WEBVTT
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Okay, here is lecture 10 in linear algebra. Two important things to do in this lecture.
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One is to correct an error from lecture 9. So the blackboard with that awful error is
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still with us. And the second the big thing to do is to tell you about the four subspaces
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that come with a matrix. We've seen two subspaces, the column space and the null space, there's
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two to go. Okay, first of all, and this is a great way to recap and correct the previous
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lecture. So you remember I was just doing R3. I couldn't have taken a simpler example
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than R3. And I wrote down the standard basis. That's the standard basis. The obvious basis
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for the whole three-dimensional space. And then I wanted to make the point that there
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was nothing special, nothing about that basis that another basis couldn't have. It could
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have linear independence. It could span a space. There's lots of other bases. So I started
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with these vectors, one, one, two, and two, two, five. And those were independent. And
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then I said three, three, seven wouldn't do because three, three, seven is the sum of
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those. So in my innocence, I put in three, three, eight. I figured probably if three,
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three, seven is on the plane, which I know it's in the plane with these two, then probably
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three, three, eight sticks a little bit out of the plane. And it's independent and it
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gives a basis. But after class, to my sorrow, student tells me, wait a minute, that third
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vector, three, three, eight is not independent. And why did she say that? She didn't actually
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take the time, didn't have to to find what combination of this one and this one gives
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three, three, eight. She did something else. In other words, she looked ahead because
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she said, wait a minute, if I look at that matrix, it's not invertible. That third column
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can't be independent of the first two because when I look at that matrix, it's got two identical
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rows. I have a square matrix. Its rows are obviously dependent. And that makes the columns
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dependent. So there's my error. When I look at the matrix A that has those three columns,
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those three columns can't be independent because that matrix is not invertible because
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it's got two equal rows. And today's lecture will reach the conclusion, the great conclusion
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that connects the column space with the row space. So those are the row spaces now going
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to be another one of my fundamental subspaces. The row space of this matrix or of this one,
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well the row space of this one is okay. But the row space of this one, I'm looking at
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the rows of the matrix. Oh, anyway, I'll have two equal rows. And the row space will be
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only two-dimensional. The rank of the matrix with these columns will only be two. So only
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two of those columns can be independent too. The rows tell me something about the columns
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in other words, something that I should have noticed and I didn't. Okay. So now let me
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pin down these four fundamental subspaces. So here are the four fundamental subspaces.
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This is really the heart of this approach to linear algebra to see these four subspaces
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how they're related. So what are they? The column space, C of A, the null space, N of A.
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And now comes the row space, something new. The row space, what's in that? It's all combinations
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of the rows. That's natural. We want a space so we have to take all combinations and we
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start with the rows. So the rows span the row space or the rows of bases for the row space,
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maybe so, maybe no. The rows are a basis for the row space when they're independent. But
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if they're dependent, as in this example, my error from last time, those three rows are
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not a basis. The row space would only be two dimensional. I only need two rows for a
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basis. So the row space, now what's in it? It's all combinations of the rows of A, all
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combinations of the rows of A. But I don't like working with row vectors. All my vectors
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have been column vectors. I'd like to stay with column vectors. How can I get to column
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vectors out of these rows? I transpose the matrix. So if that's okay with you, I'm going
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to transpose the matrix. I'm going to say all combinations of the columns of A transpose.
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And that allows me to use the convenient notation, the column space of A transpose. Nothing,
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no mathematics went on there. We just got some vectors that were lying down to stand
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up. But it means that we can use this column space of A transpose. That's telling me in
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a nice matrix notation what the row space is. Okay. And finally, is another null space.
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The fourth fundamental space will be the null space of A transpose. The fourth guy is
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the null space of A transpose. And of course, my notation is N of A transpose. That's the
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null space of A transpose. We don't have a perfect name for this space as connecting
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with A. But our usual name is the left null space. And I'll show you why in a moment. So
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often I call this the, just to write that word, the left null space of A. So just the way
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we have the row space of A, and we switch it to the column space of A transpose. So we
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have this space of guys that I call the left null space of A. But the good notation is
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it's the null space of A transpose. Okay. Those are four spaces. Where are those spaces?
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What big space are they in for when A is M by N? In that case, the null space of A, what's
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in the null space of A? Vectors within components. Solutions to A x equals zero. So the null space
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of A is N R N. What's in the column space of A? Well, columns. How many components do
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those columns have? M. So this column space is N R N. What about the column space of A
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transpose, which are just a disguised way of saying the rows of A? The rows of A in
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this three by six matrix have six components. N components. The column space is N R N. And
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the null space of A transpose, I see that this fourth space is already getting second,
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you know, second class citizen treatment and it doesn't deserve it. It should be there.
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It is there and shouldn't be squeezed. The null space of A transpose, well, if the null
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space of A had vectors within components, the null space of A transpose will be in R N.
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I want to draw a picture of the four spaces. Okay. Here are the four spaces. Okay, let
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me put N dimensional space over on this side. Then which were the subspaces in R N? The
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null space was and the row space was. So here we have the, can I make that picture of
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the row space? And can I make this kind of picture of the null space? That's just meant
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to be a sketch. To remind you that they're in this, which, you know, what type of vectors
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are in it? Vectors with N components. Over here, inside, consisting of vectors with
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N components is the column space and what I'm calling the null space of A transpose. Those
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are the ones with N components. Okay. To understand these spaces is our job now. Because
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by understanding those spaces, we know everything about this half of linear algebra. What do I
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mean by understanding those spaces? I would like to know a basis for those spaces. For each
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one of those spaces, how would I create, construct a basis? What systematic way would produce
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a basis? And what's their dimension? Okay. So for each of the four spaces, I have to answer
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those questions. How do I produce a basis? And then, which has a somewhat long answer?
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And what's the dimension? Which is just a number. So it has a real short answer. Can I
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give you the short answer first? I shouldn't do it, but here it is. I can tell you the
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dimension of the column space. Let me start with this guy. What's its dimension? I have
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an m by n matrix. The dimension of the column space is the rank r. We actually got to that
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at the end of the last lecture, but only for an example. So I really have to say, okay,
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what's going on there? I should produce a basis, and then I just look to see how many
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vectors I needed in that basis, and the answer will be r. Actually, I'll do that before
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I get on to the others. What's a basis for the column space? We've done all the work of
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row reduction, identifying the pivot columns, the ones that end up with pivots. But now,
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the pivot columns I'm interested in are columns of a, the original a. And those pivot columns,
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there are r of them, the rank r counts those. Those are a basis. So if I answer this question
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for the column space, the answer will be a basis is the pivot columns, and the dimension
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is the rank r. And there are pivot columns and everything great. Okay, so that space,
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we pretty well understand. I probably have a little going back to see that to prove
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that this is a right answer, but you know it's the right answer. Now, let me look at the
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row space. Okay, shall I tell you the dimension of the row space? Yes. Before we do even an
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example, let me tell you the dimension of the row space. Its dimension is also r. The
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row space and the column space have the same dimension. That's a wonderful fact. The
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dimension of the column space of a transpose, that's the row space, is r. That space is
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r-dimensional. And so is this one. Okay. That's the sort of insight that got used in
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this example. If those are the three columns of a matrix, let me make them the three columns
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of a matrix by just erasing some brackets. Okay, those are the three columns of a matrix.
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The rank of that matrix, if I look at the columns, it wasn't obvious to me anyway. But if
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I look at the rows, it's now it's obvious. The row space of that matrix obviously is
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two-dimensional because I see a basis for the row space, this row and that row. And of
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course, strictly speaking, I'm supposed to transpose those guys make them stand up.
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But the rank is two. And therefore, the column space is two-dimensional by this wonderful
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fact that the row space and column space have the same dimension. And therefore, there
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are only two pivot columns, not three. And the three columns are dependent. Okay. Now,
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let me bury that error and talk about the row space. Well, I'm going to give you the
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dimensions of all the spaces because that's such a nice answer. Okay. So let me come back
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here. So we have this great fact to establish that the row space, its dimension is also the
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rank. Okay. What about the null space? What's a basis for the null space? What's the dimension
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of the null space? Let me put that answer up here for the null space. Well, how have
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we constructed the null space? We took the matrix A, we did those row operations to get
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it into a form U or even further, we got it into the reduced form R, and then we read
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off special solutions, special solutions. And every special solution came from a free
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variable. And those special solutions are in the null space and the great thing is there
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are a basis for it. So for the null space, a basis will be the special solutions. And there's
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one for every free variable, right? For each free variable, we give that variable the value
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one, the other free variable zero, we get the pivot variables, we get a vector in the,
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we get a special solution. So we get all together n minus R of them because that's the number
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of free variables. If we have R, this is, this is the dimension is R is the number of pivot
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variables, this is the number of free variables. So the beauty is that those special solutions
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do form a basis and tell us immediately that the dimension of the null space is n, I'd
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better write this well because it's so nice, n minus R. And you see the nice thing that
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the two dimensions in this n dimensional space, one subspace is R dimensional to be proved,
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that's the row space. The other subspace is n minus R dimensional, that's the null space.
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And the two dimensions like together give n, the sum of R and n minus R is n. And that's
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just great. It's really copying the fact that we have n variables, R of them are pivot
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variables and n minus R are free variables and n all together. Okay. And now what's the
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dimension of this poor, misbegotten, fourth subspace? It's got to be m minus R. The dimension
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of this left null space, left out practically, is m minus R. Well, that's really just saying
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that this again the sum of that, plus that is m. And m is correct. It's the number of
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columns in a transpose. A transpose is just as good a matrix as a. It just happens to
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be m by m. It happens to have m columns. So it will have m variables when I go to ax equals
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zero and m of them and R of them will be pivot variables and m minus R will be free variables.
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It, a transpose is as good a matrix as a. It follows the same rule that the, this plus,
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the dimension, this dimension plus this dimension adds up to the number of columns. And over
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here, a transpose has m columns. Okay. Okay. So I gave you the easy answer, the dimensions.
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Now can I go back to check on a basis? We would like to think that, say the row space.
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Because we've got a basis for the column space. The pivot columns give a basis for the column
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space. Now I'm asking you to look at the row space. And I, you could say, okay, I can
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produce a basis for the row space by transposing my matrix, making those columns, then doing
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elimination, row reduction, and checking out the pivot columns in this transposed matrix.
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But that means you had to do all that row reduction on a transpose. It ought to be possible
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if we take a matrix A, let me take the matrix. Maybe we had this matrix in the last lecture.
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1, 1, 1, 2, 1, 2, 3, 2, 3, 1, 1, 1. Okay. That matrix was so easy. We spotted its pivot
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columns 1 and 2 without actually doing row reduction. But now let's do the job properly. So I
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subtract this away from this to produce a zero. So 1, 2, 3, 1 is fine. Subtracting that away
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leaves me minus 1 minus 1, 0, right? And subtracting that from the last row, oh, well, that's easy.
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Okay, I'm doing row reduction. Now I, the first column is all set. The second column I
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now see the pivot. And I can clean up if I actually, okay, why don't I make the pivot
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into a 1? I'll multiply that row through by minus 1, and then I have 1, 1. That was
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an elementary operation. I'm allowed multiply a row by a number. And now I'll do elimination.
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2 of those away from that will knock this guy out and make this into a 1. So that's now
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a zero, and that's a 1. Okay, done. That's our. I'm seeing the identity matrix here. I'm
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seeing zeros below, and I'm seeing f there. Okay. What about its row space? What happened
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to its row space? Well, what happened for me first asked? Just because this is sometimes
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something does happen. Its column space changed. The column space of r is not the column space
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of a, right? Because 1, 1, 1 is certainly in the column space of a, and certainly not
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in the column space of r. I did row operations. Those row operations preserved the row space.
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So the rows, so the column spaces are different, different column spaces, different column
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spaces. But I believe that they have the same row space. Same row space. I believe that
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the row space of that matrix and the row space of this matrix are identical. They have exactly
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the same vectors in them. Those vectors are vectors with four components, right? They're
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all combinations of those rows. Or I believe you get the same thing by taking all combinations
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of these rows. And if true, what's a basis? What's a basis for the row space of r? And
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it will be a basis for the row space of the original a. But it's obviously a basis for the
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row space of r. What's a basis for the row space of that matrix? The first two rows.
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So a basis for the row space of a or of r is the first r rows of r. Not a way. Sometimes
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it's true for a, but not necessarily. But r, we definitely have a matrix here whose
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row space we can identify. The row space is spanned by the three rows. But if we want a basis,
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we want independence. So out goes row three. The base, the row space is also spanned by
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the first two rows. This guy didn't contribute anything. And of course over here this one,
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two, three, one in the bottom didn't contribute anything. We had it already. So this, here's
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a basis. One, zero, one, one, and zero, one, one, zero. I believe those are in the row space.
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I know they're independent. Why are they in the row space? Why are those two vectors in
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the row space? Because all those operations we did, which started with these rows and
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took combinations of them, I took this row minus this row. That gave me something that's
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still in the row space. That's the point. When I took a row minus a multiple of another
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row, I'm staying in the row space. The row space is not changing. My little basis for
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it is changing. And I've ended up with sort of the best basis. If the columns of the identity
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matrix are the best basis for R3 or Rn, the rows of this matrix are the best basis for the
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row space. Best in the sense of being as clean as I can make it. Starting off with the
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identity and then finishing up with whatever has to be in there. Do you see then that the
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dimension is R? For sure. Because we've got R pivots, R non-zero rows. We've got the
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right number of vectors. R. They're in the row space. They're independent. That's it.
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They are a basis for the row space. And we can even pin that down further. How do I know
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that every row of A is a combination? How do I know they span the row space? Well, somebody
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says I've got the right number of them so they must. That's true. But let me just say, how
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do I know that this row is a combination of these by just reversing the steps of row
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reduction? If I just reverse the steps and go from A from R back to A, then what am I
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doing? I'm starting with these rows. I'm taking combinations of them after a couple of
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steps undoing the subtractions that I did before. I'm back to these rows. So these rows are
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combinations of those rows. Those rows are combinations of those rows. The two rows
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are the same. The bases are the same. And the natural basis is this guy. Is that all right
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for the row space? The row space is sitting there in R in its cleanest possible form.
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Now what about the fourth guy, the null space of A transpose? First of all, why do I call
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that the left null space? Let me say that and bring that down. So the fourth space is
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the null space of A transpose. It has in it vectors. Let me call them Y so that A transpose
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Y equals zero. If A transpose Y equals zero, then Y is in the null space of A transpose,
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of course. So this is matrix times a column equaling zero. And now, because I want Y to
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sit on the left and I want A instead of A transpose, I'll just transpose that equation.
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Can I just transpose that? On the right, it makes the zero vector lie down. And on the
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left, it's a product, A transpose times Y. If I take the transpose, then they come in opposite
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order, right? So it's Y transpose times A transpose, but nobody's going to leave it like
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that. That's A transpose transpose is just A, of course. When I transpose A transpose,
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I get back to A. Now do you see what I have now? I have a row vector Y transpose multiplying
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A and multiplying from the left. That's why I call it the left null space. But by putting
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it on the left, I had to make it into a row instead of a column vector. And so my convention
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is I usually don't do that. I usually stay with A transpose Y equals zero. And you might
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ask, how do we get a basis, or I might ask, how do we get a basis for this fourth space,
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this left null space? Okay, I'll do it the example. As always, ah, not that one. The left
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null space is not jumping out at me here. The, I know which are the free variables, the
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special solutions, but those are special solutions to A x equals zero. And now I'm looking
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at A transpose and I'm not seeing it here. So, but somehow you feel that the work that
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you did which simplified A to R should have revealed the left null space two. And it's
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slightly less immediate, but it's there. So, from A to R, I took some steps and I guess
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I'm interested in what were those steps, or what were all of them together. I don't
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, I'm not interested in what particular ones they were. I'm interested in what was the
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whole matrix that took me from A to R. How would you find that? Do you remember Gouste
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Jordan, where you tack on the identity matrix? Let's do that again. So, I'll do it above
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here. So, this is now, this is now the idea of, I take the matrix A, which is m by n. In
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Gouste Jordan, when we saw him before, A was a square invertible matrix and we were finding
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its inverse. Now the matrix isn't square, it's probably rectangular. But I'll still tack
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on the identity matrix and of course, since these have length m, it better be m by m. And
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now I'll do the reduced row echelon form of this matrix. And what do I get? The reduced
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row echelon form starts with these columns, starts with the first columns, works like
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mad, and produces R. Of course, still at the same size m by m. And we did it before. And
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then, whatever it did to get R, something else is going to show up here. Let me call
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it E, m by m. It's whatever, you see that E is just going to contain a record of what
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we did. We did whatever it took to get A to become R. And at the same time, we were doing
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it to the identity matrix. So we started with the identity matrix, we buzzed along. So
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we took some, all this row reduction amounted to multiplying on the left by some matrix,
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some series of elementary matrices that all together gave us one matrix. And that matrix
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is E. So all this row reduction stuff amounted to multiplying by E. How do I know that?
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It certainly amounted to multiply it by something. And that something took I to E. So that's
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something was E. So now look at the first part, E, A is R. No big deal. All I've said
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is that the row reduction steps that we all know well, taking A to R, are in some matrix.
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And I can find out what that matrix is by just tacking I on and seeing what comes out.
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What comes out is E. Let's just review the invertible square case. What happened then?
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Because I was interested in chapter 2 also. When A was square in invertible, I took A,
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I, I did row elimination. What was the R that came out? It was I. So in chapter
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2, in chapter 2, R was I. The reduced row echelon form of a nice invertible square matrix
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is the identity. So if R was I, in that case, then E was then E was A inverse because E
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A is I. Good. That was good and easy. Now what I'm saying is that there still is an E.
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It's not A inverse anymore because A is a rectangle that hasn't got an inverse. But
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there is still some matrix E that connected this to this. Oh, I should have figured out
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in advance what it was. Shoot. I didn't, I did those steps in sort of a race as I went.
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And I should have done them to the identity too. Can I do that? Can I do that? I'll keep
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the identity matrix like I'm supposed to do. And I'll do the same operations on it and
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see what I end up with. So I'm starting with the identity which I'll write in light.
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Light enough. But what did I do? I subtracted that row from that one and that row from that
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one. I'll do that to the identity. So I subtract that first row from row 2 and row 3. Good.
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And I think I multiply. Do you remember I multiplied row 2 by minus 1? Let me just do that.
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Then what did I do? I subtracted 2 of row 2 away from row 1. I better do that. Subtract
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2 of this away from this. That's minus 1. 2 of these away leaves a plus 2 and 0. I
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believe that's E. The way to check is to see. Multiply that E by this A. Just to see.
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Did I do it right? So I believe E was minus 1, 2, 0, 1 minus 1, 0 and minus 1, 0, 1. Okay.
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That's my E. That's my A and that's R. All right. All I'm struggling to do is write the
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reason I wanted this blasted E was so that I could figure out the left null space. Not
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only its dimension, which I know, actually what is the dimension of the left null space?
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Here's my matrix. What's the rank of the matrix? 2. And the dimension of the left null
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space is supposed to be M minus R, 3 minus 2, 1. I believe that the left null space is one
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dimensional. There is one combination of those three rows that produces the zero row.
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There is a basis for the left null space. It's only got one vector in it. And what is that
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vector? It's here in the last row of E. But I could have seen it earlier. What combination
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of those rows gives the zero row minus one of that plus one of that? So a basis for the
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left null space of this matrix, I'm looking for combinations of rows that give the zero
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row. If I'm looking at the left null space. For the null space, I'm looking at combinations
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of columns to get the zero column. Now I'm looking at combinations of these three rows
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to get the zero row. And of course, there is my zero row. And here is my vector that produced
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it. Minus one of that row and one of that row. Obvious. Okay. So in that example, and actually
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in all examples, we have seen how to produce a basis for the left null space. I won't
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ask you that all the time. Because it didn't come out immediately from R, we had to keep
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track of E for that left null space. But at least it didn't require us to transpose
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the matrix and start all over again. Okay. Those are the four subspaces. Can I review
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them? The row space and the null space are in R n. Their dimensions add to n. The column
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space and the left null space are in R m. And their dimensions add to n. Okay. So let me
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close these last minutes by pushing you a little bit more to a new type of vector space.
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All our vector spaces, all the ones that we took seriously have been subspaces of some
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real three or n dimensional space. Now I'm going to write down another vector space, a
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new vector space. Say all three by three matrices. My matrices are the vectors. Is that all
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right? I'm just naming them. You can put quotes around vectors. Every three by three matrix
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is one of my vectors. Now how am I entitled to call those things vectors? They look very
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much like matrices. But they're vectors in my vector space because they obey the rules.
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All I'm supposed to be able to do with vectors is add them. I can add matrices. I'm supposed
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to be able to multiply them by scalar numbers, like seven. Well I can multiply a matrix
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by seven. And I can take combinations of matrices. I can take three of one matrix minus five
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of another matrix. And those combinations, there's a zero matrix. The matrix that has all
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zeroes in it. If I add that to another matrix it doesn't change it. All the good stuff.
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If I multiply a matrix by one it doesn't change it. All those eight rules for a vector space
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that we never wrote down all easily satisfied. Now we have a different... Now of course you
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can say you can multiply those matrices. I don't care. For the moment I'm only thinking
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of these matrices as forming a vector space. So I'm only doing a plus b and c times a.
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I'm not interested in a b for now. The fact that I can multiply is not relevant to a vector
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space. Okay, so I have three by three matrices. And how about subspaces? What's... tell me a
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subspace of this matrix space. Let me call this matrix space M. That's my matrix space.
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My space of all three by three matrices. Tell me a subspace of it. Okay. What about the
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upper triangular matrices? So subspaces, subspaces of M. All upper triangular matrices. Another
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subspace. All symmetric matrices. The intersection of two subspaces is supposed to be a subspace.
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We gave a little effort to the proof of that fact. If I look at the matrices that are in
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this subspace, there's a metric. And they're also in this subspace, they're upper triangular.
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What do they look like? Well, if they're symmetric, but they have zeros below the diagonal,
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they better have zeros above the diagonal. So the intersection would be diagonal matrices.
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That's another subspace. Smaller than those. How can I use the word smaller? Well, I'm
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now entitled to use the word smaller. I mean, well, one way to say is, okay, these are contained
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in those. These are contained in those. But more precisely, I could give the dimension
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of these spaces. So I could... we can compute... let's compute it next time. The dimension
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of all upper... of the subspace of upper triangular three by three matrices. The dimension of
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symmetric three by three matrices. The dimension of diagonal three by three matrices.
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Well, to produce dimension, that means I'm supposed to produce a basis. And then I just
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count how many I needed in the basis. Let me give you the answer for this one. What's
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the dimension? The dimension of this... say this subspace, let me call it D, all diagonal
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matrices. The dimension of this subspace is, as I write, you're working it out, three.
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Because here's a matrix in this... it's a diagonal matrix. Here's another one. Here's
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another one. Better make a diagonal. Let me put a seven there. That was not a very great
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choice, but it's three diagonal matrices. And I believe that they're a basis. I believe
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that those three matrices are independent. And I believe that any diagonal matrix is a
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combination of those three. So they span the subspace of diagonal matrices. Do you see
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that idea? It's like stretching the idea from R n to R n by n, three by three. But we can
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still add... we can still multiply by numbers, and we just ignore the fact that we can multiply
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two matrices together. Okay, thank you. That's lecture ten.