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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 1: Functions and linear transformations- A more formal understanding of functions
- Vector transformations
- Linear transformations
- Visualizing linear transformations
- Matrix from visual representation of transformation
- Matrix vector products as linear transformations
- Linear transformations as matrix vector products
- Image of a subset under a transformation
- im(T): Image of a transformation
- Preimage of a set
- Preimage and kernel example
- Sums and scalar multiples of linear transformations
- More on matrix addition and scalar multiplication

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# Preimage and kernel example

Example involving the preimage of a set under a transformation. Definition of kernel of a transformation. Created by Sal Khan.

## Want to join the conversation?

- Is there a difference between kernel and nullspace? They seem to be defined in the same way (Av=0)(37 votes)
- yes. the nullspace is something general to matrices. the kernel is related to transformations of vectors. the kernel of the transformation is the same as the nullspace of the transformation matrix(68 votes)

- If you are one of those who has a hard time understanding what kernel or nullspace REALLY means, I found a more intuitive understanding.

We all know that null space of a matrix is the set of solutions to the homogeneous equation Ax = 0. But I found it really helpful when I saw it this way:

Suppose T(x) = Ax, then the kernel of this transformation is the set of vectors that are SENT to the zero vector by T. In other words, it is the set of vectors that LOSE THEIR IDENTITY because of T.(20 votes) - If you struggle to understand what is the Kernel and the Image, let me show you the mnemonic I use to do so.

Imagine a linear space V is a formation of spaceships, each having its own unique coordinates. Linear operator (transformation) T is the voyage near a black hole. Then the Kernel of T is all the ships that fell into the black hole and the Image of T is the ships that survived, but now have their coordinates distorted by the black hole.

Hope that helps :)(7 votes) - I still don't get it... How do I visually think of images and preimages??(2 votes)
- Say I have an object, and I transform it with T. This creates a new object, which we call the image.

Physically, T might be a carnival mirror which makes you look tall and skinny, and the object in question would be yourself. In the mirror you see a transformed version of yourself which is tall and skinny. This transformed version of yourself is the image of yourself under T.

A preimage takes finds all the possible objects which can create a given image (or set of images) when placed under the transformation T.

Physically, consider T as casting a shadow, with shadows being the codomain. If you look at a shadow, you can then consider what objects might cast that shadow. The set of all the possible objects which can create a given shadow is the preimage of that shadow under T.(8 votes)

- Regarding the two blue and orange lines that map to two points: isn't it like you'd be looking in a computer game at a line that basically maps onto the screen in a single pixel, because it's "perpendicular to the screen"?

Of course, that would be a mapping of R^3 -> R^2, but still, the idea is the same, right?(5 votes) - At0:10Sal says that all linear transformations can be written as matrix multiplication problems, but my linear algebra professor says that this is only the case when you're going from Rn to Rm. My professor says that, technically, the derivative and the integral are linear transformations that can't be written as matrix multiplication. Who's right?(5 votes)
- At11:48, wouldn't the solution to [x1,x2]=[1,0]+t[-3,1] be the set of vectors that, in the graph, fill the space between the two parallel lines Sal drew? We are adding the vector t[-3,1] to [1,0], which graphically would mean that the sum of these two vectors would start at the origin and end on a point on the 'orange' line (i.e. t[-3,1] shifted by 1 unit on the horizontal axis.). What am I missing?(3 votes)
- Yes, all those vectors go from the origin to the line y = 1/3 - x/3, as you say.

What I noticed, though, is that all the end points of those vectors "(1, 0) + t(-3, 1)" are actually on that line. He doesn't seem to realise this, but I think that the lines he's talking about are actually the loci of the endpoints of these vectors.

And in general, all the geometric entities that he describes with vectors (like the definition of a plane we studied in a previous video) are loci of endpoints of vectors, not spans of vectors.(2 votes)

- How is a Kernel, as used in linear algebra, related to the usage in other branches of mathematics (statistics - kernel functions, kernel density, etc.) or even programming? I've seen the word used in all of the above, but the connection isn't all that clear to me(4 votes)
- In mathematics, a word can have several different meanings based on the context. See "homogeneous" - it has multiple meanings in differential equations and linear algebra.(0 votes)

- Sal at14:50says that the kernel is the set of all vectors x in R^2 where T(x)=0. Does x have to be in R^2 or could it be in any R^n?(2 votes)
- In this example, x had to be in R^2 because the matrix that underwent transformation was a 2x2 matrix. In order to map an mxn matrix A (in this example, a 2x2 matrix) from a vector space V to a vector space W, we have to multiply the mxn matrix by an nx1 vector. The number of rows in the vector x must match the number of columns in the matrix A.(1 vote)

- how can I determine the dimension of the kernel and the Image?(1 vote)
- The dimension of the Kernel would be the dimension of N(A), which would be the number of free variables in your N(rref(A)). The dimension of the Image under T would be the dimension of C(A), which would be the number of pivot variables in your C(rref(A)). I believe this is correct. So Total variables in A = dim(N(A)) + dim(C(A)).(3 votes)

## Video transcript

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.