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  • Hi. This is the first lecture in MIT's course 18.06,

  • linear algebra, and I'm Gilbert Strang.

  • The text for the course is this book, Introduction to Linear Algebra.

  • And the course web page, which has got a lot of exercises

  • from the past, MatLab codes, the syllabus for the

  • course, is web.mit.edu/18.06. And this is the first

  • lecture, lecture one. So, and later we'll give the web

  • address for viewing these, videotapes. Okay, so what's in the

  • first lecture? This is my plan.

  • The fundamental problem of linear algebra, which is to solve a system

  • of linear equations. So let's start with a case when we

  • have some number of equations, say n equations and n unknowns.

  • So an equal number of equations and unknowns.

  • That's the normal, nice case. And what I want to do is

  • -- with examples, of course -- to describe,

  • first, what I call the Row picture. That's the picture of one equation

  • at a time. It's the picture you've seen before in two by two equations

  • where lines meet. So in a minute,

  • you'll see lines meeting. The second picture, I'll put a star

  • beside that, because that's such an important one.

  • And maybe new to you is the picture -- a column at a time.

  • And those are the rows and columns of a matrix.

  • So the third -- the algebra way to look at the

  • problem is the matrix form and using a matrix that I'll call A.

  • Okay, so can I do an example? The whole semester will be examples

  • and then see what's going on with the example. So,

  • take an example. Two equations, two unknowns.

  • So let me take 2x -y =0, let's say. And -x +2y=3.

  • Okay. let me -- I can even say right away -- what's the matrix,

  • that is, what's the coefficient matrix? The matrix that involves

  • these numbers -- a matrix is just a rectangular array

  • of numbers. Here it's two rows and two columns, so 2 and --

  • minus 1 in the first row minus 1 and 2 in the second row,

  • that's the matrix. And the right-hand -- the,

  • unknown -- well, we've got two unknowns.

  • So we've got a vector, with two components, x and x,

  • and we've got two right-hand sides that go into a vector 0 3.

  • I couldn't resist writing the matrix form, right -- even

  • before the pictures. So I always will think of this as

  • the matrix A, the matrix of coefficients, then there's a vector

  • of unknowns. Here we've only got two unknowns. Later we'll have any

  • number of unknowns. And that vector of unknowns,

  • well I'll often -- I'll make that x -- extra bold.

  • A and the right-hand side is also a vector that I'll always call b.

  • So linear equations are A x equal b and the idea now is to solve this

  • particular example and then step back to see the bigger picture.

  • Okay, what's the picture for this example, the Row picture?

  • Okay, so here comes the Row picture. So that means I take one row at a

  • time and I'm drawing here the xy plane and I'm going to plot all the

  • points that satisfy that first equation. So I'm looking at all the

  • points that satisfy 2x-y =0. It's often good to start with which

  • point on the horizontal line -- on this horizontal line,

  • y is zero. The x axis has y as zero and that -- in this case,

  • actually, then x is zero. So the point, the origin --

  • the point with coordinates (0, ) is on the line. It solves that

  • equation. Okay, tell me in -- well, I guess I have to tell you

  • another point that solves this same equation. Let me suppose x is one,

  • so I'll take x to be one. Then y should be two,

  • right? So there's the point one two that also solves this equation.

  • And I could put in more points. But, but let me put in all the

  • points at once, because they all lie on a straight

  • line. This is a linear equation and that word linear got the letters for

  • line in it. That's the equation -- this is the line that ... of

  • solutions to 2x-y=0 my first row, first equation.

  • So typically, maybe, x equal a half, y equal one will

  • work. And sure enough it does. Okay, that's the first one. Now

  • the second one is not going to go through the origin. It's

  • always important. Do we go through the origin or not?

  • In this case, yes, because there's a zero over there.

  • In this case we don't go through the origin, because if x and y are

  • zero, we don't get three. So, let me again say suppose y is

  • zero, what x do we actually get? If y is zero, then I get x is minus

  • three. So if y is zero, I go along minus three. So there's

  • one point on this second line. Now let me say, well, suppose x is

  • minus one -- just to take another x. If x is minus one, then this is a

  • one and I think y should be a one, because if x is minus one,

  • then I think y should be a one and we'll get that point.

  • Is that right? If x is minus one, that's a one. If y is a one, that's

  • a two and the one and the two make three and that point's

  • on the equation. Okay. Now, I should just draw the

  • line, right, connecting those two points at -- that will give me the

  • whole line. And if I've done this reasonably well,

  • I think it's going to happen to go through -- well,

  • not happen -- it was arranged to go through that point.

  • So I think that the second line is this one, and this is the

  • all-important point that lies on both lines. Shall we just check

  • that that point which is the point x equal one and y was two,

  • right? That's the point there and that, I believe, solves

  • both equations. Let's just check this.

  • If x is one, I have a minus one plus four equals three,

  • okay. Apologies for drawing this picture that you've seen before.

  • But this -- seeing the row picture -- first of all, for n equal 2,

  • two equations and two unknowns, it's the right place to start. Okay.

  • So we've got the solution. The point that lies on both lines.

  • Now can I come to the column picture? Pay attention,

  • this is the key point. So the column picture.

  • I'm now going to look at the columns of the matrix.

  • I'm going to look at this part and this part. I'm going to say that

  • the x part is really x times -- you see, I'm putting the two -- I'm

  • kind of getting the two equations at once --

  • that part and then I have a y and in the first equation it's multiplying

  • a minus one and in the second equation a two,

  • and on the right-hand side, zero and three. You see, the

  • columns of the matrix, the columns of A are here and the

  • right-hand side b is there. And now what is the equation asking

  • for? It's asking us to find -- somehow to combine that vector and

  • this one in the right amounts to get that one. It's asking us to find

  • the right linear combination -- this is called a linear combination.

  • And it's the most fundamental operation in the whole course.

  • It's a linear combination of the columns. That's what we're seeing

  • on the left side. Again, I don't want to write down a

  • big definition. You can see what it is.

  • There's column one, there's column two.

  • I multiply by some numbers and I add. That's a combination --

  • a linear combination and I want to make those numbers the right numbers

  • to produce zero three. Okay. Now I want to draw a picture

  • that, represents what this -- this is algebra.

  • What's the geometry, what's the picture that goes with it?

  • Okay. So again, these vectors have two components,

  • so I better draw a picture like that. So can I put down these columns?

  • I'll draw these columns as they are, and then I'll do a combination of

  • them. So the first column is over two and down one, right?

  • So there's the first column. The first column. Column one.

  • It's the vector two minus one. The second column is --

  • minus one is the first component and up two. It's here.

  • There's column two. So this, again, you see what its

  • components are. Its components are minus one,

  • two. Good. That's this guy. Now I have to take a combination.

  • What combination shall I take? Why not the right combination, what

  • the hell? Okay. So the combination I'm going

  • to take is the right one to produce zero three and then we'll see it

  • happen in the picture. So the right combination is to take

  • x as one of those and two of these. It's because we already know that

  • that's the right x and y, so why not take the correct

  • combination here and see it happen? Okay, so how do I picture this

  • linear combination? So I start with this vector that's

  • already here -- so that's one of column one,

  • that's one times column one, right there. And now I want to add on --

  • so I'm going to hook the next vector onto the front of the arrow will

  • start the next vector and it will go this way. So let's see,

  • can I do it right? If I added on one of these vectors,

  • it would go left one and up two, so we'd go left one and up two, so

  • it would probably get us to there. Maybe I'll do dotted line for that.

  • Okay? That's one of column two tucked onto the end,

  • but I wanted to tuck on two of column two. So that --

  • the second one -- we'll go up left one and up two also.

  • It'll probably end there. And there's another one. So what

  • I've put in here is two of column two.

  • Added on. And where did I end up? What are the coordinates of this

  • result? What do I get when I take one of this plus two of that?

  • I do get that, of course. There it is, x is zero, y is three, that's b.

  • That's the answer we wanted. And how do I do it? You see I do

  • it just like the first component. I have a two and a minus two that

  • produces a zero, and in the second component I have a

  • minus one and a four, they combine to give the three.

  • But look at this picture. So here's our key picture.

  • I combine this column and this column to get this guy.

  • That was the b. That's the zero three. Okay. So that idea of

  • linear combination is crucial, and also -- do we want to think

  • about this question? Sure, why not. What are all the combinations?

  • If I took -- can I go back to xs and ys? This is a question for

  • really -- it's going to come up over and over, but why don't we see it

  • once now? If I took all the xs and all the ys, all the combinations,

  • what would be all the results? And, actually,

  • the result would be that I could get any right-hand side at all.

  • The combinations of this and this would fill the whole plane.

  • You can tuck that away. We'll, explore it further. But this idea

  • of what linear combination gives b and what do all the linear

  • combinations give, what are all the possible,

  • achievable right-hand sides be -- that's going to be basic. Okay.

  • Can I move to three equations and three unknowns?

  • Because it's easy to picture the two by two case.

  • Let me do a three by three example. Okay, I'll sort of start it the

  • same way, say maybe 2x-y and maybe I'll take

  • no zs as a zero and maybe a -x+2y and maybe a -z as a --

  • oh, let me make that a minus one and, just for variety let me take,

  • -3z, -3ys, I should keep the ys in that line, and 4zs is, say, 4.

  • Okay. That's three equations. I'm in three dimensions, x, y,

  • z. And, I don't have a solution yet. So I want to understand the

  • equations and then solve them. Okay. So how do I you understand

  • them? The row picture one way. The column picture is another very

  • important way. Just let's remember the matrix form,

  • here, because that's easy. The matrix form -- what's our matrix A?

  • Our matrix A is this right-hand side, the two and the minus one and

  • the zero from the first row, the minus one and the two and the

  • minus one from the second row, the zero, the minus three and the

  • four from the third row. So it's a three by three matrix.

  • Three equations, three unknowns. And what's our right-hand side?

  • Of course, it's the vector, zero minus one, four.

  • Okay. So that's the way, well, that's the short-hand to write

  • out the three equations. But it's the picture that I'm

  • looking for today. Okay, so the row picture.

  • All right, so I'm in three dimensions, x,

  • y and z. And I want to take those equations one at a time and ask --

  • and make a picture of all the points that satisfy --

  • let's take equation number two. If I make a picture of all the

  • points that satisfy -- all the x, y, z points that solve

  • this equation -- well, first of all,

  • the origin is not one of them. x, y, z -- it being 0, 0, 0 would

  • not solve that equation. So what are some points that do

  • solve the equation? Let's see, maybe if x is one,

  • y and z could be zero. That would work, right? So there's one point.

  • I'm looking at this second equation, here, just, to start

  • with. Let's see. Also, I guess,

  • if z could be one, x and y could be zero,

  • so that would just go straight up that axis. And,

  • probably I'd want a third point here. Let me take x to be zero,

  • z to be zero, then y would be minus a half, right?

  • So there's a third point, somewhere -- oh my -- okay.

  • Let's see. I want to put in all the points that satisfy that

  • equation. Do you know what that bunch of points will be?

  • It's a plane. If we have a linear equation, then,

  • fortunately, the graph of the thing, the plot of all the points that

  • solve it are a plane. These three points determine a

  • plane, but your lecturer is not Rembrandt and the art is going to be

  • the weak point here. So I'm just going to draw a plane,

  • right? There's a plane somewhere.

  • That's my plane. That plane is all the points that solves this guy.

  • Then, what about this one? Two x minus y plus zero z.

  • So z actually can be anything. Again, it's going to be another

  • plane. Each row in a three by three problem

  • gives us a plane in three dimensions. So this one is going to be some

  • other plane -- maybe I'll try to draw it like this.

  • And those two planes meet in a line. So if I have two equations,

  • just the first two equations in three dimensions,

  • those give me a line. The line where those two planes

  • meet. And now, the third guy is a third plane.

  • And it goes somewhere. Okay, those three things meet in a point.

  • Now I don't know where that point is, frankly. But --

  • linear algebra will find it. The main point is that the three

  • planes, because they're not parallel, they're not special.

  • They do meet in one point and that's the solution.

  • But, maybe you can see that this row picture is getting a little hard to

  • see. The row picture was a cinch when we looked at two lines meeting.

  • When we look at three planes meeting, it's not so clear and in

  • four dimensions probably a little less clear.

  • So, can I quit on the row picture? Or quit on the row picture before

  • I've successfully found the point where the three planes meet?

  • All I really want to see is that the row picture consists of three

  • planes and, if everything works right,

  • three planes meet in one point and that's a solution.

  • Now, you can tell I prefer the column picture.

  • Okay, so let me take the column picture. That's x times --

  • so there were two xs in the first equation minus one x is,

  • and no xs in the third. It's just the first column of that.

  • And how many ys are there? There's minus one in the first equations,

  • two in the second and maybe minus three in the third.

  • Just the second column of my matrix. And z times no zs minus one zs and

  • four zs. And it's those three columns,

  • right, that I have to combine to produce the right-hand side,

  • which is zero minus one four. Okay. So what have we got on this

  • left-hand side? A linear combination.

  • It's a linear combination now of three vectors,

  • and they happen to be -- each one is a three dimensional

  • vector, so we want to know what combination

  • of those three vectors produces that one. Shall I try to draw the column

  • picture, then? So, since these vectors have three

  • components -- so it's some multiple -- let me draw in the first

  • column as before -- x is two and y is minus one.

  • Maybe there is the first column. y -- the second column has maybe a

  • minus one and a two and the y is a minus three, somewhere,

  • there possibly, column two. And the third column has --

  • no zero minus one four, so how shall I draw that?

  • So this was the first component. The second component was a minus

  • one. Maybe up here. That's column three,

  • that's the column zero minus one and four.

  • This guy. So, again, what's my problem?

  • What this equation is asking me to do is to combine these three vectors

  • with a right combination to produce this one. Well,

  • you can see what the right combination is, because in

  • this special problem, specially chosen by the lecturer,

  • that right-hand side that I'm trying to get is actually one of these

  • columns. So I know how to get that one. So what's the solution?

  • What combination will work? I just want one of these and none of these.

  • So x should be zero, y should be zero and z should be one.

  • That's the combination. One of those is obviously the right

  • one. Column three is actually the same as b in this particular problem.

  • I made it work that way just so we would get an answer,

  • (0,0,1), so somehow that's the point where those three planes met and I

  • couldn't see it before. Of course, I won't always be able

  • to see it from the column picture, either. It's the next lecture,

  • actually, which is about elimination,

  • which is the systematic way that everybody -- every bit of software,

  • too -- production, large-scale software would solve the equations.

  • So the lecture that's coming up. If I was to add that to the syllabus,

  • will be about how to find x, y, z in all cases.

  • Can I just think again, though, about the big picture?

  • By the big picture I mean let's keep this same matrix on the left

  • but imagine that we have a different right-hand side.

  • Oh, let me take a different right-hand side.

  • So I'll change that right-hand side to something that actually is also

  • pretty special. Let me change it to --

  • if I add those first two columns, that would give me a one and a one

  • and a minus three. There's a very special right-hand

  • side. I just cooked it up by adding this one to this one.

  • Now, what's the solution with this new right-hand side?

  • The solution with this new right-hand side is clear.

  • took one of these and none of those. So actually, it just changed around

  • to this when I took this new right-hand side.

  • Okay. So in the row picture, I have three different planes,

  • three new planes meeting now at this point. In the column picture,

  • I have the same three columns, but now I'm combining them

  • to produce this guy, and it turned out that column one

  • plus column two which would be somewhere -- there is the right

  • column -- one of this and one of this would give me the new b.

  • Okay. So we squeezed in an extra example. But now think about all bs,

  • all right-hand sides. Can I solve these equations for

  • every right-hand side? Can I ask that question?

  • So that's the algebra question. Can I solve A x=b for every b? Let

  • me write that down. Can I solve A x =b for every

  • right-hand side b? I mean, is there a solution?

  • And then, if there is, elimination will give me a way to find it.

  • I really wanted to ask, is there a solution for every right-hand side?

  • So now, can I put that in different words -- in this linear

  • combination words? So in linear combination words,

  • do the linear combinations of the columns fill three dimensional space?

  • Every b means all the bs in three dimensional space.

  • Do you see that I'm just asking the same question in different words?

  • Solving A x -- A x -- that's very important.

  • A times x -- when I multiply a matrix by a vector,

  • I get a combination of the columns. I'll write that down in a moment.

  • But in my column picture, that's really what I'm doing.

  • I'm taking linear combinations of these three columns and I'm trying

  • to find b. And, actually, the answer for this matrix

  • will be yes. For this matrix A -- for these columns, the answer is yes.

  • This matrix -- that I chose for an example is a good matrix.

  • A non-singular matrix. An invertible matrix.

  • Those will be the matrices that we like best. There could be other --

  • and we will see other matrices where the answer becomes,

  • no -- oh, actually, you can see when it would become no.

  • What could go wrong? How could it go wrong that out of these --

  • out of three columns and all their combinations --

  • when would I not be able to produce some b off here?

  • When could it go wrong? Do you see that the combinations --

  • let me say when it goes wrong. If these three columns all lie in

  • the same plane, then their combinations will lie in

  • that same plane. So then we're in trouble.

  • If the three columns of my matrix -- if those three vectors happen to lie

  • in the same plane -- for example, if column three is just the sum of

  • column one and column two, I would be in trouble. That would

  • be a matrix A where the answer would be no, because the combinations --

  • if column three is in the same plane as column one and two,

  • I don't get anything new from that. All the combinations are in the

  • plane and only right-hand sides b that I could get would be the ones

  • in that plane. So I could solve it for some

  • right-hand sides, when b is in the plane,

  • but most right-hand sides would be out of the plane and unreachable.

  • So that would be a singular case. The matrix would be not invertible.

  • There would not be a solution for every b. The answer would become no

  • for that. Okay. I don't know -- shall we take just

  • a little shot at thinking about nine dimensions? Imagine that we have

  • vectors with nine components. Well, it's going to be hard to

  • visualize those. I don't pretend to do it.

  • But somehow, pretend you do. Pretend we have -- if this was nine

  • equations and nine unknowns, then we would have nine columns,

  • and each one would be a vector in nine-dimensional space and we would

  • be looking at their linear combinations.

  • So we would be having the linear combinations of nine vectors in

  • nine-dimensional space, and we would be trying to find the

  • combination that hit the correct right-hand side b.

  • And we might also ask the question can we always do it?

  • Can we get every right-hand side b? And certainly it will depend on

  • those nine columns. Sometimes the answer will be yes --

  • if I picked a random matrix, it would be yes, actually.

  • If I used MatLab and just used the random command,

  • picked out a nine by nine matrix, I guarantee it would be good. It

  • would be non-singular, it would be invertible,

  • all beautiful. But if I choose those columns so

  • that they're not independent, so that the ninth column is the same

  • as the eighth column, then it contributes nothing new and

  • there would be right-hand sides b that I couldn't get.

  • Can you sort of think about nine vectors in nine-dimensional space an

  • take their combinations? That's really the central thought --

  • that you get kind of used to in linear algebra.

  • Even though you can't really visualize it, you sort of think you

  • can after a while. Those nine columns and all their

  • combinations may very well fill out the whole nine-dimensional space.

  • But if the ninth column happened to be the same as the eighth column and

  • gave nothing new, then probably what it would fill out

  • would be -- I hesitate even to say this --

  • it would be a sort of a plane -- an eight dimensional plane inside

  • nine-dimensional space. And it's those eight dimensional

  • planes inside nine-dimensional space that we have to work with eventually.

  • For now, let's stay with a nice case where the matrices work,

  • we can get every right-hand side b and here we see how to

  • do it with columns. Okay. There was one step which I

  • realized I was saying in words that I now want to write in letters.

  • Because I'm coming back to the matrix form of the equation,

  • so let me write it here. The matrix form of my equation,

  • of my system is some matrix A times some vector x equals some right-hand

  • side b. Okay. So this is a multiplication.

  • A times x. Matrix times vector, and I just want to say how do you

  • multiply a matrix by a vector? Okay, so I'm just going to create a

  • matrix -- let me take two five one three --

  • and let me take a vector x to be, say, 1and 2. How do I multiply a

  • matrix by a vector? But just think a little bit about

  • matrix notation and how to do that in multiplication.

  • So let me say how I multiply a matrix by a vector.

  • Actually, there are two ways to do it. Let me tell you my favorite way.

  • It's columns again. It's a column at a time.

  • For me, this matrix multiplication says I take one of that column and

  • two of that column and add. So this is the way I would think of

  • it is one of the first column and two of the second column and let's

  • just see what we get. So in the first component I'm

  • getting a two and a ten. I'm getting a twelve there.

  • In the second component I'm getting a one and a six, I'm

  • getting a seven. So that matrix times that vector is

  • twelve seven. Now, you could do that another way.

  • You could do it a row at a time. And you would get this twelve --

  • and actually I pretty much did it here -- this way.

  • Two -- I could take that row times my vector. This is the idea of a

  • dot product. This vector times this vector, two times one plus five

  • times two is the twelve. This vector times this vector --

  • one times one plus three times two is the seven.

  • So I can do it by rows, and in each row times my x is what

  • I'll later call a dot product. But I also like to see it by

  • columns. I see this as a linear combination of a column.

  • So here's my point. A times x is a combination of the

  • columns of A. That's how I hope you will think of A times x when we need

  • it. Right now we've got -- with small ones, we can always do it

  • in different ways, but later, think of it that way.

  • Okay. So that's the picture for a two by two system.

  • And if the right-hand side B happened to be twelve seven,

  • then of course the correct solution would be one two.

  • Okay. So let me come back next time to a systematic way,

  • using elimination, to find the solution, if there is one,

  • to a system of any size and find out -- because if elimination fails,

  • find out when there isn't a solution. Okay, thanks.

Hi. This is the first lecture in MIT's course 18.06,

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Lec 1 | 麻省理工學院 18.06 線性代數,2005年春季。 (Lec 1 | MIT 18.06 Linear Algebra, Spring 2005)

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