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 y, 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

(0,0) 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 Okay, thanks. 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, find out when there

isn’t a solution. 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? find out

— because if elimination fails, 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

Here to learn linear algebra for iit prep:p

미국이 않풀리는 루트는 닫아버리고 다른것은 열었다. 그래서 화웨이를 ….. 시스템을 그렇게 했다. 미국이 한국 국정원과 NSS에서 무엇을 하는지 다 안다!

like gilbert strang, and thank mit.

"the whole semester will be examples" I already love him, examples are so underrated….

Great !!!

joh xD i laughed my ass off at 20:00, totally get how this guy feels, he explained everything amazingly well for me but when he had 2 illustarte it, "your lecturer is not rembrandt" xD. good video thank you

Lol ! In India ,we use to study it in school.

Thank you so much Professor! By the way I wish the camera is not moving around.

Thanks professor. 🙂

Endless applications of linear algebra, some great lectures. Other lectures I found useful was the UCI Math for Econ lectures, which has linear algebra from the very basics and is great for total beginners

MIT 💓

Lol in India, we learn this in 12th grade.

راءع

RIP Sir. Gone too soon.

The principal at the very beginning of the video….. Is called "when the meme becomes aware of itself"…yeah…. Sorry 😅😅😂😂

wtf？

His teaching method is awesome. I understand every step of that

Truly Amazing! Thankyou M.I.T and Gilbert Sir!

nice

why it feels so uncomfortable with only right ear !

unbelievable. Gilbert Strang's courses online for free? Brave world it is!

GreatProfessor!