Let's say I have some linear transformation
T from R2 to R2.
So that is R2, and then this is R2.
T just maps from any member of R2 to another member of R2,
just like that.
And I'm going to define T.
It's a linear transformation.
I'm going to define it.
When I take the transformation of some member of R2, it's
equivalent to multiplying it by this matrix, by the matrix
1 minus 3, minus 1, 3.
So just to kind of understand this transformation a little
bit more, let's think about all of the values that it can
take on in our codomain.
So if we say that 1 minus 3, minus 1, 3, times any vector
in our domain-- so x1, x2-- it's going to be equal to some
other vector in our codomain.
Let me call that vector b.
B is a member of R2.
So it's going to be equal to b1, b2.
All I'm doing here is I'm essentially-- if I call this
right here, I call that matrix A-- I'm just trying to find
out what are all the possible values for Ax is equal to b?
I'm trying to find out all of the possible b's in this case.
So if I were to solve this equation for any particular b,
what I would do is I would put this in
reduced row echelon form.
Well, first I would make an augmented matrix of it.
So I would put a 1, a minus 3, a minus 1, and a 3.
And I would augment it with the member of our codomain
we're trying to be equal to.
So b1, b2.
And then, I would put it in reduced row echelon form.
So how can we do that?
How can we put it in reduced row echelon form?
I'll keep my first row the same.
1 minus 3, and then b1.
And then my second row, I'll replace it with my second row
plus my first row.
So minus 1 plus 1 is 0.
3 plus minus 3 is 0.
And then b2 plus b1 is-- well I can just write
that as b1 plus b2.
So this is only going to have solutions.
I have an interesting situation here.
You've seen it before.
I have a row of 0's.
And the only way that this is going to actually have
solutions is if this thing right here is also going to be
equal to 0.
So the only members b that are members of Rm that have
solutions are the ones where if you add up their two
terms-- so let's say b is equal to b1 and b2-- are the
ones where the two terms added up have to be to 0.
b1 plus b2 have to be equal to 0.
Or another way you could write it is that b2 has to be equal
to minus b1.
So if we were to actually draw our codomain-- let's do it.
We always stay in the abstract, but sometimes it's
useful to draw an actual example.
Let's say that our codomain is R2.
Let me draw some axes right here.
Let's say that this is my b1-axis,
and this is my b2-axis.
I could have called that x and y, but I
called that b1 and b2.
What are all of the members of my codomain that have a
solution, that have a mapping?
Well, b2 has to be equal to minus b1.
So it's going to look like this.
It's just going to be a line with a slope of negative 1.
This is all the b's that have a solution.
Because if you're not on this line.
If you're a member of your codomain-- this is the
codomain right here, R2.
R2 is also our domain, but let me make it very clear that
this is the codomain that I've drawn.
This is what we're mapping into.
It's very clear that if we're not on this line, if you pick
somebody whose two terms don't add up to equalling 0, or
aren't the negative of each other, if you pick someone
over here in our codomain, and then you try to solve the
equation over here, 0 is going to equal some non-zero number
here and you're not going to have a solution.
And we touched on this in the last video.
And so in this case we could say that this right here is
the image of our transformation.
Or even another way of thinking about it.
This is our codomain.
Let me draw a domain.
So if you take any member of R2, and it's always mapping
onto something onto that line, clearly each point on that
line is going to be mapped to by more than one vector.
So we're not dealing with an onto transformation.
We saw that in the last video.
In order for something to be onto, when you put it in
reduced row echelon form, you cannot have all 0's
in one of the rows.
Or another way to say it in reduced row echelon form,
every row has to have a pivot entry.
Well, let's focus on the b's that
actually do have a solution.
So let's focus on these b's, so that when you take a b1
plus b2, it actually is going to be equal to 0.
So we could have the b, I don't know, that could be the
b5 minus 5.
Obviously the b0, 0 is going to work.
You could have 1, negative 1.
Maybe that's right there.
Let's focus on those for a second and see how many
members of our domain map into them.
So if we take this right here and then we apply this
equation up here, we only have one constraint.
We're assuming that this is going to be equal to 0.
So let's assume that we're dealing with
the b's in our image.
Let's assume that we're dealing with something where
we can get a solution.
That b1 plus b2 is equal to 0.
Then what is our constraints?
What will map to our vector b that we're dealing with?
If we just take this topic equation right here, we have 1
times x1-- let me switch colors-- minus 3 times x2 is
equal to b1.
And then this row will give us no constraints, because it's
just going to be a bunch of 0's.
So this is the only constraint for a member of our domain
that will map to some particular b now that we are
picking, some particular b that satisfies this
constraint.
So we could write this solution set-- let me rewrite
it as x1 is equal to b1, plus 3x2.
Or if we wanted to write the entire solution set, it would
look like this.
x1, x2 is equal to the vector b1, 0, plus-- x1 is b1, plus
3, times x2, so plus x2 times 3.
And x2 is just going to be equal to x2.
That's a free variable.
So x2 is equal to 0 plus x2, times 1.
So what we're saying here is this transformation just maps
to this line here for all of the vectors in our codomain,
where their two entries add up to each other.
Let's assume that we actually have one of those vectors.
And so first of all, this is definitely not an onto
transformation.
But let's assume that we are dealing with
one of those guys.
So if we are picking a particular one of those guys
for a particular b-- let me write this down-- that has a
solution to Ax is equal to b, the solution set will be equal
to this thing right here.
It'll be equal to x1, x2 is equal to the first
entry of your b.
b1, 0 plus x2, times the vector 3,1.
And if you think about this, if you pick a particular b--
so let's say we pick our-- let me draw this out, because I
think it's nice to visualize it all.
Actually, maybe I'll draw it like this.
I don't want to draw our blobs anymore.
So let me draw my axes.
So my axes look like this.
We know that the image of our transformation is the line
with a negative 1 slope, because the two entries have
to be equal to the negative of each other.
Let's pick a particular b that has a solution.
So let's say we pick that b right there.
In order for it to have a solution, its entries have to
be the negative of each other.
So let's say that's the entry 5, minus 5.
That is our b.
So what we just showed is that the solution set, if we want
to say, hey, what in our domain maps to this guy?
So let's think about what in our domain maps
to this right here.
Let's think about what in our domain maps to this point
right here, to this particular b.
So that's going to be all of the x's that satisfy Ax is
equal to 5, minus 5.
So x1, x2 is going to be equal to b1, it's going to be equal
to 5, 0 plus x2, plus any scale or multiple of the
vector 3, 1.
So our solution set is going to be, you take the vector 5,
0-- so maybe the vector 5, 0, specifies this position right
here-- and then you're going to add to it multiples of the
vector 3, 1.
The vector 3, 1 looks like this, 1, 2, 3 and you go up 1.
The vector 3, 1 is going to look like this.
So if you add multiples of this-- multiples of that could
stretch out like that or could go negative like that-- to
this vector 5, 0, you're essentially going to-- let me
see if I can draw this neatly-- you're going to end
up with a solution set that looks like that right there.
So if you pick a particular b right there that has a
solution, we just said that everything on this line will
map to that point in our solution set.
And in fact, if you picked another point, let's say you
picked the point minus 5, 5.
Then the solution set that maps to that would actually
be-- this first term would be the minus 5.
It would be here.
And all of these guys would map to that.
Well, this all interesting.
I mean, we've been doing a lot of abstract things.
And I think it might be satisfying that you're
actually seeing something more concrete in this example.
But I'm doing all of this for a reason.
Because I want to understand what the solution set is to a
general non-homogeneous equation like this.
And to understand it a little bit better, let's imagine what
is the solution set if I were to pick
this guy, the 0 vector?
Then what's going to be the solution set?
So if we say that Ax is equal to 0, then our solution set is
going to the vector 0, 0, plus x2, times 3, 1.
So it's going to be what?
This is just the 0 vector, so it's going to be here.
It's going to be here, and it's going to just be
multiples of 3, 1.
It's going to look something like that.
What's this?
What is the solution set to the equation of Ax equal to 0.
This is the null space.
This, by definition, is the null space of A.
So notice-- and this is the big takeaway from this video--
that for any solution-- we're picking b's that actually do
have solutions, because we're picking them on this line.
We're picking them in the image of our codomain.
The solution set to any Ax is equal to some b where b does
have a solution, it's essentially equal to a shifted
version of the null set, or the null space.
This right here is the null space.
That right there is the null space for any real number x2.
Any scale or multiple of 3, 1 is the null space.
I just showed it right there.
It's going to be that.
And so all of these other solution sets are just some
particular vector, some x particular,
plus the null space.
Obviously, this vector by itself would also be a
solution to Ax is equal to b, because you can just set x2 to
be equal to 0.
So in general-- and I haven't proven this to you rigorously,
but hopefully you kind of get the intuition behind it.
The solution-- and I'll do this in the next video, just
because I realize I'm running long on time.
Assuming Axb has a solution-- in the example we just did, we
can assume it has, if we pick one of these points here.
So assuming it has a solution, if we pick a point off of our
image, then we're not going to have a solution.
But assuming Axb has a solution, the solution set is
going to be equal to some particular vector-- so you
could just think of it as a set with just one vector right
there-- with, or combined with, or the union of that
set, with your null space of this matrix right here.
I haven't proven this to you yet, but hopefully you get the
intuition of why this is true.
We just solved it for particular cases
that do have solutions.
We say, hey, it's going to take this form.
And I just showed you that this is the
form of the null space.
And the reason why we're doing that is because we've been
talking about invertibility.
And in order to be invertible you have to be
onto, and 1 to 1.
And for something to be 1 to 1, you have to have at most
one solution that maps to a particular vector.
You might have none.
But you might have, at most, one.
So in order to have, at most, one solution-- and the
solution set is always going to be equal to this, so you're
always going to have this solution-- so in order to
have, at most, one solution, your null space can't have
anything in it.
Or it can just have the zero vector.
So that means that your null space of A has to be trivial,
or to be empty, or to just have the zero vector.
I'll do this a little bit more rigorously in the next video,
but I think when you do it sometimes with the rigor, you
don't necessarily get the intuition.
But this is going to be a very interesting takeway, and I
think you already understand it's leading up to conditions
for invertibility.