OK. Good morning, everyone. I think maybe we'll get started. So
welcome to Linear Algebra two. If you weren't expecting announceable to either
at least one of your eyes in the one game. So I'll start off just with a few
practical bits of information. So this is your.
Algebra two, which is booting on from
an out. For one that you saw last term. Say
we're meeting Monday is in Wednesdays
from 10 to 11 a.m. and that's always in this room alone.
So since you're here, you've probably on the set the timetable. Okay.
And one thing to note is that this is just a short course. So we only have nachas
between weeks. One in four. As you saw in
many outbreed one, the idea of leaning out there.
Is really to study maybe
functions which have the property, the
F of X plus B, Y is
A times X plus B times
Y and hey maybe I'm thinking of X and Y as being
factors and and B, it's been scalars, but you grew so much in spite more
general situations where this is true and any function that satisfies this, we call in Libya.
And then linear algebra is the outback study
of these sorts of functions. And it turns
out that this is one of the most important and well understood areas of my Hartwick's
say there's a very beautiful theory of this. And
one quick summary is you can understand
a lot about this by looking at matrices
and vector spaces. So objects that we've encountered in Article one
and actually a huge amount of modern mathematics is often
to take some very complicated situation and to try and maybe
approximate it in some way by some linear system. So when you do calculus,
your approximating a more general function by something that looks linear on a small scale.
And now it's really lies at the very heart of a huge amount of mathematics.
You've maybe already seen that linear algebra has various different applications. If you want
to solve simultaneous equations, you can cause this is a problem in linear algebra,
just as you can understand linear algebra as looking at vector spaces and more geometric problems.
And one philosophy for the way that I like to think
about giving out that I think would be useful in this course.
Is to when you're thinking about statements
to you, always imagine them geometrically as questions
about two by two matrices and transmissions of the plane.
But to always write down proofs and arguments much more abstractly
using out there. So think
about it around the.
Geometrically, in particular, as
I say, transformations
of the plane. So I'm not too
good at visualising things in three dimensions, but I can visualise things okay in two dimensions.
And often it's very easy to visualise things if you just imagine that everything is
some operation on the plane. So I like to think about what's going
on in a sort of geometric way. But we're going to prove the more general statements.
And one of the big Paphos, if they bring out the powerful things about an out there is you can take this
geometric intuition, then write down formally that work in arbitrary dimensions in a very
nice way. And often the outbreak of doing the calculations is much, much more convenient
than the geometric way. Even maybe we were actually motivated by the geometry, so
on to do calculations
and poufs. Abstractly.
I using outerwear.
So if you do it this way. The proof's in the statements often hold in much more
generality, but also it's much more convenient to do the calculations in this way.
And they say, I've baby said this
in very vague terms. Maybe we can sort of think about a concrete
example as to how this might work. And this gets us to
the first thing that I'd like to talk about, a many algebra, which is determinants.
But I'd like to demonstrate it to this sort of philosophy. So
think of just some linear transformation of the plain.
So what do I mean by this? Well, some combination of rotation
or stretching in one direction, or maybe I could skew
it if I could shift things, but a linear transformation is
going to preserve Rhines. And I can think about just taking the
unit box. And this will then be matched
to some maybe skewed box. So be some sort of parallelogram.
And maybe the axes won't be
Zonta anymore. And I want to understand transformations like this.
So this is just, uh. The Unit Square.
An. Squares. And then this gets mapped to some poll.
And. The Union Square has area.
One. And this parallelogram, the area
where Masoli represent so have some different area. See,
but I can think about the linear transmission acting on other shapes.
So I could have maybe a triangle and then
set out or get map to some. Strange, skewed Trango or something.
And maybe I can draw some
try and go with area, too.
And it's a fact that
the image of the triangle would then have.
A.C. So you couldn't. It's an exercise if you want to
work this out. And in fact. Well, maybe we would guess
just from doing a couple of examples like this,
that every.
Shape. I'm not going to be precise about what I mean by shape.
Has the area. Scaled
by a factor of say.
So often it's unwise to make a general guess just from doing two examples, but you can do a few more examples
if you like. And
it turns out this guess is true. And this
constancy is going to depend on the linear transformation. And this constancy
is essentially the determinant of delay and transformation. So
it's clearly going to be some important property of any in transmission that
it turns out that even if you work in many dimensions,
any reasonable shape is has its volume scaled by a constant factor.
And this constant factor is called the determinant. So this is very much
what I mean by thinking about things geometrically. But it turns out that
it's really quite difficult to make lots of this precise in the calculations get a bit horrible if you
try and think about things just in terms of volumes. So very
much in the spirit of this philosophy. I want to do calculations
trying to abstract out properties of this. So let's think about a few
properties of this scaling areas would
have in R-squared and try and think of
if I can write these down as abstract properties, maybe I can understand the properties
of this Lydian transmission just from these abstract ideas
and say. Note
that the
area or volume.
Of a linear transformation.
Uh. Is limited
in terms of the entries of the linear transformation. So maybe let's
consider a matrix. I say that
a b a matrix
with columns.
Two a n. Say, Lisa, if we can think
of these are just factors in RVN, say Isaan and by a Matrix
and the full name of the image of maybe
the unique Cheban, RCN is
living in the columns.
Say the for you. Of.
So maybe a nice way for me to write this is I whenever Matrixes columns,
I want to I n. All right. This is a sequel to square brackets.
A one of two I. N. And the volume
of some columns and even the great column
I have fact I've a to B.J. procedure Jay
of the Cube is equal to the volume.
Well, I have the same columns with B.J.
Plus, the volume when I have C.J. is the 28th
column, and this
is basically based on the idea that
the if I have to vote is.
Here, A and B something, and this is or maybe I
say consistent, I call this B and C. The area of this.
It is equal to just the air.
Well, I have to be policy here.
So that was high then I'd have just a hair.
So that's one property. A
second and easy policy is that
if a E.J. is equal to
ACA for some. Jay, not equal
to Kay. So to. The columns are the same. Then
the image of the uniquely gave to cops. So maybe it's easiest
if we just think about aside for two consecutive columns.
And then we don't need to worry about. Yes.
So if I have two consecutive columns, which are exactly the same,
then the volume of. Of the
image. Of the Unit Q.
Is equal to zero. And
so, again, I'm really thinking about.
This and if I have two vectors being the same, then I would just have.
Two columns. Ajanta J plus one and
the whole unit Kubel just degenerate down to a line.
And then a third property that we get from just looking at transmission's on
R-squared, is that the.
If you see I didn't see Matrix,
then it preserves the unique Q.
In the end and the volume.
Equals walking. So I haven't necessarily proved any of these statements, but maybe they're all believable.
And so what I want to say then is that these are properties
that this determine I think should satisfy. And these are all outback properties.
And therefore, I want to study the determinant, just
the outback properties. So this is going to.
To study the term, we we're then going to define what it means for any map
to be determined. The term mental, which means it's going to satisfy the equivalent
of these properties one, two and three. So we're not going to start
doing some concrete mathematics maybe before with some geometric motivation.
So we're going to say if I take any function.
So I'm thinking of A Function B and it's going to take his input.
And by and matrices with real entries. And it's going to output some
real value. And I'm going to call this map.
I'm going to say it has the property been determined, determined until meaning
they pay you smoke. What I expect this the tenants behave like if it satisfies the charges
properties. So if I think about my
map acting on some matrix while I like it in column form and in the
JF column, it has the
Vector B.J. Policy J.
Then I want it to be equal to the determinant
B.J., plus the determinant at.
C.J.
For any choice column in the exchange between one
man and simile, I wanted to scale
so servi I scale one column. If I factor Alanda
then I just wanted this to scale the determining factor. Let's
say this is going to be the first popsy and I want this to be D
is linear in each column.
So this is basically the poverty that we guessed, the volume of the
image of the unit Bush should satisfy. Funny late in translation.
The second thing is that.
I want. These have. According to the second property here,
that if.
I evaluated some matrix
and two of the columns are equal to each other,
maybe two consecutive columns. Then I want the determined
to be equal to zero.
And then the final thing is, if I evaluate
the discernment of the identity matrix, I just want this to be equal to what?
So this is a completely abstract definition of what it means for any function
on matrices to be determined to two, and at least based on this motivation,
we expect that the volume of the image of the unit cube should satisfy this. And
so we think it should be an important property of
any living man. But now this applies in general to matrices, that outright call.
So the first thing I want to do is to say that
if any linear maps are strikes, these properties that are actually satisfies slightly stronger
versions of these properties. So we can automatically upgrade.
So proposition.
If the is a determined mental maths.
Then one.
I can swap two columns if I
switch the sign of the determinates, say.
There's alternating and the columns
say the
name of. Some
matrix with columns, B.J. and B.J. plus one
is equal to minus the determent of the same matrix if I swapped the columns.
Say, swapping to.
Just change is a sign.
A second property is that.
OK. I said the deterrent is equal to zero. Two consecutive columns the same. But actually they don't have to be consecutive.
So if I evaluated it some matrix
with two columns, B.J. and B.I, this is zero
if any two of the different columns the same.
And then a final property is a stronger version
of it being alternating.
The first one we said, and it's also 18 consecutive columns. But actually I can do it for
general columns, so if I swap any two columns.
I say whenever I switch and into columns, I just get exactly the same answer. But with the
opposite side.
So let's go ahead and prove this proposition. Based on the properties that we've defined
for the term mentally.
So. Let's start off with part one, because that really holds the
key to the whole thing, and what I want to do is to
think about being evaluated where I'm evaluating
the JF Common column and the J plus one column, APJ plus PJ
one.
So here this is the JF column.
And this is the the past first column.
Some say by many
Anstee, I can expand the sounds.
Say, what do I get? Well, I'll get one term
with B.J. and B.J.
I'll get one term with B.J.
and B.J. plus one. I'll get one term
with Jerry Jay column, B.J. plus one and J plus first column B J.
And then I'll get a final term
with J. Colin, B.J. Paswan and J. Plus first columns. Also,
B.J. plus one.
So this is probably one.
But by property, too,
we have. That,
uh, the left hand side.
Is equal to zero. Because it has the J. Column in the J plus first column being the same.
So I'm just using this property, but moreover, we also have
that in my expansion into four terms on the right hand side. The first time in the last term
must also be zero because these also have two columns which are the same.
And so if I substitute these building codes zero, I therefore find.
Yes, this is it. This time is. This time you.
And so I just got some.
Well, I have great column B, Jane Deprez, first column B.J. plus one
is equal to the negative of exactly the same thing when I stopped the two columns.
And so this gives the first property.
So for part two. I can
now just repeatedly use part one.
To move column, Jay.
Next to. Column
I and I changed the determine just by a factor of plus or minus one.
But then the determined has to be zero, because I have. We'll have two consecutive columns being the same.
And so that instantly gives the final property safe and now
for the third part of part three. There's
a few different ways I could do it, but I can essentially follow exactly the same
proofers. Part one. I can
think about a valley right in. My matrix,
where I are now evaluating it with columns, B.I
and. B.J. in. Position.
I and P.J. plus B.I also
in position, Jay.
And I can follow exactly the same argument.
You men as four Part
II. I can expand this out four terms. This
left hand side must vanish because two of the columns are the same.
And by part two, I've seen that whenever two columns are the same, the whole thing vanishes.
But then when I think trying to get out the first term, the last term must also vanish.
Using. Part two.
OK. So so far, what we've seen is that we've maybe
had this geometric guess that there's an important property of any
matrix or linear transformation, which is how it scales volumes. And this
is a important concept that you associate to any matrix. I've then
defined this abstract property of. Any
function being determinant, too, if it satisfies first of the factors that we expect this
genuine constant to satisfy. And I've seen
that if it satisfies these properties, then feel this proposition I actually to slightly stronger
versions of those properties. So we've now
done enough of the basic important properties of the determinant.
But we haven't seen that actually there's any such map that satisfies these properties.
And so that's what I'd like to do now. I want to show you that there is at
least some function which satisfies all these properties in the back of my mind. I'm
thinking that a function that satisfies these properties is the volume of the image
of the unit cube. But again, I don't want to do any calculations geometrically. I want to do them all in
an outback way because that makes everything much, much clean. And so this is maybe the first
important result of the cool. So let's call it a film.
Are you determined to map exists?
On and by and matrices
exists for each
and beginning to one
say.
It's not totally obvious that something should exist from this. And
to prove this, I want to prove by induction where I'm going to induct
on the science, the matrices.
OK, so let's first of all, start thinking about the case when N equals one.
So we just we at one by one matrices.
And I can define the of just a one by one matrix, a
to be maybe the most obvious thing, just essentially a. And since
the function, the identity function is certainly linear, it's very easy to cheque
that dissatisfies the properties one, two and three here.
OK. So you're saying that something exists for
one by one waitress's? And now we want to
consider the case when this began one. And we're going to assume
that something exists. Some determine mental map exists for and minus one by
minus one seeds.
So we somehow want to construct a function on and by and matrices
out of this function that we're assuming to exist on and minus one Byan minus one matrices.
So I want to come up with some. I
want to cook up some way of coming up with a determined to map the satisfies these three properties.
But I want to somehow use this and minus one by minus one.
So I want to think of some Matrix A.
which is an end by an all matrix. So let's say
it has entry's little a J.
And if I'm going to use this Dearne minus one, which is clearly gonna be important in the argument,
I have to think about how can I construct a minus one by minus one
matrix out of this and by an matrix? Maybe there's one easy way to do this.
Which is just to fill out some loans, some colleton.
So I want a IJA capley IJA to be
the N minus one and minus one matrix formed by removing the.
I threw a and and. Colin.
OK. So this is one way
of constructing various different and minus one by minus one matrices for my.
And by Matrix A and are now
going to consider. So let's choose
some. I between
one and N and I'm going to define my attempt.
And a determined man to function on and by and make sees
as follows.
So I would put minus one to the five plus one a
I won and then I'm going to use the N minus one. Five minus one function
on the Matrix. I won.
But maybe say this is some function that would work on environment sees,
but is ignoring lots of the elements
in the i3. So instead, I want to add lots of these similar versions of this together
to get something that sensitive to every single element in The Matrix.
And it turns out that a good way to do this is to keep on multiplying by
minus one. And just going through all the different possible choices
in the i3.
OK, so I've just plucked from thin air this matrix here. But I think if you play
around with a little bit and you decide that you want to come up with some way of defining
a function on environmental crises in terms of function by minus one by one and minus one makes these.
That depends on each element. You see that actually this isn't such an unnatural thing
that I just picked out of the air. Maybe you wouldn't have guessed these minus one factors,
but it should be clear once you go through a bit more the. Why these Manasquan factors turn off.
Anyway, I've defined this function and this is a function on and by and matrices.
And I want to show you that it's now the term mental. So I want to verify that it satisfies properties one,
two and three. And if I done this, I have completed the proof.
OK, so I'm just going to verify the properties in order. I got to move
on to the middle.
I guess maybe would be useful to keep the sport up so that we can see for ourselves. And maybe I'll move
after I say I start off with property
one.
So we'd like to show you these línea in each column.
So since the
is a sum of terms
ahj the N minus one. Hey ija
it is sufficient
to show each of these.
Is Minear. In the columns.
OK. So I just need to show you that each of these different.
Quantity is AIG, the minus one of Charolais IJA is
in the columns. And so let's concentrate
on the case column.
Of. And biometrics. And I want to consider
these functions. So if Jay
is equal to Kay, then
AIG doesn't depend.
On the case column of A,
because this is formed by removing the case column. But
little IJA. It's just the
ice elements of the cave column. And so certainly Línea.
And so I think these two things together, we see the AIG.
The N minus one of A.J. is living the case. Him.
When James equal to Kay.
What if Jay is not equal? OK.
Well, then we have the
AIG does not depend on the cave column at all because it's just some other entry
in the J column.
But. We have the the minus one
AHJ is linnear. In the cave column.
Of a sense, we we're assuming that this function is the term mental and socially
and the continent's.
And therefore, in this case, we have the AIG deal,
minus one of AIG is Línea
in the Case column.
And therefore, regardless of what are your charges, these functions are always living in the cave column. And
so D William Lillian column.
Hmm hmm hmm hmm,
OK. So we've therefore verified Property One for this function
D that I've written down. So we're left to verify
properties. Two and three.
So maybe we have just enough space here.
So let's go for probably to.
Imagine that
in my matrix, hey, I have a veejays column is equal
to the Japes first column. And so I want to show you that this determined
has to be zero then. So, again, it's going to look a little bit
similar to this.
Hey, I. J will have
two columns, the same. Two consecutive columns, the same
because it essentially has the columns in Hatchet from A.J., and I just put one unless we removed
one of these two. OK. So
my notations slightly panned. Let's imagine that AK is equal to
a K plus one just because otherwise I'll get confused by Wiggins's apologies.
So I imagine that the 28th column is equal to the Kapos first column. And now I'm looking at one
of these matrices I ija we are moving the Arthur in the great column. If I
haven't removed either the case column or the K plus first column, then whatever corresponds to these two
will still be the same. And so we'll have two consecutive columns, the same. And therefore
our have the D minus one of AIG is
equal to zero. But this is only true if I haven't removed
one of those two columns. So if I didn't remove either those two columns, then two columns would be the same.
And so the determinate will vanish on these and minus one by minus one matrices.
And so this means that when I'm evaluating DNA, I would have
loads and I see terms. But virtually all of them say I'm only gonna get the terms
from a K. And I keep us warm
so I can write out precisely what this is.
And I just get the two terms from when I have a K and
I keep us one.
But then. If the case column
was equal to the Kapos first column, I have that
a K is equal to a I keep us one and
little I k is equal to a little a. I keep Wasswa.
And so take these two terms are equal to each other. And so the determining functions.
So now, finally, I just need to cheque property, see?
So if A is equal to I can
then. A IJA is equal to zero
unless Jay Eco's by. And
so all but one term in this huge expansion vanishes.
And so I have the dysfunction valuated eight is minus one
to the I plus I, I, I, I the
minus one of a I. But this is equal
to one because minus one to the iPod, sorry, is one I
is one whenever I is the identity matrix and by assumption
of the N minus one being the term rental. This also evaluates to one at the identity
matrix, which is what I get when I remove the fifth column and the i3.
Yes.
And so this means the function de
satisfies poverties one, two and three. And so I've shown it's determined to.
And so, therefore, by induction, we know that
there exists at least one different determinant to map on environment. Sources
say. Therefore, these term rental.
And therefore determine from exit tests.
Okay, so I'll stop here. What we've
seen is the this we guess that there's this important geometric
property of matrices, which is how they scale volumes. But it turns out that is
very messy to work with. So we've abstract this out to some outback properties and then working
with those outback properties. We've shown that there's something that exists that behaves
the exact same way. And next time will show that this is, in fact, a unique map that has all these properties.
OK. Thanks a lot.