Okay, this is linear
algebra, lecture four.
And, the first
thing I have to do
is something that was on
the list for last time,
but here it is now.
What's the inverse of a product?
If I multiply two
matrices together
and I know their
inverses, how do I
get the inverse of A times B?
So I know what inverses
mean for a single matrix A
and for a matrix B.
What matrix do I multiply by to
get the identity if I have A B?
Okay, that'll be
simple but so basic.
Then I'm going to use that to --
I will have a product of
matrices and the product
that we'll meet will be
these elimination matrices
and the net result
of today's lectures
is the big formula
for elimination,
so the net result
of today's lecture
is this great way to look
at Gaussian elimination.
We know that we get from
A to U by elimination.
We know the steps -- but now
we get the right way to look
at it, A equals L U.
So that's the high
point for today.
Okay.
Can I take the easy part,
the first step first?
So, suppose A is invertible --
and of course it's going to be
a big question, when is
the matrix invertible?
But let's say A is invertible
and B is invertible,
then what matrix gives
me the inverse of A B?
So that's the direct question.
What's the inverse of A B?
Do I multiply those
separate inverses?
Yes.
I multiply the two matrices
A inverse and B inverse,
but what order do I multiply?
In reverse order.
And you see why.
So the right thing to put
here is B inverse A inverse.
That's the inverse I'm after.
We can just check that A B times
that matrix gives the identity.
Okay.
So why -- once again, it's
this fact that I can move
parentheses around.
I can just erase them all and
do the multiplications any way
I want to.
So what's the right
multiplication to do first?
B times B inverse.
This product here
I is the identity.
Then A times the
identity is the identity
and then finally A times A
inverse gives the identity.
So forgive the dumb
example in the book.
Why do you, do the inverse
things in reverse order?
It's just like -- you
take off your shoes,
you take off your socks, then
the good way to invert that
process is socks back
on first, then shoes.
Sorry, okay.
I'm sorry that's on the tape.
And, of course, on the other
side we should really just
check -- on the other side
I have B inverse, A inverse.
That does multiply A
B, and this time it's
these guys that
give the identity,
squeeze down, they give the
identity, we're in shape.
Okay.
So there's the inverse.
Good.
While we're at it,
let me do a transpose,
because the next lecture
has got a lot to --
involves transposes.
So how do I -- if I
transpose a matrix,
I'm talking about square,
invertible matrices right now.
If I transpose one,
what's its inverse?
Well, the nice formula
is -- let's see.
Let me start from A, A
inverse equal the identity.
And let me transpose both sides.
That will bring a
transpose into the picture.
So if I transpose the identity
matrix, what do I have?
The identity, right?
If I exchange rows and
columns, the identity
is a symmetric matrix.
It doesn't know the difference.
If I transpose these guys,
that product, then again
it turns out that I have
to reverse the order.
I can transpose them
separately, but when I multiply,
those transposes come
in the opposite order.
So it's A inverse
transpose times A transpose
giving the identity.
So that's -- this equation is
-- just comes directly from that
one.
But this equation tells
me what I wanted to know,
namely what is the inverse
of this guy A transpose?
What's the inverse of that
-- if I transpose a matrix,
what'ss the inverse
of the result?
And this equation tells
me that here it is.
This is the inverse
of A transpose.
Inverse of A transpose.
Of A transpose.
So I'll put a big
circle around that,
because that's the
answer, that's the best
answer we could hope for.
That if you want to know
the inverse of A transpose
and you know the inverse of A,
then you just transpose that.
So in a -- to put
it another way,
transposing and inversing
you can do in either order
for a single matrix.
Okay.
So these are like basic
facts that we can now use,
all right -- so now
I put it to use.
I put it to use by thinking
-- we're really completing,
the subject of elimination.
Actually, -- the thing about
elimination is it's the right
way to understand what
the matrix has got.
This A equal L U is the
most basic factorization
of a matrix.
I always worry
that you will think
this course is all elimination.
It's just row operations.
And please don't.
We'll be beyond that, but it's
the right algebra to do first.
Okay.
So, now I'm coming
near the end of it,
but I want to get
it in a decent form.
So my decent form
is matrix form.
I have a matrix A, let's
suppose it's a good matrix,
I can do elimination,
no row exchanges --
So no row exchanges for now.
Pivots all fine, nothing
zero in the pivot position.
I get to the very
end, which is U.
So I get from A to U.
And I want to know
what's the connection?
How is A related to U?
And this is going to
tell me that there's
a matrix L that connects them.
Okay.
Can I do it for a
two by two first?
Okay.
Two by two, elimination.
Okay, so I'll do it under here.
Okay.
So let my matrix A be --
We'll keep it simple,
say two and an eight,
so we know that the
first pivot is a two,
and the multiplier's
going to be a four
and then let me put a
one here and what number
do I not want to put there?
Four.
I don't want a four there,
because in that case,
the second pivot would not --
we wouldn't have a second pivot.
The matrix would be
singular, general screw up.
Okay.
So let me put some other
number here like seven.
Okay.
Okay.
Now I want to operate on that
with my elementary matrix.
So what's the elementary matrix?
Strictly speaking,
it's E21, because it's
the guy that's going to produce
a zero in that position.
And it's going to
produce U in one shot,
because it's just a
two by two matrix.
So two one and I'm going to take
four of those away from those,
produce that zero and
leave a three there.
And that's U.
And what's the
matrix that did it?
Quick review, then.
What's the elimination
elementary matrix E21 --
it's one zero, thanks.
And -- negative four one, right.
Good.
Okay.
So that -- you see the
difference between this
and what I'm shooting for.
I'm shooting for A on one
side and the other matrices
on the other side
of the equation.
Okay.
So I can do that right away.
Now here's going to
be my A equals L U.
And you won't have any
trouble telling me what --
so A is still two
one eight seven.
L is what you're going to
tell me and U is still two one
zero three.
Okay.
So what's L in this case?
Well, first -- so how is
L related to this E guy?
It's the inverse, because
I want to multiply through
by the inverse of this, which
will put the identity here,
and the inverse will show
up there and I'll call it L.
So what is the inverse of this?
Remember those elimination
matrices are easy to invert.
The inverse matrix for
this one is 1 0 4 1,
it has the plus sign because
it adds back what this removes.
Okay.
Do you want -- if we did the
numbers right, we must --
this should be correct.
Okay.
And of course it is.
That says the first row's
right, four times the first row
plus the second
row is eight seven.
Good.
Okay.
That's simple, two by two.
But it already shows the
form that we're headed for.
It shows -- so what's
the L stand for?
Why the letter L?
If U stood for upper
triangular, then of course L
stands for lower triangular.
And actually, it has ones on the
diagonal, where this thing has
the pivots on the diagonal.
Oh, sometimes we may want
to separate out the pivots,
so can I just mention that
sometimes we could also write
this as -- we could have
this one zero four one --
I'll just show you how I
would divide out this matrix
of pivots -- two three.
There's a diagonal matrix.
And I just -- whatever
is left is here.
Now what's left?
If I divide this first row
by two to pull out the two,
then I have a one
and a one half.
And if I divide the second row
by three to pull out the three,
then I have a one.
So if this is L U, this is
maybe called L D or pivot U.
And now it's a
little more balanced,
because we have ones on
the diagonal here and here.
And the diagonal
matrix in the middle.
So both of those...
Matlab would produce either one.
I'll basically stay with L U.
Okay.
Now I have to think about
bigger than two by two.
But right now, this was
just like easy exercise.
And, to tell the truth,
this one was a minus sign
and this one was a plus sign.
I mean, that's the
only difference.
But, with three
by three, there's
a more significant difference.
Let me show you how that works.
Let me move up to
a three by three,
let's say some matrix A, okay?
Let's imagine it's
three by three.
I won't write
numbers down for now.
So what's the first
elimination step
that I do, the first
matrix I multiply it by,
what letter will I use for that?
It'll be E two one,
because it's --
the first step will be to get a
zero in that two one position.
right?
And then the next step will be
to get a zero in the three one
position.
And the final step will be
to get a zero in the three
two That's what elimination is,
and it produced U. position.
And again, no row exchanges.
I'm taking the nice case,
now, the typical case, too --
when I don't have to
do any row exchange,
all I do is these
elimination steps.
Okay.
Now, suppose I want that stuff
over on the right hand side,
as I really do.
That's, like, my point here.
I can multiply these
together to get a matrix E,
but I want it over on the right.
I want its inverse over there.
So what's the right
expression now?
If I write A and
U, what goes there?
Okay.
So I've got the
inverse of this, I've
got three matrices in a row now.
And it's their
inverses that are going
to show up, because each
one is easy to invert.
Question is, what
about the whole bunch?
How easy is it to
invert the whole bunch?
So, that's what
we know how to do.
We know how to invert, we should
take the separate inverses,
but they go in the
opposite order.
So what goes here?
E three two inverse,
right, because I'll
multiply from the left
by E three two inverse,
then I'll pop it up next to U.
And then will come
E three one inverse.
And then this'll be
the only guy left
standing and that's gone when
I do an E two one inverse.
So there is L.
That's L U.
L is product of inverses.
Now you still can ask why is
this guy preferring inverses?
And let me explain why.
Let me explain why is this
product nicer than this one?
This product turns out to
be better than this one.
Let me take a typical case here.
Let me take a typical case.
So let me --
I have to do three by three
for you to see the improvement.
Two by two, it was
just one E, no problem.
But let me go up to this case.
Suppose my matrices E21 --
suppose E21 has a minus two
in there.
Suppose that -- and
now suppose -- oh,
I'll even suppose
E31 is the identity.
I'm going to make the point
with just a couple of these.
Okay.
Now this guy will have
-- do something --
now let's suppose
minus five one.
Okay.
There's typical.
That's a typical case in
which we didn't need an E31.
Maybe we already had a zero
in that three one position.
Okay.
Let me see -- is that going to
be enough to, show my point?
Let me do that multiplication.
So if I do that multiplication
it's like good practice
to multiply these matrices.
Tell me what's above
the diagonal when
I do this multiplication?
All zeroes.
When I do this
multiplication, I'm
going to get ones on the
diagonal and zeroes above.
Because -- what does that say?
That says that I'm subtracting
rows from lower rows.
So nothing is
moving upwards as it
did last time in Gauss Jordan.
Okay.
Now -- so really, what I have to
do is check this minus two one
zero, now this is --
what's that number?
This is the number that
I'm really have in mind.
That number is ten.
And this one is
-- what goes here?
Row three against column two,
it looks like the minus five.
What it's that ten.
How did that ten get in there?
I don't like that ten.
I mean -- of course, I
don't want to erase it,
because it's right.
But I don't want it there.
It's because -- the ten got in
there because I subtracted two
of row one from row two, and
then I subtracted five of that
new row two from row three.
So doing it in that order, how
did row one effect row three?
Well, it did, because two of
it got removed from row two
and then five of those got
removed from row three.
So altogether ten of row one
got thrown into row three.
Now my point is in the
reverse direction --
so now I can do it -- below
it I'll do the inverses.
Okay.
And, of course, opposite order.
Reverse order.
Reverse order.
Okay.
So now this is going to -- this
is the E that goes on the left
side.
Left of A.
Now I'm going to do the
inverses in the opposite order,
so what's the --
So the opposite order means
I put this inverse first.
And what is its inverse?
What's the inverse of E21?
Same thing with a
plus sign, right?
For the individual matrices,
instead of taking away two
I add back two of row one
to row two, so no problem.
And now, in reverse order,
I want to invert that.
Just right?
I'm doing just this, this.
So now the inverse is
again the same thing,
but add in the five.
And now I'll do
that multiplication
and I'll get a happy result.
I hope.
Did I do it right so far?
Yes, okay.
Let me do the multiplication.
I believe this comes out.
So row one of the
answer is one zero zero.
Oh, I know that all this
is going to be left, right?
Then I have two one zero.
So I get two one
zero there, right?
And what's the third row?
What's the third
row in this product?
Just read it out to
me, the third row?
0 5 1 Because one
way to say is this
is saying take one of the
last row and there it is.
And this is my matrix L.
And it's the one that
goes on the left of U.
It goes into --
what do I mean here?
Maybe rather than saying
left of A, left of U,
let me right down
again what I mean.
E A is U, whereas A is L U.
Okay.
Let me make the
point now in words.
The order that the matrices
come for L is the right order.
The two and the
five don't sort of
interfere to produce
this ten one.
In the right order,
the multipliers
just sit in the matrix L.
That's the point -- that
if I want to know L,
I have no work to do.
I just keep a record of
what those multipliers were,
and that gives me L.
So I'll draw the
-- let me say it.
So this is the A=L U.
So if no row exchanges, the
multipliers that those numbers
that we multiplied
rows by and subtracted,
when we did an
elimination step --
the multipliers go
directly into L.
Okay.
So L is -- this is the way,
to look at elimination.
You go through the
elimination steps,
and actually if you
do it right, you
can throw away A
as you create L U.
If you think about it,
those steps of elimination,
as when you've finished
with row two of A,
you've created a new row two
of U, which you have to save,
and you've created the
multipliers that you used --
which you have to save,
and then you can forget A.
So because it's all
there in L and U.
So that's -- this moment
is maybe the new insight
in elimination that
comes from matrix --
doing it in matrix form.
So it was -- the
product of Es is --
we can't see what
that product of Es is.
The matrix E is not a
particularly attractive one.
What's great is when we put
them on the other side --
their inverses in
the opposite order,
there the L comes
out just right.
Okay.
Now -- oh gosh, so today's a
sort of, like, practical day.
Can we think together how
expensive is elimination?
How many operations do we do?
So this is now a kind of new
topic which I didn't list as --
on the program,
but here it came.
Here it comes.
How many operations
on an n by n matrix A.
I mean, it's a very
practical question.
Can we solve systems of
order a thousand, in a second
or a minute or a week?
Can we solve systems of order a
million in a second or an hour
or a week?
I mean, what's the
-- if it's n by n,
we often want to take n bigger.
I mean, we've put
in more information.
We make the whole thing is more
accurate for the bigger matrix.
But it's more expensive,
too, and the question
is how much more expensive?
If I have matrices
of order a hundred.
Let's say a hundred
by a hundred.
Let me take n to be a hundred.
Say n equal a hundred.
How many steps are we doing?
How many operations are we
actually doing that we --
And let's suppose there
aren't any zeroes,
because of course if a matrix
has got a lot of zeroes
in good places, we don't
have to do those operations,
and, it'll be much faster.
But -- so just think for a
moment about the first step.
So here's our matrix A,
hundred by a hundred.
And the first step
will be -- that column,
is got zeroes down here.
So it's down to 99 by 99, right?
That's really like the
first stage of elimination,
to get from this hundred
by hundred non zero
matrix to this stage where the
first pivot is sitting up here
and the first row's okay
the first column is okay.
So, eventually -- how
many steps did that take?
You see, I'm trying
to get an idea.
Is the answer proportional to n?
Is the total number of steps in
elimination, the total number,
is it proportional to n -- in
which case if I double n from
a hundred to two hundred --
does it take me twice as long?
Does it square, so it would
take me four times as long?
Does it cube so it would
take me eight times as long?
Or is it n factorial, so it
would take me a hundred times
as long?
I think, you know, from a
practical point of view,
we have to have some
idea of the cost, here.
So these are the
questions that I'm --
let me ask those
questions again.
Is it proportional -- does
it go like n, like n squared,
like n cubed -- or
some higher power of n?
Like n factorial where every
step up multiplies by a hundred
and then by a hundred and one
and then by a hundred and two
-- which is it?
Okay, so that's
the only way I know
to answer that is to think
through what we actually
had to do.
Okay.
So what was the cost here?
Well, let's see.
What do I mean by an operation?
I guess I mean, well
an addition or -- yeah.
No big deal.
I guess I mean an
addition or a subtraction
or a multiplication
or a division.
Okay.
And actually, what operation
I doing all the time?
When I multiply row one
by multiplier L and I
subtract from row six.
What's happening
there individually?
What's going on?
If I multiply --
I do a multiplication by
L and then a subtraction.
So I guess operation --
Can I count that for the
moment as, like, one operation?
Or you may want to
count them separately.
The typical operation is
multiply plus a subtract.
So if I count those together, my
answer's going to come out half
as many as if --
I mean, if I count
them separately,
I'd have a certain
number of multiplies,
certain number of subtracts.
That's really want to do.
Okay.
How many have I got here?
So, I think -- let's see.
It's about -- well,
how many, roughly?
How many operations to
get from here to here?
Well, maybe one
way to look at it
is all these numbers
had to get changed.
The first row
didn't get changed,
but all the other rows
got changed at this step.
So this step -- well,
I guess maybe --
shall I say it cost
about a hundred squared.
I mean, if I had
changed the first row,
then it would have been exactly
hundred squared, because --
because that's how
many numbers are here.
A hundred squared numbers is
the total count of the entry,
and all but this insignificant
first row got changed.
So I would say about
a hundred squared.
Okay.
Now, what about the next step?
So now the first row is fine.
The second row is fine.
And I'm changing these
zeroes are all fine,
so what's up with
the second step?
And then you're with me.
Roughly, what's the cost?
If this first step
cost a hundred squared,
about, operations
then this one, which
is really working on this guy to
produce this, costs about what?
How many operations to fix?
About ninety nine squared, or
ninety nine times ninety eight.
But less, right?
Less, because our
problem's getting smaller.
About ninety-nine squared.
And then I go down and down
and the next one will be ninety
eight squared, the next ninety
seven squared and finally
I'm down around
one squared or --
where it's just like
the little numbers.
The big numbers are here.
So the number of operations
is about n squared plus that
was n, right? n was a hundred?
n squared for the first step,
then n minus one squared,
then n minus two squared,
finally down to three squared
and two squared and
even one squared.
No way I should have written
that -- squeezed that in.
Let me try it so the count
is n squared plus n minus one
squared plus -- all the
way down to one squared.
That's a pretty decent count.
Admittedly, we didn't catch
every single tiny operation,
but we got the right
leading term here.
And what do those add up to?
Okay, so now we're coming
to the punch of this,
question, this operation count.
So the operations on the
left side, on the matrix A
to finally get to U.
And anybody -- so which of
these quantities is the right
ballpark for that count?
If I add a hundred squared to
ninety nine squared to ninety
eight squared --
ninety seven squared,
all the way down to two
squared then one squared,
what have I got, about?
It's just one of these --
let's identify it first.
Is it n?
Certainly not.
Is it n factorial?
No.
If it was n factorial, we
would -- with determinants,
it is n factorial.
I'll put in a bad mark against
determinants, because that --
okay, so what is it?
It's n -- well,
this is the answer.
It's this order -- n cubed.
It's like I have n terms, right?
I've got n terms in this sum.
And the biggest
one is n squared.
So the worst it could
be would be n cubed,
but it's not as bad as
-- it's n cubed times --
it's about one third of n cubed.
That's the magic
operation count.
Somehow that one third
takes account of the fact
that the numbers
are getting smaller.
If they weren't
getting smaller, we
would have n terms
times n squared,
but it would be exactly n cubed.
But our numbers are getting
smaller -- actually,
row two and row one
moves down to row three.
do you remember where does
one third come in this --
I'll even allow a
mention of calculus.
So calculus can be
mentioned, integration
can be mentioned now in the next
minute and not again for weeks.
It's not that I don't like
18.01, but18.06 is better.
Okay.
So, -- so what's -- what's the
calculus formula that looks
like?
It looks like -- if we were
in calculus instead of summing
stuff, we would integrate.
So I would integrate
x squared and I
would get one third x cubed.
So if that was like an
integral from one to n,
of x squared b x, if the answer
would be one third n cubed --
and it's correct for the sum
also, because that's, like,
the whole point of calculus.
The whole point of
calculus is -- oh,
I don't want to tell
you the whole --
I mean, you know the
whole point of calculus.
Calculus is like sums
except it's continuous.
Okay.
And algebra is discrete.
Okay.
So the answer is
one third n cubed.
Now I'll just -- let me say one
more thing about operations.
What about the right-hand side?
This was what it cost
on the left side.
This is on A.
Because this is A that
we're working with.
But what's the cost on
the extra column vector b
that we're hanging around here?
So b costs a lot
less, obviously,
because it's just one column.
We carry it through
elimination and then
actually we do
back substitution.
Let me just tell you
the answer there.
It's n squared.
So the cost for every right
hand side is n squared.
So let me --
I'll just fit that in here --
for the cost of b turns out
to be n squared.
So you see if we have,
as we often have,
a a matrix A and several
right hand sides, then we
pay the price on A, the
higher price on A to get it
split up into L and U
to do elimination on A,
but then we can process every
right hand side at low cost.
Okay.
So the -- We really have
discussed the most fundamental
algorithm for a
system of equations.
Okay.
So, I'm ready to
allow row exchanges.
I'm ready to allow -- now
what happens to this whole --
today's lecture if
there are row exchanges?
When would there
be row exchanges?
There are row -- we need to do
row exchanges if a zero shows
up in the pivot position.
So moving then into the final
section of this chapter,
which is about
transposes -- well,
we've already seen
some transposes, and --
the title of this section is,
"Transposes and Permutations."
Okay.
So can I say, now, where
does a permutation come in?
Let me talk a little
about permutations.
So that'll be up
here, permutations.
So these are the matrices that
I need to do row exchanges.
And I may have to do
two row exchanges.
Can you invent a
matrix where I would
have to do two row exchanges
and then would come out fine?
Yeah let's just, for the heck
of it -- so I'll put it here.
Let me do three by threes.
Actually, why don't
I just plain list
all the three by three
permutation matrices.
There're a nice
little group of them.
What are all the matrices
that exchange no rows at all?
Well, I'll include the identity.
So that's a permutation matrix
that doesn't do anything.
Now what's the permutation
matrix that exchanges --
what is P12?
The permutation matrix that
exchanges rows one and two
would be --
0 1 0 --
1 0 0, right.
I just exchanged those rows of
the identity and I've got it.
Okay.
Actually, I'll -- yes.
Let me clutter this up.
Okay.
Give me a complete list of
all the row exchange matrices.
So what are they?
They're all the ways I can
take the identity matrix
and rearrange its rows.
How many will there be?
How many three by three
permutation matrices?
Shall we keep going
and get the answer?
So tell me some more.
STUDENT: Zero one
STRANG: Zero What
one are you going to do now?
STUDENT: I'm going to switch the
STRANG: Switch rows one and --
One and three, okay.
One and three,
leaving two alone.
Okay.
Now what else?
Switch -- what would
be the next easy one --
is switch two and three, good.
So I'll leave one zero zero
alone and I'll switch --
I'll move number three
up and number two down.
Okay.
Those are the ones that
just exchange single --
a pair of rows.
This guy, this guy and this
guy exchanges a pair of rows,
but now there are
more possibilities.
What's left?
So tell -- there is
another one here.
What's that?
It's going to move -- it's
going to change all rows, right?
Where shall we put them?
So -- give me a first row.
STUDENT: Zero one zero?
STRANG: Zero one zero.
Okay, now a second row -- say
zero zero one and the third guy
One zero zero.
So that is like a cycle.
That puts row two moves up to
row one, row three moves up
to row two and row one
moves down to row three.
And there's one more,
which is -- let's see.
What's left?
I'm lost.
STUDENT: Is it zero zero one?
STRANG: Is it zero zero one?
Okay.
STUDENT: One zero zero.
STRANG: One zero zero, okay.
Zero one zero, okay.
Great.
Six.
Six of them.
Six P. And they're sort
of nice, because what
happens if I write, multiply
two of them together?
If I multiply two of
these matrices together,
what can you tell
me about the answer?
It's on the list.
If I do some row
exchanges and then I
do some more row exchanges,
then all together
I've done row exchanges.
So if I multiply --
but, I don't know.
And if I invert, then I'm
just doing row exchanges
to get back again.
So the inverses are all there.
It's a little family
of matrices that --
they've got their
own -- if I multiply,
I'm still inside this group.
If I invert I'm inside
this group -- actually,
group is the right
name for this subject.
It's a group of six matrices,
and what about the inverses?
What's the inverse of
this guy, for example?
What's the inverse -- if I
exchange rows one and two,
what's the inverse matrix?
Just tell me fast.
The inverse of that matrix is --
if I exchange rows one and two,
then what I should do to get
back to where I started is
the same thing.
So this thing is
its own inverse.
That's probably its own inverse.
This is probably
not -- actually,
I think these are
inverses of each other.
Oh, yeah, actually -- the
inverse is the transpose.
There's a curious fact
about permutations matrices,
that the inverses
are the transposes.
And final moment -- how
many are there if I --
how many four by
four permutations?
So let me take four by
four -- how many Ps?
Well, okay.
Make a good guess.
Twenty four, right.
Twenty four Ps.
Okay.
So, we've got these
permutation matrices,
and in the next
lecture, we'll use them.
So the next lecture,
finishes Chapter 2
and moves to Chapter 3.
Thank you.
