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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR STRANG: Just to give
an overview in three lines: the text is the book of that
name, Computational Science and Engineering. That was completed
just last year, so it really ties pretty
well with the course. I don't cover everything
in the book, by all means. And I don't, certainly, don't
stand here and read the book. That would be no good. But you'll be able, if
you miss a class -- well, don't miss a class. But if you miss a class,
you'll be able, probably, to see roughly what we did. OK, so the first
part of the semester is applied linear algebra. And I don't know how many of
you have had a linear algebra course, and that's
why I thought I would start with a quick review. And you'll catch on. I want matrices to
come to life, actually. You know, instead
of just being a four by four array of numbers,
there are four by four, or n by n or m by n
array of special numbers. They have a meaning. When they multiply a
vector, they do something. And so it's just, part
of this first step is just, like,
getting to recognize, what's that matrix doing? Where does it come from? What are its properties? So that's a theme at the start. Then differential equations,
like Laplace's equation, are beautiful examples. So here we get, especially,
to numerical methods, finite differences, finite
elements, above all. So I think in this
class you'll really see how finite elements
work, and other ideas. All sorts of ideas. And then the last part of
the course is about Fourier. That's Fourier series,
that you may have seen, and Fourier integrals. But also, highly important,
Discrete Fourier Transform, DFT. That's a fundamental
step for understanding what a signal contains. Yeah, so that's
great stuff, Fourier. OK, what else should
I say before I start? I said this was my
favorite course, and maybe I'll
elaborate a little. Well, I think what I want
to say is that I really feel my life is here to teach
you and not to grade you. I'm not going to spend this
semester worrying about grades, and please don't. They come out fine. We've got lots to learn. And I'll do my very best
to explain it clearly. And I know you'll do your best. I know from experience. This class goes for
it and does it right. So that's what makes it so good. OK. Homeworks, by the way,
well, the first homework will simply be a way to get a
grade list, a list of everybody taking the course. They won't be graded
in great detail. Too large a class. And you're allowed to talk
to each other about homework. So homework is not
an exam at all. So let me just leave any
discussion of exams and grades for the future. I'll tell you, you'll see how
informally the first homework will be. And I hope it'll go
up on the website. The first homework
will be for Monday. So it's a bit early, but
it's pretty open-ended. If you could take three
problems from 1.1, the first section of
the book, any three, and any three problems from
1.2, and print your name on the homework -- because we're
going to use that to create the grade list -- I'll
be completely happy. Well, especially if
you get them right and do them neatly and so on. But actually we won't know. So that's for Monday. OK. And we'll talk more about it. I'll announce the TA on the
website and the TA hours, the office hours,
and everything. There'll be a Friday
afternoon office hour, because homeworks will
typically come Monday. OK. Questions about the course
before I just start? OK. Another time for questions, too. OK, so can we just
start with that matrix? So I said about matrices, I'm
interested in their properties. Like, I'm going to
ask you about that. And then, I'm interested
in their meaning. Where do they come from? You know, why that matrix
instead of some other? And then, the numerical part
is how do we deal with them? How do we solve a linear system
with that coefficient matrix? What can we say
about the solution? So the purpose. Right. OK, now help me out. So I guess my plan
with the video taping is, whatever you
say, I'll repeat. So say it as clearly
as possible, and it's fantastic to have discussion,
conversation here. So I'll just repeat it so that
it safely gets on the tape. So tell me its properties. Tell me the first property that
you notice about that matrix. Symmetric. Symmetric. Right. I could have slowed
down a little and everybody probably would
have said that at once. So that's a symmetric matrix. Now we might as well pick
up some matrix notation. How do I express the fact
that this a symmetric matrix? In simple matrix
notation, I would say that K is the
same as K transpose. The transpose, everybody
knows, it comes from -- oh, I shouldn't say this --
flipping it across the diagonal. That's not a very
"math" thing to do. But that's the way
to visualize it. And let me use a
capital T for transpose. So it's symmetric. Very important. Very, very important. That's the most important
class of matrices, symmetric matrices. We'll see them all
the time, because they come from equilibrium problems. They come from all sorts
of -- they come everywhere in applications. And we will be
doing applications. The first week or
week and a half, you'll see pretty
much discussion of matrices and the reasons,
what their meaning is. And then we'll get to
physical applications: mechanics and more. OK. All right. Now I'm looking for
properties, other properties, of that matrix. Let me write "2" here so that
you got a spot to put it. What are you going to tell
me next about that matrix? Periodic. Well, okay. Actually, that's
a good question. Let me write periodic down here. You're using that word,
because somehow that pattern is suggesting something. But you'll see I
have a little more to add before I would
use the word periodic. So that's great
to see that here. What else? Somebody else was
going to say something. Please. Sparse! Oh, very good. Sparse. That's also an obvious
property that you see from looking at the matrix. What does sparse mean? Mostly zeros. Well that isn't
mostly zeros, I guess. I mean, that's got what,
out of sixteen entries, it's got six zeros. That doesn't sound like sparse. But when I grow the matrix --
because this is just a four by four. I would even call this one K_4. When the matrix
grows to 100 by 100, then you really
see it as sparse. So if that matrix was 100
by 100, how many non-zeros would it have? So if n is 100, then the
number of non-zeros -- wow, that's the first MATLAB
command I've written. A number of non-zeros
of K would be -- anybody know what it would be? I'm just asking to go
up to five by five. I'm asking you to keep
that pattern alive. Twos on the diagonal,
minus ones above and below. So yeah, so 298, would it be? A hundred diagonal entries,
99 and 99, maybe 298? 298 out of 100 by
100 would be what? It's been a long summer. Yeah, a lot of zeros. A lot. Right. Because the matrix has got what
100 by 100, 10,000 entries. Out of 10,000. So that's sparse. But we see those all the
time, and fortunately we do. Because, of course, this
matrix, or even 100 by 100, we could deal with
if it was dense. But 10,000, 100,000,
or 1 million, which happens all the time
now in scientific computation. A million by
million dense matrix is not a nice thing
to think about. A million by million matrix
like this is a cinch. OK. So sparse. What else do you want to say? Toeplitz. Holy Moses. Exactly right. But I want to say,
before I use that word, so that'll be my
second MATLAB command. Thanks. Toeplitz. What's that mean? So this matrix has
a property that we see right away, which is? I want to stay with Toeplitz
but everybody tell me something more about
properties of that matrix. Tridiagonal. Tridiagonal, so that's almost
a special subcase of sparse. It has just three diagonals. Tridiagonal matrices
are truly important. They come in all
the time, we'll see that they come from second order
differential equations, which are, thanks to
Newton, the big ones. Ok, now it's more than
tridiagonal and what more? So what further, we're
getting deeper now. What patterns do you see
beyond just tridiagonal, because tridiagonal would
allow any numbers there but those are not, there's more
of a pattern than just three diagonals, what is it? Those diagonals are constant. If I run down each of
those three diagonals, I see the same number. Twos, minus ones,
minus ones, and that's what the word Toeplitz means. Toeplitz is constant diagonal. Ok. And that kind of
matrix is so important. It corresponds, yeah,
if we were in EE, I would use the words
time-invariant filter, linear, time-invariant. So it's linear because
we're dealing with a matrix. And it's time-invariant,
shift-invariant. I just use all these
equivalent words to mean that we're
seeing the same thing row by row, except of course, at,
shall I call that the boundary? That's like, the end
of the system and this is like the other end and
there it's chopped off. But if it was ten by ten I
would see that row eight times. 100 by 100 I'd see it 98 times. So it's constant diagonals
and the guy who first studied that was Toeplitz. And we wouldn't need that
great historical information except that MATLAB created a
command to create that matrix. K, MATLAB is all set to
create Toeplitz matrices. Yeah, so I'll have to put
what MATLAB would put. I realize I'm already
using the word MATLAB. I think that MATLAB language
is really convenient to talk about linear algebra. And how many know
MATLAB or have used it? Yeah. You know it better than I. I
talk a good line with MATLAB but I -- the code never runs. Never! I always forget some
stupid semicolon. You may have had
that experience. And I just want to say it now
that there are other languages, and if you want to
do homeworks and want to do your own work in other
languages, that makes sense. So the older
established alternatives were Mathematica and
Maple and those two have symbolic-- they can
deal with algebra as well as numbers. But there are newer languages. I don't know if you know them. I just know my friends
say, yes they're terrific. Python is one. And R. I've just
had a email saying, tell your class
about R. And others. Ok, so but we'll use MATLAB
language because that's really a good common language. Ok, so what is a
Toeplitz matrix? A Toeplitz matrix is one
with constant diagonals. You could use the
word time-invariant, linear time-invariant filter. And to create K, this
is an 18.085 command. It's just set up for us. I can create K by telling
the system the first row. Two, minus one, zero, zero. That would, then if
it wasn't symmetric I would have to give
the first column also. Toeplitz would be
constant diagonal, it doesn't have to be symmetric. But if it's symmetric, then
the first row and first column are the same vector, so I
just have to give that vector. Okay, so that's the
quickest way to create K. And of course, if it
was bigger then I would, rather than writing 100 zeros,
I could put zeros of 98 and one. Wouldn't I have to say that? Or is it one and 98? You see why it doesn't run. Well I guess I'm thinking
of that as a row. I don't know. Anyway. I realize getting this
videotaped means I'm supposed to get things right! Usually it's like, we'll
get it right later. But anyway, that might work. Okay. So there's a command
that you know. "zeros", that creates a matrix
of this size with all zeros. Okay. That would create
the 100 by 100. Good. Ok. Oh, by the way, as long as
we're speaking about computation I've gotta say something more. We said that the
matrix is sparse. And this 100 by 100 matrix
is certainly sparse. But if I create it this way,
I've created all those zeros and if I ask MATLAB to work
with that matrix, to square it or whatever, it would
carry all those zeros and do all those
zero computations. In other words, it would
treat K like a dense matrix and it would just, it wouldn't
know the zeros were there until it looked. So I just want to say that if
you have really big systems sparse MATLAB is the way to go. Because sparse MATLAB keeps
track only of the non-zeros. So it knows-- and their
locations, of course. What the numbers are
and their location. So I could create a
sparse matrix out of that, like KS for K sparse. I think if I just
did sparse(K) that would create a sparse matrix. And then if I do
stuff to it, MATLAB would automatically know
those zeros were there and not spend it's time
multiplying by zero. But of course,
this isn't perfect because I've created the big
matrix before sparsifying it. And better to have created
it in the first place as a sparse matrix. Ok. So those were properties
that you could see. Now I'm looking
for little deeper. What's the first question I
would ask about a matrix if I have to solve a
system of equations, say KU=F or something. I got a 4 by 4 matrix, four
equations, four unknowns. What would I want to know next? Is it invertible? Is the matrix invertible? And that's an important
question and how do you recognize an
invertible matrix? This one is invertible. So let me say K is invertible. And what does that mean? That means that
there's another matrix, K inverse such that K times K
inverse is the identity matrix. The identity matrix in
MATLAB would be eye(n) and it's the diagonal
matrix of ones. It's the unit matrix; it's the
matrix that doesn't do anything to a vector. So this K has an inverse. But how do you know? How can you recognize that
a matrix is invertible? Because obviously that's a
critical question and many, many-- since our
matrices are not-- a random matrix would
be invertible, for sure, but our matrices have
patterns, they're created out of a
problem and the question of whether that matrix is
invertible is fundamental. I mean finite
elements has these, zero-energy modes that you have
to watch out for because, what are they? They produce non-invertible
stiffness matrix. Ok. So how did we know,
or how could we know that this K is invertible? Somebody said invertible
and I wrote it down. Yeah? Well ok. Now I get to make a
speech about determinants. Don't deal with them! Don't touch determinants. I mean this particular
four by four happens to have a
nice determinant. I think it's five. But if it was a 100
by 100 how would we show that the
matrix was invertible? And what I mean by this is the
whole family is invertible. All sizes are invertible. K_n is invertible
for every n, not just this particular guy, whose
determinant we could take. But as five by
five, six by six, we would be up in the-- but
you're completely right. The determinant is a test. Alright. But I guess I'm saying that it's
not the test that I would use. So what I do? I would row reduce. That's the default
option in linear algebra. If you don't know what
to do with a matrix, if you want to see what's
going on, row reduce. What does that mean? That means-- shall I try it? So let me just start it,
just so I'm not using a word that we don't need. Ok. And actually, maybe the third
lecture, maybe next Monday we'll come back to row reduce. So I won't make heavy weather
of that, certainly not now. So what is row reduce? Just so you know. I want to get that
minus one to be a zero. I'm aiming for a
triangular matrix. I want to clean out
below the diagonal because if my matrix
is triangular then I can see immediately
everything. Right? Ultimately I'll reach a matrix
U that'll be upper triangular and that first row won't change
but the second row will change. And what does it change to? How do I clean out, get
a zero in that, where the minus one is right now? Well I want to use the first
row, the first equation. I want to add some
multiple of the first row to the second row. And what should
that multiple be? I want to multiply
that row by something. And I'll say "add" today. Later I'll say "subtract." But what shall I do? Just tell me what
the heck to do. I've got that row
and I want to use it, I want to take a combination
of these two rows. This row and some multiple
of this one that'll produce a zero. This is called the pivot. That's the first pivot
P-I-V-O-T. Pivot. And then that's the pivot row. And what do I do? Tell me what to do. Add half this row to this one. When I add half of that row
to that one, what do I get? I get that zero. What do I get here
for the second pivot? What is it? 1.5, 3/2. Because half of that is, so 3/2. And the rest won't change. So I'm happy with that zero. Now I've got a couple more
entries below that first pivot, but they're already zero. That's where the
sparseness pays off. The tridiagonal really pays off. So those zeros say the
first column is finished. So I'm ready to go on
to the second column. It's like I got to this smaller
problem with the 3/2 here. And a zero there. What do I do now? There is the second pivot, 3/2. Below it is a non-zero. I gotta get rid of it. What do I multiply by now? 2/3. 2/3 of that new second
row added to the third row will clean out the third row. This was already cleaned out. This is already a zero. But I want to have 2/3 of
this row added to this one so what's my new third row? Starts with zero and
what's the third pivot now? You see the pivots appearing? The third pivot will be 4/3
because I've got 2/3 this -1 and 2 is 6/3 so I have 6/3,
I'm taking 2/3 away, I get 4/3 and that -1 is still there. So you see that
I'm-- this is fast. This is really fast. And the next step, maybe you
can see the beautiful patterns that are coming. Do you want to just
guess the fourth pivot? 5/4, good guess, right. 5/4. Now this is actually how MATLAB
would find the determinant. It would do elimination. I call that
elimination because it eliminated all those
numbers below the diagonal and got zeros. Now what's the determinant? If I asked you for
the determinant, and I will very rarely
use the word determinant, but I guess I'm into it now,
so tell me the determinant. Five. Why's that? I guess I did say five earlier. But how do you know it's five? Whatever the determinant of
that matrix is, why is it five? Because it's a
triangular matrix. Triangular matrices,
you've got all these zeros. You can see what's happening. And the determinant
of a triangular matrix is just the product
down the diagonal. The product of these pivots. The determinant is the
product of the pivots. And that's how MATLAB would
compute a determinant. And it would take 2 times
3/2 times 4/3 times 5/4 and it would give
the answer five. My friend Alan Edelman told
me something yesterday. MATLAB computes
in floating point. So 4/3, that's 1.3333, etc. So MATLAB would not, when
it does that multiplication, get a whole number. Right? Because in MATLAB that
would be 1.333 and probably it would make that last pivot
a decimal, a long decimal. And then when it multiplies
that it gets whatever it gets. But it's not exactly
five I think. Nevertheless MATLAB will
print the answer five. It's cheated actually. It's done that
calculation and I don't know if it takes the
nearest integer when it knows that the-- I
shouldn't tell you this, this isn't even interesting. If the determinant of an
integer matrix, whole number is a whole number,
so MATLAB says, better get a whole number. And somehow it gets one. Actually, it doesn't
always get the right one. So maybe later I'll know
the matrix whose determinant might not come out right. But ours is right, five. Now where was this going? It got thrown off track
by the determinant. What's the real test? Well so I said there
are two ways to see that a matrix is invertible. Or not invertible. Here we're talking
about the first way. How do I know that this matrix--
I've got an upper triangular matrix. When is it invertible? When is an upper triangular
matrix invertible? Upper triangular is great. When you've got it
in that form you should be able to see stuff. So this key question
of invertible, which is not obvious
for a typical matrix is obvious for a
triangular matrix. And why? What's the test? Well, we could do
the determinant but we can say it without
using that long word. The diagonal is non-zero. K as invertible because
the diagonal-- no, it's got a full set of pivots. It's got four non-zero pivots. That's what it takes. That's what it's going
to take to solve systems. So this is the first step
in solving this system. In other words, to decide
if a matrix is invertible, you just go ahead and use it. You don't stop first necessarily
to check invertibility. You go forward, you
get to this point and you see non-zeros
there and then you're practically got
to the answer here. I'll leave for another day the
final back-- going back upwards that gives you the answer. So K is invertible. That means full set of
pivots. n non-zero pivots. And here they are,
two, 3/2, 4/3 and 5/4. Worth knowing because this
matrix K is so important. We'll see it over
and over again. Part of my purpose today
is to give some matrices a name because we'll see them
again and you'll know them and you'll recognize them. While I'm on this invertible
or not invertible business I want to ask you to change
K. To make it not invertible. Change that matrix. How could I change that matrix? Well, of course, many ways. But I'm interested
in another matrix and this'll be among
my special matrices. And it will start out the same. It'll have these same diagonals. It'll be Toeplitz. I'm going to call
it C and I want to say the reason I'm talking
about it now is that it's not going to be invertible. And I'm going to tell you a
C and see if you can tell me why it is not invertible. So here's the difference:
I'm going to put minus one in the corners. Still zeros there. So that matrix C still
has that pattern. It's still a Toeplitz
matrix, actually. That would still be the matrix
Toeplitz of 2, -1, 0, -1. I claim that matrix
is not invertible and I claim that we can see that
without computing determinants, we can see it without
doing elimination, too. MATLAB would see it
by doing elimination. We can see it by just
human intelligence. Now why? How do I recognize a matrix
that's not invertible? And then, by converse, how
a matrix that is invertible. I claim-- and let
may say first, let me say why that letter C. That
letter C stands for circulant. It's because-- This word
circulant, why circulant, it's because that diagonal
which only had three guys circled around to the fourth. This diagonal that
only had three entries circled around to
the fourth entry. This diagonal with two
zeros circled around to the other two zeros. The diagonal are not only
constant, they loop around. And you use the word periodic. Now for me, that's
the periodic matrix. See, a circulant matrix comes
from a periodic problem. Because it loops around. It brings numbers,
zero is the same as number four or something. And why is that not invertible? The thing is can
you find a vector? Because matrices
multiply vectors, that's their whole point. Can you see a vector
that it takes to zero? Can you see a solution to Cu=0? I'm looking for a
u with four entries so that I get four zeros. Do you see it? All ones. All ones. That will do it. So that's a nice, natural
entry, a constant. And do you see why
when I-- we haven't spoken about multiplying
matrices times vectors. And most people
will do it this way. And let's do this one this way. You take row one times that,
you get two, minus one, zero, minus one. You get the zero because
of that new number. Here we always got zero from
the all ones vector and now over here that minus one,
you see it's just right. If all the rows add to zero
then this vector of all ones will be, I would use the
word "in the null space" if you wanted a fancy word,
a linear algebra word. What does that mean? It solves Cu=0. And why does that show that
the matrix isn't invertible? Because that's our point here. I have a solution to Cu=0. I claim that the existence
of such a solution has wiped out the possibility
that the matrix is invertible because if it was invertible,
what would this lead to? If invertible, if C
inverse exists what would I do to that equation
that would show me that C inverse can't exist? Multiply both
sides by C inverse. So you're seeing, just
this first day you're seeing some of the natural
steps of linear algebra. Row reduction,
multiply-- when you want to see what's happening,
multiply both sides by C inverse. That's the same as
in ordinary language, do the same thing to
all the equations. So I multiply both sides
by the same matrix. And here I would get
C^(-1) C u = C^(-1) 0. So what does that tell me? I made it long, I threw
in this extra step. You were going to jump
immediately to C^(-1) C is I, is the identity matrix and when
the identity matrix multiplies a vector u, you get u. And on the right
side, C inverse, whatever it is, if it
existed, times zero would have to be zero. So this would say that
if C inverse exists, then the only solution
is u equals zero. That's a good way to
recognize invertible matrices. If it is invertible then the
only solution to Cu=0 is u=0. And that wasn't true here. So we conclude C
is not invertible. C is therefore not invertible. Now can I even jump in. I've got two more
matrices that I want to tell you about that are
also close cousins of K and C. But let me just explain
physically a little bit about where these
matrices are coming from. So maybe next to K-- so I'm not
going to put periodic there. Right? That's the one that I
would call periodic. This one is fixed at the ends. Can I draw a little picture
that aims to show that? Aims to show where
this is coming from. It's coming from I think
of this as controlling like four masses. Mass one, mass two,
mass three and mass four with springs attached
and with endpoints fixed. So if I put some weights on
those masses-- we'll do this; masses and springs is going to
be the very first application and it will connect
to all these matrices. And all I'm doing now is just
asking to draw the system. Draw the mechanical system. Actually I'll usually
draw it vertically. But anyway, it's got
four masses and the fact that this minus one
here got chopped off, what would I call that end? I'd call that a fixed end. So this is a fixed,
fixed matrix. Both ends are fixed. And it's the matrix
that would govern-- and the springs and
masses all the same is what tells me that
the thing is Toeplitz. Now what's the picture
that goes with C? What's the picture with C? Do you have an instinct of that? So C is periodic. So again we've got four
masses connected by springs. But what's up with those masses
to make the problem cyclic, periodic, circular,
whatever word you like. They're arranged in a ring. The fourth guy comes
back to the first one. So the four masses would be,
so in some kind of a ring, the springs would connect them. I don't know if that's
suggestive, but I hope so. And what's the point
of, can we just speak about
mechanics one moment? How does that system differ
from this fixed system? Here the whole system
can't move, right? If there no force, then
nothing can happen. Here the whole system can turn. They can all displace
the same amount and just turn without any
compression of the springs, without any force
having to do anything. And that's why the solution
that kills this matrix is [1, 1, 1, 1]. So [1, 1, 1, 1] would
describe a case where all the displacements were equal. In a way it's like the
arbitrary constant in calculus. You're always adding
plus C. So here we've got a solution of all ones
that produces zero the way the derivative of a constant
function is the zero function. So this is just
like an indication. Yes, perfect. I've got two more matrices. Are you okay for two more? Yes okay, what are they? Okay a different blackboard
for the last two. So one of them is going to
come by freeing up this end. So I'm going to take
that support away. And you might imagine like a
tower oscillating up and down or you might turn it upside
down and like a hanging spring, or rather four springs with
four masses hanging onto them. But this end is fixed and
this is not fixed anymore, this is now free. And can I tell you the
matrix, the free-fixed matrix. Free-fixed. Because it's the top
end that I changed, I'm going to call it T.
So all the other guys are going to be the same but
the top one, the top row, the boundary row,
boundary conditions are always the tough
part, the tricky part, the key part of
a model, and here the natural boundary condition
is to have a 1 there. That two changed to a one. Now if I asked you for the
properties of that matrix-- so that's the third.
shall I do the fourth one? So you have them all, you'll
have the whole picture. The fourth one,
well you can guess. What's the fourth? What am I going to do? Free up the other end. So this guy had one free
end and the other guy has B for both ends. B for both ends are
going to be free. So this is free-fixed. This'll be free-free. So that means I
have this free end, the usual stuff in
the middle, no change, and the last row is what? What am I going to put
in the last row? -1, 1. -1, 1. So I've changed the diagonal. There I put a single one in
because I freed up one end. With B I freed both ends
and I got two minus ones. Now what do you think? So we've drawn the free-fixed
one and what's your guess? They're all symmetric. That's no accident. They're all tridiagonal,
no accident again. Why are they tridiagonal? Physically they're tridiagonal
because that mass is only connected to its
two neighbors, it's not connected to that mass. That's why we get a zero
in the two, four position. Because two is not
connected to four. So it's tridiagonal. And it's not Toeplitz
anymore, right? Toeplitz says constant
diagonals and these are not quite constant. I would create K, I
would take T equal K, if I was going to create this
matrix and then I would say T(1, 1) = 1. That command would fix
up the first entry. Yeah, that's a serious question. Maybe, can I hang on
until Friday, and even maybe next week. Because it's very important. When I said boundary conditions
are the key to problems, I'm serious. If I had to think okay, what
do people come in my office ask about questions,
I say right away, what's the boundary condition? Because I know that's
where the problem is. And so here we'll see
these guys clearly. Fixed and free, very important. But also let me say two more
words, I never can resist. So fixed means the
displacement is zero. Something was set to zero. The fifth guy, the fifth over
here, that fifth column was knocked out. Free means that in here it
could mean that the fifth guy is the same as the fourth. The slope is zero. Fixed is u is zero. Free is slope is zero. So here I have a slope
of zero at that end, here I have it at both ends. So maybe that's a
sort of part answer. Now I wanted to get to the
difference between these two matrices. And the main properties. So what are we see? Symmetric again,
tridiagonal again, not quite Toeplitz, but almost,
sort of morally Toeplitz. But then the key question
was invertible or not. Key question was
invertible or not. Right. And what's your
guess on these two? Do you think that one's
invertible or not? Make a guess. You're allowed to guess. Yeah it is. Why's that? Because this thing has
still got a support. It's not free to shift forever. It's held in there. So that gives you a
hint about this guy. Invertible or not for B? No. And now prove that it's not. Physically you were
saying, well this free guy with this thing gone now,
this is now free-free. Physically we're saying
the whole thing can move, there's nothing holding it. But now, for linear algebra,
that's not the proper language. You have to say something
about that matrix. Maybe tell me
something about Bu=0. What are you going
to take for u? Yeah. Same u. We're lucky in this course, u =
[1, 1, 1, 1] is the guilty main vector many times. Because again the rows are all
adding to zero and the all ones vector is in the null space. If I could just close
with one more word. Because it's the most important. Two words, two words. Because they're the
most important words, they're the words that we're
leading to in this chapter. And I'm assuming that for most
people they will be new words, but not for all. It's a further property
of this matrix. So we've got, how many? Four properties, or five? I'm going to go for one more. And I'm just going to
say that name first so you know it's coming. And then I'll say,
I can't resist saying a tiny bit about it. I'll use a whole
blackboard for this. So I'm going to say that K
and T are -- here it comes, take a breath -- positive
definite matrices. So if you don't know what
that means, I'm happy. Right? Because well, I can
tell you one way to recognize a positive
definite matrix. And while we're at it, let
me tell you about C and B. Those are positive semi-definite
because they hit zero somehow. Positive means up there,
greater than zero. And what is greater than
zero that we've already seen? And we'll say more. The pivots were. So if I have a symmetric matrix
and the pivots are all positive then that matrix is not
only invertible, because I'm in good shape, the
determinant isn't zero, I can go backwards
and do everything, those positive numbers are
telling me that more than that, the matrix is positive definite. So that's a test. We'll say more about
positive definite, but one way to recognize
it is compute the pivots by elimination. Are they positive? We'll see that all the
eigenvalues are positive. The word positive definite
just brings the whole of linear algebra together. It connects to pivots, it
connects to eigenvalues, it connects to least squares,
it's all over the place. Determinants too. Questions or discussion. It's a big class and we're
just meeting for the first time but there's lots of time
to, chance to ask me. I'll always be here after class. So shall we stop today? I'll see you Friday
or this afternoon. If this wasn't familiar, this
afternoon would be a good idea. Thank you.
