Lec 1 | MIT 18.06 Linear Algebra, Spring 2005

Lec 1 | MIT 18.06 Linear Algebra, Spring 2005


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

21 thoughts to “Lec 1 | MIT 18.06 Linear Algebra, Spring 2005”

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

  2. 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

  3. 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

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

Leave a Reply

Your email address will not be published. Required fields are marked *