our matrix first that we are going to be talking about.
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So, we will let a equal the following matrix... a11... a12... a1N... and a21... a22... so on.
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a31 all the way down to aM1... and of course down here it is going to be aN... the subscript is mn.
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Of course this is an m by n matrix. Okay.
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Okay. So, this is an m by n matrix, and these entries of course are the entries of that matrix. Now, m by n. There are m rows.
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So, 1, 2, 3, 4... m this way. n columns... that way.
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Alright. If I take the rows of this matrix, I have m of them, right? So, let me write -- just write R1, which is equal to a11, a12, and so on, to a1n.
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Let us say... let us just go ahead and write out row 2 also. a21, a22, all the way to a2N, so the first row, the second row, the third row.
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If I treat them as just individual vectors, well, these rows... considered as vectors in RN.
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The reason they are considered as vectors in RN is because they have n entries in them, right?
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1, 2, 3, 4, 5... all the way to n. That is why they are vectors in RN.
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A little bit better of an r here -- that one was a little bit odd.
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Considered as vectors in RN... they span a subspace of RN called the row space.
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So, let us stop and think about that again. If I take the rows of this m by n matrix, I have m rows.
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Well, the rows have n entries in them, because they are n columns.
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Well, if n entries, that means it is a vector in RN, right? Just like if I have 3 vectors, it is a vector in R3.
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Well, those rows, let us say if I have 6 of them, they actually span a subspace of RN.
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That subspace, we call the row space... so, let us differentiate the rows are the individual vectors from the row space, which is the space that the rows actually span.
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I can define something called the column space analogously.
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So, the columns... I will call them c1, c2, so forth, and I will write them as column vectors, in fact... a11, a21, all the way down to am1.
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And, c2, I will put the second one also... a21, a22, a23, all the way down to am2.
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Wait, I think my a2... have my indices incorrect here.
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This is going to be a12, a22, a32, there we go... am2... it is very confusing.
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Okay. So, if I take columns, notice now the columns have m entries in them, because they have m rows. If I take them as columns, they have m entries.
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Well, if they have m entries, then they are vectors in RM.
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So, the columns, considered as vectors in RM span another subspace called exactly what you think, the column space... subspace of RM, called the column space.
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Okay. So, once again, if I just take any random matrix, any random rectangular array... m by n... if I take the rows and treat them as vectors in RN, those vectors span a space, a sub-space of RN as it turns out.
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So, they are not just a subset, they actually span a subspace.
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We call that the row space. If I take the columns of that matrix, any random matrix, and if I... those are going to be vectors in m-space... RM.
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They span a space, a subspace of RM and we call that the column space.
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So, any matrix has 2 subspaces associated with it automatically, by virtue of just picking the numbers -- a row space and a column space.
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We are going to investigate the structure of that row space and that column space.
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Okay. Now, let me... quick theorem that we are going to use before our first example.
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If the 2 matrices a and b are 2 m by n matrices, which are row equivalent, and again row equivalence just means I have converted one to the other.
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Like, for example, when I convert to reduced row echelon, those 2 matrices that I get are row equivalent.
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Then, the row spaces of a and b are equal.
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In other words, excuse me, simply by converting it by a series of those operations that we do... converting a matrix to, let us say reduced row echelon.
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The reduced row echelon matrix still has the same row space. I have not changed anything. I added one equation to the other, I have not changed anything at all. That is what this theorem is saying.
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So, now we are actually going to use this theorem. So, let us do our first example. Let us do it in red, here... oops, there we go.
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Okay. We will let s equal the set of vectors v1, v2, v3... and v4.
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Here is what the vectors are: v1 is the vector (1,-2,0,3,-4).
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Let me write that down here, actually.
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v2... so I do not feel so crushed in... (3,2,8,1,4).
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v3 is (2,3,7,2,3).
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v4 is equal to (-1,2,0,4,-3).
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So, these vectors are vectors in R5, right? because they have 1, 2, 3, 4, 5 elements in them.
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I have four vectors in R5. Okay.
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We want to find a basis for a span of s, so these four vectors, they span a space.
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I do now know what space, but I know they span a space.
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Well, we want to find a basis for that space... for that subspace, not just any old space. Okay.
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Let us do the following. Let us write these vectors, 1, 2, 3, 4 of them... and they are vectors in R5, so let us write them as the rows of a matrix.
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Let us just take these random vectors and put them in matrix form.
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So, we will let a equal to... we want to take the first vector (1,-2,0,3,-4), and I just write it our as my first row. (1,-2,0,3,-4).
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I take vector 2, and that is (3,2,8,1,4).
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I take vector 3, which is (2,3,7,2,3).
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And, I take (-1,2,0,4,-3).
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So, all I have done is I have taken these vectors and I have arranged them as the rows of a matrix.
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Well, if I subject this matrix to reduced row echelon form, the matrix... nothing augmented... just take the vectors, stick it in matrix form and convert it to reduced row echelon form. Here is what I get.
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(1,0,2,0,1), (0,1,1,0,1), (0,0,0,1,-1), and (0,0,0,0,0).
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This is my reduced row echelon form. Well, as it turns out, the non-zero rows of this reduced row echelon form, in other words row 1 row 2, and row 3, they are as vectors linearly independent.
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So we have these vectors that span a space, which means that we need to find vectors that are linearly independent.
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As it turns out, there is a particular theorem which I did not mention here, but it is something... a result that we are using.
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When I take a matrix and I reduce it to reduced row echelon form, the non-zero rows are linearly independent as vectors.
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Therefore, I can take this vector as my first one, I will just call it w1. w1 = (1,0,2,0,1).
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w2 = (0,1,1,0,1).
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w3 = (0,0,0,1,-1).
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This is a basis for the span of s. I was given this set of vectors.
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I arranged these vectors as rows of a matrix. I reduced that matrix to reduced row echelon form, and the non-zero rows I just read them straight off.
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Those non-zero rows, they form a basis for the span of those vectors.
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Well, how many vectors do we have? 1, 2, 3.
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So, the dimension of that space, of that span, is 3... because we have 3 linearly independent vectors that span the same space.
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Notice, these vectors, (1,0,2,0,1)... they are vectors in R5, they have 5 entries in them.
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(0,1,1,0,1), (0,0,0,1,-1), these are not the same vectors as these, okay? They are not from the original set.
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They are completely different, and yet they span the same space. That is what makes this kind of extraordinary... that you can take these original vectors, arrange them in a matrix, reduced row echelon, whatever vectors are left over, they give you a vector that spans the same space.
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Okay. So, now let us say we actually want to find a basis for the span of this space, but this time, we want vectors that are actually v1, v2, v3, v4.
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We want vectors from this subset, whether it is v1 and v3, or v2, v3, v4, we do not know... I mean we do know, because we do know it is going to be at least 3 of them.
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But, the idea is here we came up with 3 vectors that span this that are not from this original set, but there is actually a way to find a basis for this, consisting of vectors that are from this set.
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Let us actually go ahead and do that. Before we do that, we are going to define this thing called the rank.
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Well, we just calculated the dimension of that sub-space which was 3. Well, the dimension of the row space is called the row rank.
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In a minute, we are going to talk about column rank, so I will just go ahead and write that too... the dimension of the column space is called column rank.
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Okay. So, now let us do the same problem, but let us find a basis that actually consists of vectors from the original set.
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This time, we take those vectors that we had... the v1, v2, v3, v4, and instead of writing them as rows, we are going to write them as columns and we are going to solve the associated homogeneous system for that.
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Notice, for the last one, we just wrote them as rows and we converted that matrix to reduced row echelon form. Now we are going to write those as columns and we are going to augment it.
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We are going to actually solve the associated system.
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So, it turns out to be the following.
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Yes... so, when I write v1, v2, v3, v4 as columns this time instead of rows, it looks like this.
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(1,-2,0,3,-4), (3,2,8,1,4), (2,3,7,2,3), (-1,2,0,4,3)... and the augmented system.
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Now, we have done this before, actually. If you remember a couple of lessons back when we were looking for the basis of a particular subspace, this is what we did.
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We solved the associated homogeneous system, and then when we converted it to reduced row echelon, which we will do in a minute, the columns that have leading entries, the corresponding vectors are a basis from the original set.
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So, let us go ahead and do that.
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We convert to reduced row echelon and we get the following... (1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (11/24,0,-49/24,0,7/3), (0,0,0,0,0).
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Okay, here we go. Now we have this column, this column, and this column, the first, second and third columns have leading entries in them.
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Because they have leading entries in them, the original vectors corresponding to them.
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In other words, that vector, that vector and that vector... which are part of the original set, they form a basis for the span of that original set of four vectors, right?
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We had four vectors, but we do not need four vectors, we need just three, and that is what this procedure does.
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It allows us to find a basis consisting of vectors from the original set.
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Whereas the thing we did before allowed us to find a basis that had nothing to do with the set. Notice, we have three of them. It is not a coincidence.
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So, in this particular case, we could take v1, v2, v3, the original, that is a basis for a span of s.
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In other words, the span of s which is the row space.
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Originally, we wrote them as rows... row vectors.
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Here, we wrote them as columns in order to solve it in a certain way so that we end up with vectors from the original set.
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It is still a row space, or the span of s, if you will.
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Okay... and again, the row rank, well the row rank is a number of linearly independent vectors in the basis, or the number of vectors in the basis.
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They are linearly independent by definition. The basis is 3.
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Okay. So, now, let us write this out.
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So, given a set of vectors, what it is that we just did.
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Given a set of vectors, s, v1, v2, v3, and so on... to vk.
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One. If we want a basis for the span of s consisting -- that is fine -- consisting of vectors from s, from the original set, then, set up the vectors as columns.
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Augment the 0 vector, 0's, okay.
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Convert to reduced row echelon, and then the vectors corresponding to columns with leading entries form the basis you are looking for.
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The basis of span s. That was the second example that we did.
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Two. Well, if we want a basis for span s consisting of vectors not necessarily in s...
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Vectors completely different from that, but that still form a basis for that space.
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Then, set up the vectors in s as rows of a matrix.
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Convert to reduced row echelon.
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Then, the non-zero rows, the actual non-zero rows of the reduced row echelon matrix form a basis which span s.
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So, there you go. If you are given a set of vectors, and if you want to find a basis for the span of those vectors and if you want this particular basis to actually consist of vectors from the original set, then you set those vectors up as columns.
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You solve the associated homogeneous system. You just add a zero, you augment the matrix, and then once you get the columns that have leading entries in them -- excuse me.
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Let us say it is the first, third and fifth column, you reduce row echelon.
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Then you go back to the original matrix and you pick the first, the third, and the fifth column. That is your basis.
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If you want the basis for the span that has nothing to do with the original set of vectors, the set of the vectors as a row... reduced row echelon form... know worries about augmentation.
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The non-zero rows give you the basis for the same space, and you will always end up with the same number.
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If it is 3 for 1, it will be three for the other, if it is 2 for 1, it will be 2 for the other.
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Okay. Let us do another example here. We will let -- this time, we will just throw out the matrix itself as opposed to given the vectors, we will just give you the matrix.
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Often, it is done that way. You remember the definition at the beginning of this lesson was some row space and some column space. That is what we do. We just sort of often given a random matrix, and we have to treat the rows and columns as vectors.
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So we will let a, let me go back to blue here, let a equal (1,2,-1), (1,9,-1), (-3,8,3), and (-2,8,2). That is a.
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We want to find a basis for the row space of this matrix.
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So for the row space of a, first part, we want this basis to consist of vectors which are not rows of a.
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Part b, we want it to consist of vectors, the basis consisting of vectors which are rows of a.
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Okay. So, we have a row space, so find the basis for the row space.
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Well ,the row space is just these treated as vectors, and they are vectors in R3... 1, 2, 3... 1, 2, 3... 1, 2, 3... 1, 2, 3.
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We have four vectors in R3. So, let us set them up.
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Let us do part a first. Now, they want us to find a basis for this row space consisting of vectors, which are not actually these vectors at all.
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These rows... So, for that, we need to set these vectors up as columns and solve the associated homogeneous system.
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So, let us go ahead and do that.
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Not -- no sorry -- we are consisting of vectors that are not rows of a. Okay.
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Since we are doing it that are not rows of a, we are actually just going to set them up as rows. They are already set up as rows. My apologies.
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It is hard to keep these straight sometimes. So, not rows of a. We set them up as this.
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So, we are just going to go ahead and solve this matrix. Convert to reduced row echelon form.
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So, let us rewrite it... (1,2,-1), so we write them as rows... yes. (1,2,-1).
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(1,9,-1), (-3,8,3), and (-2,3,2).
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Okay. We convert to reduced row echelon form, and we end up with the following: (1,0,-1), (0,1,0), (0,0,0), (0,0,0).
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Okay. That, oops I wanted blue. So, that is a non-zero row, that is a non-zero row.
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Therefore, I can take my basis as the set (1,0,-1)... that and that.
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(1,0,-1)... and (0,1,0)... there two vectors form a basis for the row space of this matrix.
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In other words, there are four vectors, 1, 2, 3, 4 vectors in R3. Three entries.
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Well, they span a space, it is called the row space.
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By simply leaving them as rows or arranging these as rows, row echelon, I get 2 non-zero vectors.
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These two non-zero vectors, they span the same space.
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So, dimension is, in other words, the row rank is equal to 2, because I have 2 vectors.
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Okay. Now let us do part b. We want to find a basis for this row space that consists of vectors which are rows of a.
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So, if they are rows of a, that means I have to solve a homogeneous system.
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So, I will set up these rows as actual columns.
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Okay. So, that is going to look like this... (1,2,-1), (1,9,-1), (-3,8,3), (-2,3,2)... and setting up a homogeneous system... (0,0,0).
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Now, I need to convert this to reduced row echelon form, and when I do that, I get the following.
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I get (1,0,0), (0,1,0), (-5,2,0), (-3,1,0), (0,0,0).
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Go back to red... this column has a leading entry, this column has a leading entry... there they are.
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Therefore, my basis consists of the original vectors corresponding to those two columns.
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Now, my basis is the vector (1,2,-1) and the vector (1,9,-1).
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This basis spans the same space as the other basis that I just found.
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The rank is, well, 2.
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It is pretty amazing, is it not?
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Now, think about this for a second. I take some random vectors, and I arrange them in rows, and I convert to reduced row echelon form and I get the dimension of 2... i get a row rank of 2... two non-zero vectors.
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Then, I write them as columns, and I solve the homogeneous system and I end up with, even though I end up with no non-zero columns, I mean I end up with two columns with leading entries.
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I still end up with 2... that is kind of extraordinary. At least it is to me.
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That you can just arrange these things as columns or rows and still end up essentially in the same place.
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Now, this space, this row space, this space that is spanned by the original vectors that we had, this is a perfectly good basis for it, and the basis that we got otherwise is also a perfectly good basis for it.
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Notice, both of them. They have 2 vectors in them. Okay.
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So, row rank, row space. Profoundly important concepts.
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We will continue this discussion in the next lesson. Thank you for joining us here at Educator.com. Take care.