The following content is
provided under a Creative Commons license. Your support will help
MIT OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. GILBERT STRANG: Shall we start? Let me just say, this is
a great adventure for me to be here all on my own,
teaching a course that involves learning from data. So it's an exciting subject,
and a lot of linear algebra goes into it. So it's a second course
on linear algebra. Can I just-- so there is a
Stellar site established, and that will be the
basic thing that we use. This is a public site-- math.mit.edu/learningfromdata. So a book is coming pretty
quickly as we speak, or after we speak, and that
site has the table of contents of the book, which would
give you an idea of what could be in the course. And I printed out a copy for
everybody just of that one page. This is probably the final-- first and last handout-- maybe-- with a
table of contents, which you'll see there. And also, you'll see there
the first two sections of the book, which is
what I'll talk about today and a little bit into Friday. So that's linear algebra, of
course, because the course begins with linear algebra-- actually, things that you
would know from 18.06, but this is a way to say this
is really important stuff. So that's what I'll do today. I'd like to start on the
linear algebra today. Here's a great fact
about the course. So we taught it last
year, several of us together, and we knew there
wouldn't be a final exam, but we imagined there might
be quizzes along the way. But then we couldn't think of
anything to put on the quizzes. So we canceled those. But you do we learn
a lot, nevertheless. And so I guess we base the
grades on the homeworks. So the homeworks will be
partly linear algebra questions and partly online, like
recognizing handwriting, stitching images together,
many other things. And I'll talk about
those as we go. Good. So that's the general picture. And I'll say more
about it today. And I could answer
any question about it. So we're getting videotaped. So if anybody's bashful,
sit in the far back. But it'll be fun. You may know the
videos for 18.06. So this is the
next step, 18.065. It's pretty exciting for me. So any question, or
shall I do a little math? Why not? And then I'll say a little
more about the course, just so you have an
idea of what's ahead. And it looks like this room is
exactly the right size to me, so I'm pleased. So what's the deal
in linear algebra? Forgive me if I
start with what I do the very first day
in 18.06, which is multiply a matrix by a vector. And then I'll graduate
to multiplying a matrix by a matrix. And you will say,
I know that stuff. But do you know
it the right way? Do you think of the
multiplication the right way? So let me tell you what I
believe to be the right way. So let me take a matrix, say,
2, 3, 5, 1, 1, 7, and 3, 4, 12. And I'll always call matrices A. So first step is just A
times x, A times a vector. So I multiply A by,
let's say, x1, x2, x3. And how do I look
at that answer? So the choices-- think
of the rows of the matrix or think of the columns. And if you think
of the rows, which is the standard way to multiply,
you would take the dot product. So the first way is dot
products of row dotted with x. 2x1 plus x2 plus 3x3. It gives you the answer
a component at a time. That's the low level way. The good way to see
it is vector-wise. See this as x1 multiplies
that first column, x2 times the second column,
1, 1, 7, and x3 times the third column, 3, 4, 12. Good. So it's a combination
of vectors, and of course, it
produces a vector. And here, we have a 3 by
3 matrix on our vectors are in R3. And most vectors will be in
R3 or Rn for this course. So that's the right answer. And of course, the
first component is 2x1 and 1x2 and 3x3. The same 2, 1, 3-- the same dot
product, it comes out right. But you see it all at
once instead of piecemeal. Part of the course-- I guess part of what
I hope to get across is thinking of a matrix
as a whole thing, not just a bunch of nine
or m times n numbers. But thinking of it as a thing. A matrix multiplies a vector
to give another vector. So when I say Ax, you
immediately think that-- you immediately think, OK,
Ax has a clear meaning, it's a combination
of the columns of A. So now let me take
that next step. And the next step is think
about all combinations of the columns of A.
We take the matrix A, and we take all x's and we
imagine all the outputs. And what I want to ask you
is, what does that look like? If I just take 1 vector
x, I get a vector output. It takes a vector to a vector. But now I take all x's-- all of vectors x in 3D and
I get all these answers, and I think of
them all together. So I've got a bunch of vectors-- infinitely many
vectors, actually. And the question is, if I plot
those infinitely many vectors, what do I have? And the beauty of linear algebra
is that questions like this-- you can answer them and
you intuitively see it. You certainly see it in 3D
and you have the right idea in 10 dimensions, even. Most of us don't see
too well in 10d-- in R10. But here we got three. So do you see what I'm saying? I'm taking all x's. So all Ax gives us a
big bunch of vectors. And that collection
of vectors is called the column space of A.
It's a space, in other words. That's the key word there,
the column space of A. And I'll just write it as C of
A when I need letters for it. So I'm going to ask you,
what does this column space look like? And that depends on the matrix. Sometimes that column space
would be the whole of R3. Sometimes it's a smaller
set in R3, a space. Do you know what
it is in this case? Do I get all of 3D
out of these guys by choosing all
x1 and x2 and x3? It seems like if it
was a random matrix, the answer would
be, surely, yes. If a random three
by three matrix is-- it's column space is going
to be all of our three, its columns are going
to be independent, its rows are going
to be independent, it's going to be invertible,
it's going to be great. But is this matrix-- what's up with that matrix? No, it's not. So what do I get
instead of all of R3? I get a plane, yeah. If you get that insight. So by taking x's-- everybody's with
me-- all x's here. So that means that
it fills in whatever. And, well, because
it's linear algebra, it's going to be likely all of
R3 or a plane, or even a line. Let me just bat over
here for a moment. Give me a matrix whose column
space would only be a line. All 1s? OK. Wow, that's a-- let me
liven it up a little. 3, 3, 3, 8, 8, 8. So I think that the
combinations of those columns are all on the same line. That says that the column
space is just a line. And I would say then
that the matrix-- so column space of
this A is a line. Another way I could say this
is that the rank of the matrix is 1. The rank is sort of the
dimension of the column space. Well, not sort,
that's what it is. The rank is the
dimension of the column-- everybody sees that
you get a line? Because any combination
x1 of that plus x2 of that plus x3 of that is
going to go along that line. Here's the first column,
here's the second column. They're all on a line. So I'll never get off that line. If I allow all x's,
I'll get the whole line. Now here, you said not a line. What was the column
space for this guy? Plane. Now, why isn't it all of R3? What do you see
that's special there-- because it is special-- that's making the column
space special, a plane instead of the whole thing? Yeah, what's up with
these three columns? AUDIENCE: The third
column is the sum of it. GILBERT STRANG: The third
column is the sum of these two. So the first column
is fine, 2, 3, 5. Second column is in a
different direction. And when I take combinations
of the first two columns, what will I get? Sorry to keep asking you
questions, but that's-- anyway, that's what I always do. So the combinations of those
first two columns are--? AUDIENCE: A plane. GILBERT STRANG: A
plane because they're in different-- everybody
sees that picture. We've got column one going
that way and column two going that way. And then if I take any
multiple of column one, I've got a whole line, any
multiple of column two. And then when I put
the two together, it fills in the plane. Yeah. Your intuition
says that's right. So this is a matrix of rank. What's the rank of this matrix? AUDIENCE: Two. GILBERT STRANG: Two. Because it's got two
independent columns, but the third
column is dependent. The third column is a
combination of the others. So matrices like this are
really the building blocks of linear algebra, they're
the building blocks of data science. They're rank one matrices. And let me show you a special
way to write those rank one matrices. I think of this matrix as
the column vector 1, 1, 1 times the row vector 1, 3, 8. So it's a column times a row. That's a rank one matrix. Do you see-- that's a
true multiplication there. It looks a little weird, but
it's a 3 by 1 matrix times a 1 by 3 matrix. These numbers have
to be the same, and then the output is 3 by 3. And it's that. And do you see that it factors? So I'm going to move
on to that idea. The next idea coming up
will be that we can see-- well, that's coming, that
we're going to see matrices with two factors. Let me move to that, but back
for this original matrix. So what's with this? The column space is a plane. Think about now the key
idea of independent columns. How many independent
columns have I got here? AUDIENCE: Two. GILBERT STRANG: Two, correct. Two. The third column, if I want to
especially pick on that one-- I'll often go left to right. So I'll say the first guy
is good, second guy is good, the third guy is not
independent of the others. So I just have two
independent columns. And those two columns would be
a basis for the column space. So that's the critical
idea of linear algebra. That's what you compute
when you find a basis, and everything in
the column space is a combination of these,
including that is also-- that's already a
combination of those. But everything else
in the column space is a combination of those two. So they're a basis for it. So you have the
idea of A times x. You have the idea of
column space of A, which allows all x's. Then now, we're moving into the
idea of independent columns, and the number of independent
columns is the rank. So the rank-- shall I
write that somewhere? Maybe here. The rank is the number
of independent columns. And right now, what do I
mean by independent columns? Well, let's just see what
that means by using it. Are we good? I know I'm reviewing here. But allow me for this first
class and part of next time also to review. But you'll see something new. In fact, why don't we see
something new right away. Let me follow up on the
idea of independence in a systematic way. So here's my matrix A.
Can I write that again? 2, 3, 5, 1, 1, 7, and this
guy was the sum of those. So that's my matrix A. So let's start from scratch
and find a basis for the column space in the most natural way. So I'm going to take a basis-- what's a basis? A basis is independent columns. So all three together
would not be a basis. But they have to be
not just independent, but they have to fill the
space-- their combinations have to fill the space. So 2, 3, 5-- let's say I want
to create a basis. I'll call the matrix C, a
basis for the column space. So here's a natural
way to do it. I look at the first column. It's not 0s. If it was all 0s, I wouldn't
want that in a basis. But it's not. So I put it in. So that's the first
vector in my basis. Then I go on to
the second column. If that column was 4,
6, 10, what would I do? If that second
column was 4, 6, 10, would I put it in the basis? No. But 1, 1, 7, is OK, right? 1, 1, 7 is in a
different direction. It's not a combination of
what we've already got. So I say, OK, that
adds something new. Put it in. Then I move on to
the third column. Do I put that into a basis? You know the answer by now. No. Because I'm looking to see, is
it a combination of these guys? And it is. It's one of that
plus one of that. So it's not independent. So I've finished now. I've got a matrix C, which
was taken directly from A, and I kept only
independent columns and I worked from left to right. And I can see that right away
now that the rank is two-- the column rank, I should
say-- the column rank. The number of independent
columns is two. Good? Now comes the key step. I'm going to produce
a third matrix, R, which is going to tell me
how to get these columns from these columns. And its shape is going to be-- well, its shape-- I
don't have any choice. This is three by two, so Rx
would be two by something. So I'm like so. I guess it has to be two by
three because I want to come out this way, two by three. What am I going to do here? I'm just going to
put it in the numbers R that make this correct. So this is a first
matrix factorization. It's not-- well, it is
a famous one, actually. When we see it, we'll
recognize what it is. It's famous in teaching
linear algebra, but now, actually, C times
R, columns times rows has become very, very
important in large scale numerical linear algebra. So let's figure out
what goes into R. What am I thinking here? I'm putting in R. So
every one of these columns is a combination of these. That's the whole point. And I'm just going to put in
the numbers that you need. So what goes in the
first column of R? What goes in the
first column of R? So I want to look,
what combination of that column and that
column gives me this one? Yeah? AUDIENCE: 1, 0. GILBERT STRANG: 1, 0. Because you remember how we
multiply a matrix by a vector? When I multiply that
matrix by that vector, I take one of this
plus zero of that. I see it vector-wise, and
of course I get it right. And what about the
second column of R? So the second column
should be the combination that produces the second
column of A correctly. What will that be? AUDIENCE: 0, 1. GILBERT STRANG: 0, 1. Thanks. And finally the third column? AUDIENCE: 1, 1. GILBERT STRANG: 1, 1. Yes, right. Because one of this plus one of
this produces the third column. So all I did was put in the
right numbers there, really. And this is correct now. A is C times R. And so this is-- what I've done here
is the first two pages of section 1.1 in these notes. So 1.1. And actually, I'll
tell you literally what happened earlier this year. I had finished this. I wrote this down with
a different example. And then I realized
something, that sitting here in front of me was the
first great theorem in linear algebra, the fact that
the column rank equals the row rank. The fact that if I
have a matrix where that column plus that
column gives that one. Oh. So what am I going to say here? I'm getting nervous about it. I believe that a combination
of the rows gives 0. Do you believe that? You've got to believe it. This is linear algebra. The matrix is not
invertible, it's square. But the columns are dependent. So the rows have
to be dependent. And I don't exactly see-- there's some combination
of that row and that row that gives that one. And of course, when I
looked at the first column, I thought, OK, it's
going to be too easy. One of that and one
of that gives that. But then my eye went over
to the second column, and I realized it's
not easy at all. So you're entitled to pull out
your phone and figure this out. But there is some damn
combination of those-- [LAUGHTER] --of those two rows that
gives the third row. Otherwise, the course
is over, we stop. Well, and maybe we're
going to find somehow. So this is the theorem. So I have to tell you,
I was really pleased. So the first two pages
got two more pages to follow up on that
idea, that here, we were seeing something that-- I proved in 18.06, but not
in the first lecture, that's for sure, and not
maybe so clearly. But now I can try to prove it. There won't be a lot of
proofs in this class, but this is such
an important fact-- A equals CR is an
important factorization. And out of it, we
can connect them. So what am I saying? I am saying that all-- so what's the row rank? I have to back up here. What's the row rank? What's the row space? So the row rank is going to be
the dimension of that space. So I look at my matrix A.
What's the row space of A? I'm going to look at its rows. Now, maybe just so we
don't get whole new letters for the row space-- for me, the row space
of A of a matrix-- so first of all, tell
me in words what it is. What's the row
space of the matrix? AUDIENCE: [INAUDIBLE] GILBERT STRANG: All
combinations of the rows. All combinations of the
rows, that's the space. So I would take all
combinations of those rows. To get combinations
of the rows-- well, two ways I can get
combinations of the rows. And the way I'll do it is,
I'll just transpose the matrix. So those rows become
columns, and then I'm back to what I did. So the row space of A is the
column space of A transpose. And this has the
advantage that we don't introduce a new letter. So it'll be the column
space of A transpose. So we don't introduce
a new letter. We keep the convention
that vectors are column vectors, which
would be the MATLAB and Julia and Python convention. So is that OK? The row space is the
combination of these rows. But I'll flip-- I'll
make them stand up-- 2, 1, 3-- to be column vectors. So it's a totally
different space. And actually, I happened to
take a three by three example, so that the column
space is part of R3 and the row space
is also part of R3. Because my matrix
is three by three. A better example-- and the
whole point of data-- data doesn't come in square matrices. Fortunately for us, data very,
very often comes in matrices. But the two-- the
columns might be sample, it might be patients,
and the rows might be diseases or something. They're different spaces. So matrices are not
likely to be square. But anyway, we're good here. So the row space. Now can I come
back to the proof? Because what I want to
say is that the proof of this fundamental
fact is staring at us, but we don't quite see it yet. And I want to see it. So I claim that these rows
are a basis for the row space. And we already saw
that these columns are a basis for the column space. And two equals two, right? Two vectors here were a
basis for the column space. Now if I can see why it shows
me that these two vectors are a basis for the row
space, then my example is right, that both of
these will give two. The column rank is
two, two columns, the row rank is two, two rows. I have to explain why I
believe that these two rows are a basis for the row space. Are you with me? I have to prove-- I have to see why. First, so when I say basis,
what do I have to check? Basis is the critical idea. I have to check that
they're independent, so I haven't got too many vectors-- I haven't got any
extra vectors in there. And I also have to check that
their combinations produce--? All the rows. Should I say that again? Because that's what
I'm going to check. I'm going to check that
those guys are independent. Well, you can see that
they're independent. And I'm going to check that
their combinations produce all three of these rows. We didn't create those
numbers for this purpose, but what I'm saying
is they work. So I claim that this is a
basis, because what combination of those two rows would
produce this first row? Yeah, let me just ask you that. What numbers should I
multiply these two rows by to get the first row of A? AUDIENCE: 2, 1. GILBERT STRANG: 2 and 1. And where do you find 2 and 1? It's sitting there in
C. Will it work again? Does three of this plus
one of that give 3, 1, 4? Yes. So far, so good. Does five of this and seven of
that-- see, I'm multiplying-- I guess I'm doing matrix
multiplication a backward way or a different way. I'm taking combinations of
the rows of the second guy. The wonderful thing about
matrix multiplication is you can do it a lot
of ways, it comes out the same every way, and each
way tells you something. So five of that row plus seven
of that row, sure enough, is here. Do you see that that
is not accident? The proof is really to look
at this multiplication, C times R, two ways. Look at it first as
combinations of columns of C to give the columns. Look at it second to get as
combinations of the rows of R, and that produces the rows. So that factorization A
equals CR was the key idea. And actually, this R that
we've come up with has a name. Anybody remember enough 18.06? Have you all taken 18.06? No. I see-- how many have? Yes. OK. Good. For a while, 18.06 was taught
in a very abstract way. I said, what's going on? But anyway, so if you
took it that semester, you maybe never heard
of the column space. I'm not sure. Or by a different name. It has another name. What's its other name? The column space of a matrix? Range-- I think it's the range. Yeah. And of course, all this is
fundamental in mathematics. So of course, everything
here has different languages and different emphasis. But you see what the
emphasis is here. So you see the proof,
that A equals CR just reveals everything. So it's our first idea of a
factorization of a matrix. And we've multiplied C
times R. I could just say-- so really, you've seen now
the main point of Section 1.1 of the notes to come up
with that factorization and that conclusion. And you see why C has
the same number of-- the number of columns of C
equals the number of rows of R, and those are the column
rank and the row rank. Yeah, it's just pretty neat. And here was the special case
where those the column space is all multiples of U--
it's a line through U. The row space is
all multiples of V-- it's a line through V. And
that's the basic building block. Can I just say
another little word before I push on beyond CR? That this has become-- if you have a giant matrix,
like size 10 to the 5th, you can't put that
into fast memory. It's a mess. How do you deal with a
matrix of size 10 to the 5th, when you cannot deal
with all the entries? That's just not possible. Well, you sample it. So later in the
course, we'll be doing random sampling of a matrix. So how could you
sample a matrix? So you have a matrix. Of course you're looking
at it, but it's-- and you want to get
some typical columns. Here's the natural idea. You just look at A times x. Let x be a random vector. Rand of-- so it's got
m rows and one column. It's a vector. And what can I say about Ax? It's in the-- what
space is it in? AUDIENCE: Column. GILBERT STRANG: Column space. Thanks. That was the first
idea in this lecture. Ax is in the column space. So if you want a random
vector in the column space, I wouldn't suggest
to just randomly pick one of the columns. Better to take a
mixture of columns by taking a random vector
x, and looking at Ax. And if you wanted
100 random vectors, you'd take a 100 random
x's, and that would give you a pretty good idea, in many
cases, of what the column space looks like. That would be enough
to work with often. Can I just throw in
another question? So Ax is in the
column space of A. Let me just ask
you this question. Is ABCx-- is that in
the column space of A? Suppose I have matrices A,
B, and C, and a vector x, and I take their product. Does that give me something
in the column space of A? AUDIENCE: Yes. GILBERT STRANG:
Yes, good for you. How do you know that? AUDIENCE: [INAUDIBLE] GILBERT STRANG: Yeah,
it's A times something. Right. Putting parentheses
in the right place is the key to linear algebra. And there it is. It's a question that
just occurred to me. And I thought, well, I
wonder if you'd do it. So we have still time
to multiply matrices. Oh, I was going to
say about C and R-- so these are real columns
from A. But R is-- the rows are not taken directly
from the rows of A. Actually, there is a name for this. It's called the row reduced
echelon form of the matrix, and it's a big goal in 18.06. It has the identity there,
and then the other columns tell you the right combinations. Another big factorization would
be to take columns from A-- so this is another-- so I'll put maybe or-- and we won't be doing
this for a month. We could start to
take columns of A and put them into C, if
they were independent. And suppose I took
rows of A. Now I'm going to take literally rows
of A, and put them into R-- well, shall I call it
R twiddle or something, because it'll be a
different R. I'm not going to use those rows,
but I'm going to take two actual rows of A. Then what? So that's an important
factorization, but it's not correct. If I took two other rows
of R, it wouldn't work. So you have to put it
in the middle some two by two matrix U that
makes it correct. You'll see in that Section
1.1 that I got excited and wrote a page about CUR. Yeah, so I'll just mention that. So now I'm ready-- oh, I
wanted to say something about the course. I got excited thinking about
math, but there is this course. So what's up? So there'll be linear
algebra problems. But what makes this course
special is the other homeworks, which are online. And you would use-- let's see. In principle, you could
use any of the languages-- MATLAB, Python, which
has become the biggest-- most used for deep
learning, or Julia, which is the hot, new language. So last time, last
year, the problems-- oh, so what happened last year? Well, everything
in this course is owed to a professor who
visited from the University of Michigan, Professor Rao-- Raj Rao, who gave most of
the lectures a year ago, brought these
homework problems-- online homework
problems, so that people brought laptops to class
and we did things in class. So he had and has a
very successful course at EE in Michigan. But he was a PhD from here, and
he came back on a sabbatical, and he created this--
got us started. And we really owe a lot to him. Also, Professor Edelman was
involved with this course. And you maybe know
that he created Julia. How many know what Julia is? Oh, wonderful. That will make his day. He tells me every time I
see him, Julia is good. And I tell him, I believe it. [LAUGHTER] Anyway-- and it's become-- Professor Johnson in
18.06 has used Julia. And every semester, Steven
Johnson gives in the first week a tutorial on Julia. And so that's arranged,
and I promised to tell you where and when that is. So I think if you don't
know anything about Julia, try to go. It's in Stata. It's on Friday at 5:00-- 5:00 to 7:00. So Julia from Professor Johnson. So he's done this
multiple times. He's good at it. I don't think we know yet what-- I guess I'm hoping that
you'll have an option to use any of the three languages. But the online thing that we
give you was created in Julia. So professor Rao had to
learn Julia last spring, and the class did, too. And there was a certain
amount of bitching about it. [LAUGHTER] But I think, with
maybe one exception, who still got an
A, everybody was OK and was glad to learn Julia. And Professor Rao now
uses Julia entirely. So he's creating a new
on ramp with Julia. MATLAB, by the way, has
just issued an on ramp to deep learning that
I'll tell you about, and probably get a MATLAB-- somebody from Math Works
to say something about it. So that's what's coming, that
we don't quite know exactly how well organized those homeworks. We'll just take the first
one and see what happens. So I'll certainly say more
about homeworks when-- maybe even Friday. But are there other questions
that I should answer? Because some people
will be thinking, OK, am I going to do this or
am I going to sit in 6.036, or some other course
in deep learning? Anything on your mind? And you can email me. So we will have a Stellar
site and you'll see all the-- the TAs are still to be named. But the wonderful thing is
that the undergraduates who took this course last
year are volunteering to be graders for you guys. So they will know what those
online homeworks were about. So that's a first word
about what's coming and about the language. I'm going to finish by
a very important topic, multiplying A times B.
Oh, look, a clean board. So now I want to multiply
a matrix by a vector. Everybody knows how to do it. You take a row of A--
so you take a row of A and you take a column of B,
and you take the dot product. So you get a dot product. Row dot column. That's, again, low
level, OK for beginners. But we want to see that matrix
multiplication, AB, in a deeper way. And the deeper way
is columns times row. Columns of A, rows of
B. Columns times row. Oh, we had a column times a row. That was this rank one example. We had a column times a row,
and it produced a matrix. And that's what it looked like. And its rank was one. So those are what-- so it's a combination of. It's very like Ax. I'm really just extending
the Ax idea to AB. So this is the old way. The new way is columns. So there's column
K. It will multiply. Sure enough, it multiplies
row K. Everybody sees that it will happen that way. If you do it the
old way, when you do a dot product of something
here, something here, you're doing these
multiplications. And when you hit column K
here, you hit row K there. So these are connected. So I get things like column
K of A times row K of B. And I don't know
what notation to use, so I just wrote the words. But now that's one piece
of the final answer, AB. That's a rank one piece. That's a building block. So I add from K equals 1
column one times row one. Column one of A times
row one of B plus-- da, da, da, plus column K plus-- and of course, I stop at
column n of A times row n of B. So it's a sum of outer products. So everybody sees
a sum because I have column one times row
1, column K times row K, column n times row n, and
then I add those pieces. It's just the generalization
of Ax to a matrix B there. So it's a sum of Column K row
K of A row K of B. And maybe, should we check that that
gives us the right answer? I won't do that here,
but all we're doing is the same multiplications
in a different order. Actually, let's just
quit with one minute. We can figure out how many
multiplications are there. How many more applications
to do an m by n matrix A times an
n by p matrix B? So that's A times B. How
many individual numbers-- because this would
determine the cost of it. How many numbers would we need? Well, suppose we do it the
old way, by inner product, row times column. So how many multiplications
to do a row times column and get one entry in the answer? n, right? The row has length
n, the column has length n, n multiplications. So that's n. And now how many of
those do I have to do? AUDIENCE: mp. GILBERT STRANG: mp. Because what's the
size of this answer? The size of this
answer is m by p. So if I do it in that old
order, like n multiplication is to do a dot product. And I've got this many dot
products in the answer. So I've mnp multiplied. Now suppose I do it this way. How many multiplications
to do one of those guys? To multiply our column by a row? This is an m by 1,
and this is a 1 by p. One column, one row. How many multiplications
for that guy? mp. And how many of those
rank ones do I have to do? n. You got it? mp times n. Now, the other way
was n times mp. So it gives the same
answer, mnp multiplication. In fact, they're exactly the
same multiplications, just a different order. OK. We're at 1:55. Thanks for coming today. I'll talk more about the
class and about linear algebra on Friday. Thank you.
