Let's say I have some transformation from R2 to R2.
And it's essentially just a multiplication times a matrix.
And we know that all linear transformations can be
expressed as a multiplication of a matrix, but this one is
equal to the matrix 1, 3, 2, 6 times whatever vector you give
me in my domain.
Times x1, x2.
Now let's say I have some subset in my codomain.
So let me draw this right here.
So my domain looks like that.
It's R2.
And of course my function or my transformation maps
elements of R2 into elements of its codomain which also
happens to be R2.
I could show it mapping into itself, but for the sake of
simplicity let's draw my codomain here.
And our transformation of course maps, for any element
here, the transformation of that will be an association or
a mapping into R2.
Now what if we take some subset of R2, and let's just
say it's a set of two vectors, the zero vector in R2, and the
vector 1, 2.
So it's literally, let's say it's this point.
Let me do it in a different color.
Let's say this is my zero vector in R2.
I'm not plotting them.
I'm just showing that they're in this, they're in R2.
That's my zero vector, and let's say the
vector 1,2 is here.
1,2.
What I want to know is, what are all of the vectors in my
domain whose transformations map to this subset?
Map to this point?
So I want to know, I'm essentially, I want to know
the preimage of S.
So the preimage of S, let me be careful, the
preimage of S under T.
And I said be careful because when you say preimage of
something without saying under something else, it implies
that you're taking.
Well when you say image of something, it implies you're
taking the image of an entire transformation like I showed
you, I think, two videos ago.
But when you're taking the image or preimage of a set,
you make sure you say under what transformation.
So we want to know the preimage of this subset of our
codomain under the transformation T.
Let me write this as T inverse of S.
And we saw in the last video this is all of the x, all of
the elements in our domain where the transformation of
those x's is a member of the subset of our codomain that
we're trying to find the preimage of.
Right?
Now what is another way of writing this?
Well this is, you can write this as, we're trying to look
for all of the x's in our domain such that this, let's
call this A, that matrix A.
Such that A times x is a member of S.
So that means A times x has to be equal to this or has to be
equal to that.
So that means A times our vector x has to be equal to
the zero vector.
Or A times our vector x has to be equal to this vector 1,2.
This is the exact same statement as
this one right here.
I just made it a little bit more explicit in terms of our
actual transformation, A times x.
And in terms of what our actual set is.
Our set is just two vectors.
So if we want to determine the preimage of S.
So we write that the preimage of S under T.
That set there.
We essentially have to just find all of the x's that
satisfy these two equations right there.
So this equation, so we have to find all of the x's that
satisfy, the first one right here is the matrix 1, 3, 2, 6
times x1, x2 is equal to the zero vector.
That's this equation right there.
We need to find all of the solutions to that.
And you might already recognize it.
All of the solutions to this, all of the x's that satisfy
this, is the null space of this matrix.
I just thought I'd point that out on the side.
Now, that's not the only one.
We also have to solve this guy over here.
I'll do that in blue.
So the preimage of S under T is going to be all the
solutions to this plus all of the solutions to 1, 3, 2, 6
times x1, x2 is equal to 1, 2.
Now we can just solve this with an augmented matrix.
So my augmented matrix would look like 1, 3, 2, 6, 0, 0.
And here my augmented matrix would be 1, 3, 2, 6, 1, 2.
Let's put this in reduced row echelon form.
So let's replace our second row with our second row minus
2 times the first row.
So what do we get?
Our first row stays the same.
1, 3, 0.
Let me do them simultaneously.
Let me solve these systems in parallel.
So my first system stays, first row stays the same.
1, 3, 1.
And in both cases, because I just want to get the left hand
side of my augmented matrix into reduced row echelon form,
I can apply the same row operation.
So I'm replacing my second row with 2 times my first row.
So 2 minus 2 times 1 is 0.
6 minus 2 times 3 is 0.
And of course 0 minus 2 times 0 is 0.
Here 2 minus 2 times 1 is 0.
6 minus 2 times 3 is 0.
And 2 minus 2 times 1 is 0.
So we get all these zeroes, and we're actually done.
We have both of these augmented matrices in reduced
row echelon form.
And how do we go back to solve all of the x1's and x2's that
satisfy these?
Well you recognize that our first columns right here are
pivot columns.
And these are associated with our variable x1, so we know
that x1 is a pivot variable.
And we know that the second column is a non pivot column.
Because it has no 1's in it.
And so it's associated with x2.
And since x2 isn't a pivot column, we know that x2 is a
free variable.
Which essentially means we can set x2 to be anything.
So let's just set x2.
x2 is equal to t, where t is a member of the reals.
In this case what is x1 going to be equal to?
So here this top equation says, let me write it.
If we just kind of go back to this world right here, this
means that x1 plus 3x2 is equal to this 0 here.
This top line right here says x1 plus 3x2 is equal to 1.
And so if we say x2 is equal to t, this equation becomes x1
plus 3t is equal to 0.
Subtract 3t from both sides, you get x1 is
equal to minus 3t.
And then this equation, you get x1 is equal to, if we
substitute x2 with t, is x1 is equal to 1 minus 3t.
Now if we wanted to write the solution sets in kind of
vector notation.
The solution set for this guy right here, for this first
equation right there, is going to be x1, x2 is equal to what?
It's going to be equal to, you have your, well, x2 is just
going to be equal to t times.
It's just going to be equal to t.
So let me just write it there.
x2 is just t times 1.
It's just equal to t.
I made that definition up here.
And then what's x1 equal to?
It's equal to minus 3 times t.
If I put a t out here as a scalar, it's just
minus 3 times t.
So this is a solution for this first equation.
Some, where t is a member of the reals, So it's just scalar
multiples of the vector minus 3, 1.
And if we think of this position vector, this'll be a
line in R2.
And I'll draw that in a second.
So that's a solution for this first equation.
And then the solution for the second equation, how can we
set this up?
It's going to be, make sure you can see it, It is x1, x2,
and let's see.
x2 once again is just t times 1.
Let me just write it like this.
So it's t times 1.
That's x2.
Now what's x1?
x1 is equal to 1 minus 3 times t.
So if we do a minus 3 that give us our minus 3 times t.
But we need to do 1 minus that.
Or 1 plus minus 3 times t.
So what we can do is, we could say that all the solution set
here is equal to the vector 1, here, so now we have 1 plus
minus 3 times t for x1.
And here we could say is 0.
x2 is equal 0 plus t.
Or x2 is equal to t.
So this is the solution set for the second equation.
So our preimage of S.
Remember S was just these two points in our codomain.
The preimage of S under T is essentially all of the x's
that satisfy these two equations.
And let's actually graphs those.
Let me turn on my graphs.
That makes it look a little bit messy, but let me
graph it down here.
Let me copy and paste my two results.
Then I want to paste it.
So those are my two results.
Put my pen back on.
And now what I can do is I can graph it right here.
So let's see.
The solution set for that first equation is all the
multiples of the vector minus 3, 1.
So the vector minus 3, 1 looks like this.
Looks like that.
That's the vector minus 3, 1.
But my solution set is all the scalar
multiples of minus 3, 1.
This is a comma right here.
Right?
So if I just take all the scalar multiples of minus 3,
1, it's going to look like this.
So if you take 2 times it, you're going to
have minus 6, 2.
So you're going to get like that.
So it's going to be all of these points right here.
I wish I could draw a little bit neater, but I think you
get the idea.
It's going to be a line like that.
That is that solution set right there.
And then what is this solution set right here?
It's the vector 1, 0.
So we go out 1 and 0, so that's there.
Plus scalar multiples of minus 3, 1.
So plus scalar multiples of this.
So if we just had 1 scalar multiple of minus 3, 1 we'll
end up right there.
But we want to put all the scalar multiples of it,
because we have this t right here.
So we're going to end up with another line with the same
slope essentially, that's just shifted a little bit.
It's shifted 1 to the right.
Now what why did we do all of this?
Remember what we wanted to find out is, what were all of
the vectors, let me turn off the graph paper, in our domain
that when we apply the transformation, map to vectors
within our subset of our codomain.
Map to either 0, 0 or the vector 1, 2.
And we've figured all of those vectors out by solving these
two equations.
And we are able to see that these two lines, so when I
turn my graph paper on, they map to the points.
So when these guys, when you apply the transformation, I'll
draw it all on the same graph.
They map, when you apply the transformation, to the point
0, 0 and to the point 1, 2.
Which was right here.
So all of these points, when you apply the transformation,
and actually all the ones in the blue, they map to 0, 0.
Because we solved this top equation.
And all the ones in orange, when you apply the
transformation, map to the point 1, 2.
Now this blue line right here.
This has a special name for it.
And I actually touched on is it a little bit before, right?
Everything in this blue line, right?
If I call this set right here, I don't know, let's
call it B for blue.
This is the blue line.
That's this set of vectors right here.
Everything there, when I apply my transformation of those
blue vectors, or if I take the image of my blue set under T,
it all maps to the zero vector.
It equals the set of the zero vector right there.
We saw that right there.
And I remember when, well earlier in the video I pointed
out that, look , this set right here.
This is equivalent to the null space, right?
The null space of a matrix is all the vectors that if you
multiply it by that matrix, you get 0.
So this is a similar idea here.
This transformation is defined by a matrix.
And we're saying, what are all of the x's that when you
transform them you get the zero vector?
And so this idea, this blue thing right here, it's called
the kernel of T.
Sometimes it's written just shorthand K-E-R of T.
And this literally is all of the vectors that if you apply
the transformation, let me write it this way.
It's all of the vectors in our domain, which was R2, such
that the transformation of those vectors is equal to the
zero vector.
There this is the definition of the kernel.
And if the transformation is equal to some matrix times
some vector, and we know that any linear transformation can
be written as a matrix vector product, then the kernel of T
is the same thing as the null space of A.
And we saw that earlier in the video.
Anyway, hopefully you found that reasonably useful.